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```					   16                        Applications of Aqueous Equilibria

16.1   (a) HNO2(aq) + OHS(aq)  NO2S(aq) + H2O(l); NO2S (basic anion), pH > 7.00
(b) H3O+(aq) + NH3(aq)  NH4+(aq) + H2O(l); NH4+ (acidic cation), pH < 7.00
(c) OHS(aq) + H3O+(aq)  2 H2O(l); pH = 7.00

16.2   (a) HF(aq) + OHS(aq)  H2O(l) + FS(aq)
K     3.5 x 10 _ 4
Kn = a =           _14
= 3.5 x 1010
K w 1.0 x 10
(b) H3O+(aq) + OHS(aq)  2 H2O(l)
1        1
Kn =     =         _14
= 1.0 x 1014
K w 1.0 x 10
(c) HF(aq) + NH3(aq)  NH4+(aq) + FS(aq)
KK      (3.5 x 10 _ 4)(1.8 x 10 _ 5)
Kn = a b =                     _14
= 6.3 x 105
Kw             1.0 x 10
The tendency to proceed to completion is determined by the magnitude of Kn. The larger
the value of Kn, the further does the reaction proceed to completion.
The tendency to proceed to completion is: reaction (c) < reaction (a) < reaction (b)

16.3                    HCN(aq) + H2O(l)  H3O+(aq) + CNS(aq)
initial (M)      0.025                         ~0            0.010
change (M) Sx                                  +x             +x
equil (M)        0.025 S x                       x           0.010 + x
+      _
[H 3O ][CN ]                       x(0.010 + x) x(0.010)
Ka =                   = 4.9 x 10 _10 =                 
[HCN]                           0.025 _ x        0.025
S9                 S9
Solve for x. x = 1.23 x 10 M = 1.2 x 10 M = [H3O+]
pH = Slog[H3O+] = Slog(1.23 x 10S9) = 8.91
_14
K w = 1.0 x 10
[OHS] =          +                 _9
= 8.2 x 10S6 M
[H 3O ] 1.23 x 10
[Na ] = [CNS] = 0.010 M; [HCN] = 0.025 M
+

[HCN ]diss               1.23 x 10 _ 9 M
% dissociation =                   x 100% =                     x 100% = 4.9 x 10S6 %
[HCN ]initial                 0.025 M

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0.10 mol
16.4   From NH4Cl(s), [NH4+]initial =                = 0.20 M
0.500 L
NH3(aq) + H2O(l)  NH4+(aq) + OHS(aq)
initial (M)     0.40                      0.20        ~0
change (M) Sx                              +x         +x
equil (M)       0.40 S x                  0.20 + x     x
+     _
[ NH 4 ][OH ]                   (0.20 + x)(x) (0.20)(x)
Kb =                  = 1.8 x 10 _ 5 =                
[ NH 3]                       (0.40 _ x)      (0.40)
S              S5
Solve for x. x = [OH ] = 3.6 x 10 M
1.0 x 10 _14
[H3O+] = K w_ =                _5
= 2.8 x 10S10 M
[OH ] 3.6 x 10
pH = Slog[H3O+] = Slog(2.8 x 10S10) = 9.55

16.5   Each solution contains the same number of B molecules. The presence of BH+ from
BHCl lowers the percent dissociation of B. Solution (2) contains no BH+, therefore it
has the largest percent dissociation. BH+ is the conjugate acid of B. Solution (1) has the
largest amount of BH+ and it would be the most acidic solution and have the lowest pH.

16.6   (a) (1) and (3). Both pictures show equal concentrations of HA and AS.
(b) (3). It contains a higher concentration of HA and AS.

16.7                   HF(aq) + H2O(l)  H3O+(aq) + FS(aq)
initial (M)     0.25                      ~0        0.50
change (M) Sx                             +x        +x
equil (M)       0.25 S x                    x       0.50 + x
+   _
[H 3O ][F ]                   x(0.50 + x) x(0.50)
Ka =                = 3.5 x 10 _ 4 =            
[HF]                       0.25 _ x    0.25
S4            +
Solve for x. x = 1.75 x 10 M = [H3O ]
For the buffer, pH = Slog[H3O+] = Slog(1.75 x 10S4) = 3.76
(a) mol HF = 0.025 mol; mol FS = 0.050 mol; vol = 0.100 L
100%
F (aq) + H3O (aq)  HF(aq) + H2O(l)
S                +

before (mol) 0.050       0.002              0.025
change (mol) S0.002      S0.002            +0.002
after (mol) 0.048          0                0.027
[HF]                   0.27             S4
[H3O+] = K a      = (3.5 x 10 _ 4)        = 1.97 x 10 M
 0.48 
_
[F ]
pH = Slog[H3O ] = Slog(1.97 x 10S4) = 3.71
+

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(b) mol HF = 0.025 mol; mol FS = 0.050 mol; vol = 0.100 L
100%
HF(aq) + OHS(aq)  FS(aq) + H2O(l)
before (mol) 0.025        0.004             0.050
change (mol) S0.004      S0.004            +0.004
after (mol) 0.021          0                0.054
[HF]                   0.21             S4
[H3O+] = K a      = (3.5 x 10 _ 4)        = 1.36 x 10 M
 0.54 
_
[F ]
pH = Slog[H3O ] = Slog(1.36 x 10S4) = 3.87
+

16.8                   HF(aq) + H2O(l)  H3O+(aq) + FS(aq)
initial (M)     0.050                    ~0        0.100
change (M)       Sx                      +x         +x
equil (M)       0.050 S x                 x        0.100 + x
+   _
[H 3O ][F ]                   x(0.100 + x) x(0.100)
Ka =                = 3.5 x 10 _ 4 =             
[HF]                       0.050 _ x      0.050
+              S4
Solve for x. x = [H3O ] = 1.75 x 10 M
pH = Slog[H3O+] = Slog(1.75 x 10S4) = 3.76
mol HF = 0.050 mol/L x 0.100 L = 0.0050 mol HF
mol FS = 0.100 mol/L x 0.100 L = 0.0100 mol FS
mol HNO3 = mol H3O+ = 0.002 mol
100%
Neutralization reaction: F (aq) + H3O (aq)  HF(aq) + H2O(l)
S             +

before reaction (mol)     0.0100 0.002               0.0050
change (mol)             S0.002 S0.002               +0.002
after reaction (mol)      0.008        0              0.007
0.007 mol                             0.008 mol
[HF] =               = 0.07 M;          [FS] =            = 0.08 M
0.100 L                               0.100 L
[HF]                   (0.07)
[H3O+] = Ka      _
= (3.5 x 10 _ 4)        = 3 x 10S4 M
[F ]                  (0.08)
pH = Slog[H3O+] = Slog(3 x 10S4) = 3.5
This solution has less buffering capacity than the solution in Problem 16.7 because it
contains less HF and FS per 100 mL. Note that the change in pH is greater than that in
Problem 16.7.

16.9   When equal volumes of two solutions are mixed together, the concentration of each
solution is cut in half.
[base]                    2
[CO 3 _ ]
pH = pKa + log           = pK a + log         _
[acid]               [ HCO 3 ]
For HCO3S, Ka = 5.6 x 10S11, pKa = Slog Ka = Slog(5.6 x 10S11) = 10.25
 0.050 
pH = 10.25 + log            = 10.25 S 0.30 = 9.95
 0.10 

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[base]                     2
[CO 3 _ ]
16.10   pH = pKa + log           = pK a + log          _
[acid]               [ HCO 3 ]
For HCO3S, Ka = 5.6 x 10S11, pKa = Slog Ka = Slog(5.6 x 10S11) = 10.25
2
[CO 3 _ ]                  2
[CO 3 _ ]
10.40 = 10.25 + log           _
;      log          _
= 10.40 S 10.25 = 0.15
[ HCO 3 ]             [ HCO 3 ]
2
[CO 3 _ ]
_
= 100.15 = 1.4
[ HCO 3 ]
To obtain a buffer solution with pH 10.40, make the Na2CO3 concentration 1.4 times the
concentration of NaHCO3.

16.11   Look for an acid with pKa near the required pH of 7.50.
Ka = 10SpH = 10S7.50 = 3.2 x 10S8
Suggested buffer system: HOCl (Ka = 3.5 x 10S8) and NaOCl.

 66 
16.12   (a) serine is 66% dissociated at pH = 9.15 + log   = 9.44
 34 
 5 
(b) serine is 5% dissociated at pH = 9.15 + log   = 7.87
 95 

16.13   (a) mol HCl = mol H3O+ = 0.100 mol/L x 0.0400 L = 0.004 00 mol
mol NaOH = mol OHS = 0.100 mol/L x 0.0350 L = 0.003 50 mol
Neutralization reaction:       H3O+(aq) + OHS(aq)  2 H2O(l)
before reaction (mol)          0.004 00      0.003 50
change (mol)                   S0.003 50 S0.003 50
after reaction (mol)           0.000 50       0
0.000 50 mol
[H3O+] =                             = 6.7 x 10S3 M
(0.0400 L + 0.0350 L)
pH = Slog[H3O+] = Slog(6.7 x 10S3) = 2.17
(b) mol HCl = mol H3O+ = 0.100 mol/L x 0.0400 L = 0.004 00 mol
mol NaOH = mol OHS = 0.100 mol/L x 0.0450 L = 0.004 50 mol
Neutralization reaction:       H3O+(aq) + OHS(aq)  2 H2O(l)
before reaction (mol)          0.004 00      0.004 50
change (mol)                   S0.004 00 S0.004 00
after reaction (mol)             0           0.000 50
0.000 50 mol
[OHS] =                             = 5.9 x 10S3 M
(0.0400 L + 0.0450 L)
_14
K w = 1.0 x 10
[H3O+] =                          = 1.7 x 10S12 M
[OH _ ] 5.9 x 10 _ 3
pH = Slog[H3O+] = Slog(1.7 x 10S12) = 11.77
The results obtained here are consistent with the pH data in Table 16.1
16.14   (a) mol NaOH = mol OHS = 0.100 mol/L x 0.0400 L = 0.004 00 mol
mol HCl = mol H3O+ = 0.0500 mol/L x 0.0600 L = 0.003 00 mol

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Neutralization reaction:       H3O+(aq) + OHS(aq)  2 H2O(l)
before reaction (mol)          0.003 00      0.004 00
change (mol)                   S0.003 00 S0.003 00
after reaction (mol)             0           0.001 00
0.001 00 mol
[OHS] =                             = 1.0 x 10S2 M
(0.0400 L + 0.0600 L)
_14
K w = 1.0 x 10
[H3O+] =        _            _2
= 1.0 x 10S12 M
[OH ] 1.0 x 10
pH = Slog[H3O+] = Slog(1.0 x 10S12) = 12.00
(b) mol NaOH = mol OHS = 0.100 mol/L x 0.0400 L = 0.004 00 mol
mol HCl = mol H3O+ = 0.0500 mol/L x 0.0802 L = 0.004 01 mol
Neutralization reaction:       H3O+(aq) + OHS(aq)  2 H2O(l)
before reaction (mol)          0.004 01      0.004 00
change (mol)                   S0.004 00 S0.004 00
after reaction (mol)           0.000 01       0
0.000 01 mol
[H3O+] =                             = 8.3 x 10S5 M
(0.0400 L + 0.0802 L)
pH = Slog[H3O+] = Slog(8.3 x 10S5) = 4.08
(c) mol NaOH = mol OHS = 0.100 mol/L x 0.0400 L = 0.004 00 mol
mol HCl = mol H3O+ = 0.0500 mol/L x 0.1000 L = 0.005 00 mol
Neutralization reaction:       H3O+(aq) + OHS(aq)  2 H2O(l)
before reaction (mol)          0.005 00      0.004 00
change (mol)                   S0.004 00 S0.004 00
after reaction (mol)           0.001 00       0
0.001 00 mol
[H3O+] =                             = 7.1 x 10S3 M
(0.0400 L + 0.1000 L)
pH = Slog[H3O+] = Slog(7.1 x 10S3) = 2.15

16.15   (a) (3), only HA present           (b) (1), HA and AS present
(c) (4), only AS present           (d) (2), AS and OHS present

 0.016 mol HOCl               1 mol NaOH 
16.16   mol NaOH required =                       (0.100 L)               = 0.0016 mol
        L                     1 mol HOCl 
     1L      
vol NaOH required = (0.0016 mol)                 = 0.040 L = 40 mL
 0.0400 mol 
40 mL of 0.0400 M NaOH are required to reach the equivalence point.
(a) mmol HOCl = 0.016 mmol/mL x 100.0 mL = 1.6 mmol
mmol NaOH = mmol OHS = 0.0400 mmol/mL x 10.0 mL = 0.400 mmol
Neutralization reaction:     HOCl(aq) + OHS(aq)  OClS(aq) + H2O(l)
before reaction (mmol)        1.6         0.400            0
change (mmol)               S0.400       S0.400         +0.400
after reaction (mmol) 1.2           0            0.400

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1.2 mmol
[HOCl] =                        = 1.09 x 10S2 M
(100.0 mL + 10.0 mL)
0.400 mmol
[OClS] =                       = 3.64 x 10S3 M
(100.0 mL + 10.0 mL)
HOCl(aq) + H2O(l)  H3O+(aq) + OClS(aq)
initial (M)    0.0109               ~0           0.003 64
change (M)      Sx                           +x +x
equil (M)      0.0109 S x                     x 0.003 64 + x
+      _
[ O ][OCl ]                     x(0.003 64 + x) x(0.003 64)
Ka = H 3              = 3.5 x 10 _ 8 =                   
[HOCl]                          0.0109 _ x          0.0109
+               S7
Solve for x. x = [H3O ] = 1.05 x 10 M
pH = Slog[H3O+] = Slog(1.05 x 10S7) = 6.98
(b) Halfway to the equivalence point, [OClS] = [HOCl]
pH = pKa = Slog Ka = Slog(3.5 x 10S8) = 7.46
(c) At the equivalence point the solution contains the salt, NaOCl.
mol NaOCl = initial mol HOCl = 0.0016 mol = 1.6 mmol
1.6 mmol
[OClS] =                             = 1.1 x 10S2 M
(100.0 mL + 40.0 mL)
Kw         1.0 x 10 _14
For OClS, Kb =                 =              _8
= 2.9 x 10S7
K a for HOCl 3.5 x 10
OClS(aq) + H2O(l)  HOCl(aq) + OHS(aq)
initial (M)   0.011                         0          ~0
change (M)       Sx                        +x +x
equil (M)     0.011 S x                     x          x
[HOCl][OH _ ]                      x
2        2
Kb =           _
= 2.9 x 10 _ 7 =            x
[OCl ]                      0.011 _ x 0.011
S              S5
Solve for x. x = [OH ] = 5.65 x 10 M
_14
K w = 1.0 x 10
[H3O+] =        _            _5
= 1.77 x 10S10 = 1.8 x 10S10 M
[OH ] 5.65 x 10
pH = Slog[H3O+] = Slog(1.77 x 10S10) = 9.75

16.17   From Problem 16.16, pH = 9.75 at the equivalence point.
Use thymolphthalein (pH 9.4 S 10.6). Bromthymol blue is unacceptable because it
changes color halfway to the equivalence point.

16.18   (a) mol NaOH required to reach first equivalence point
 0.0800 mol H 2SO3               1 mol NaOH 
=                     (0.0400 L)                 = 0.003 20 mol
         L                       1 mol H 2SO3 

vol NaOH required to reach first equivalence point

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    1L       
= (0.003 20 mol)                = 0.020 L = 20.0 mL
 0.160 mol 
20.0 mL is enough NaOH solution to reach the first equivalence point for the titration of
the diprotic acid, H2SO3.
For H2SO3,
_2                                    _2
K a1 = 1.5 x 10 , pK a1 = _log K a1 = _log(1.5 x 10 ) = 1.82
_8                                    _8
K a 2 = 6.3 x 10 , pK a 2 = _log K a 2 = _log(6.3 x 10 ) = 7.20
pK a1 + pK a 2 1.82 + 7.20
At the first equivalence point, pH =                   =              = 4.51
2            2
(b) mol NaOH required to reach second equivalence point
 0.0800 mol H 2SO3                   2 mol NaOH 
=                         (0.0400 L)                 = 0.006 40 mol
           L                         1 mol H 2SO3 
vol NaOH required to reach second equivalence point
    1L       
= (0.006 40 mol)                 = 0.040 L = 40.0 mL
 0.160 mol 
30.0 mL is enough NaOH solution to reach halfway to the second equivalent point.
Halfway to the second equivalence point
pH = p K a 2 = _log K a 2 = _log(6.3 x 10 _ 8) = 7.20
(c) mmol HSO3S = 0.0800 mmol/mL x 40.0 mL = 3.20 mmol
volume NaOH added after first equivalence point = 35.0 mL S 20.0 mL = 15.0 mL
mmol NaOH = mmol OHS = 0.160 mmol/L x 15.0 mL = 2.40 mmol
Neutralization reaction:         HSO3S(aq) + OHS(aq)  SO32S(aq) + H2O(l)
before reaction (mmol)           3.20            2.40         0
change (mmol)                    S2.40          S2.40       +2.40
after reaction (mmol) 0.80                0            2.40
0.80 mmol
[HSO3S] =                             = 0.0107 M
(40.0 mL + 35.0 mL)
2.40 mmol
[SO32S] =                             = 0.0320 M
(40.0 mL + 35.0 mL)
HSO3S(aq) + H2O(l)  H3O+(aq) + SO32S(aq)
initial (M)      0.0107                  ~0          0.0320
change (M)         Sx                             +x         +x
equil (M)        0.0107 S x                       x         0.0320 + x
2
[H 3O + ][SO 3 _ ]                  x(0.0320 + x) x(0.0320)
Ka =              _
= 6.3 x 10 _ 8 =               
[HSO 3 ]                          0.0107 _ x      0.0107
+              S8
Solve for x. x = [H3O ] = 2.1 x 10 M
pH = Slog[H3O+] = Slog(2.1 x 10S8) = 7.68

16.19   Let H2A+ = valine cation
(a) mol NaOH required to reach first equivalence point

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 0.0250 mol H 2A +               1 mol NaOH 
=                       (0.0400 L)  1 mol      + 
= 0.001 00 mol
        L                              H 2A 
vol NaOH required to reach first equivalence point
    1L      
= (0.001 00 mol)               = 0.0100 L = 10.0 mL
 0.100 mol 
10.0 mL is enough NaOH solution to reach the first equivalence point for the titration of
the diprotic acid, H2A+.
For H2A+,
_3                                   _3
K a1 = 4.8 x 10 , pK a1 = _ log K a1 = _ log(4.8 x 10 ) = 2.32
_10                                         _10
K a 2 = 2.4 x 10 , pK a 2 = _ log K a 2 = _ log(2.4 x 10 ) = 9.62
pK a1 + pK a 2 2.32 + 9.62
At the first equivalence point, pH =                         =                 = 5.97
2                 2
(b) mol NaOH required to reach second equivalence point
 0.0250 mol H 2A +                     2 mol NaOH 
=                           (0.0400 L)  1 mol            + 
= 0.002 00 mol
             L                                 H 2A 
vol NaOH required to reach second equivalence point
    1L       
= (0.002 00 mol)                    = 0.0200 L = 20.0 mL
 0.100 mol 
15.0 mL is enough NaOH solution to reach halfway to the second equivalent point.
Halfway to the second equivalence point
pH = p K a 2 = _ log K a 2 = _ log(2.4 x 10 _10) = 9.62
(c) 20.0 mL is enough NaOH to reach the second equivalence point.
At the second equivalence point
mmol AS = (0.0250 mmol/mL)(40.0 mL) = 1.00 mmol AS
solution volume = 40.0 mL + 20.0 mL = 60.0 mL
1.00 mmol
[AS] =                      = 0.0167 M
60.0 mL
AS(aq) + H2O(l)  HA(aq) + OHS(aq)
initial (M)        0.0167                      0             ~0
change (M)           Sx                        +x            +x
equil (M)          0.0167 S x                   x              x
_14
Kw              K      1.0 x 10
Kb =                       = w =                      = 4.17 x 10S5
Ka    for HA          Ka2    2.4 x 10 _10
_                              2
[HA][OH ]                                 x
Kb =                     = 4.17 x 10 _ 5 =
[A _ ]                          0.0167 _ x
2                  S5                   S7
x + (4.17 x 10 )x S (6.964 x 10 ) = 0
Use the quadratic formula to solve for x.

x=

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_ (4.17 x 10 _ 5)   (4.17 x 10 _ 5) 2 _ (4)(1)(_ 6.964 x 10 _ 7) (_ 4.17 x 10 _ 5)  (1.67 x 10 _ 3)
=
2(1)                                                  2
S4                    S4
x = 8.14 x 10 and S8.56 x 10
Of the two solutions for x, only the positive value has physical meaning because x is the
[OHS].
x = [OHS] = 8.14 x 10S4 M
_14
K w = 1.0 x 10
[H3O+] =                           = 1.23 x 10S11 M
[OH _ ] 8.14 x 10 _ 4
pH = Slog[H3O+] = Slog(1.23 x 10S11) = 10.91

16.20   (a) Ksp = [Ag+][ClS]                     (b) Ksp = [Pb2+][IS]2
(c) Ksp = [Ca2+]3[PO43S]2                (d) Ksp = [Cr3+][OHS]3

16.21   Ksp = [Ca2+]3[PO43S]2 = (2.01 x 10S8)3(1.6 x 10S5)2 = 2.1 x 10S33

16.22   [Ba2+] = [SO42S] = 1.05 x 10S5 M;        Ksp = [Ba2+][SO42S] = (1.05 x 10S5)2 = 1.10 x 10S10

16.23   (a)           AgCl(s)  Ag+(aq) + ClS(aq)
equil (M)                 x             x
Ksp = [Ag+][ClS] = 1.8 x 10S10 = (x)(x)
molar solubility = x = K sp = 1.3 x 10S5 mol/L
AgCl, 143.32 amu
        _5     143.32 g 
1.3 x 10 mol x          
solubility = 
1 mol 
= 0.0019 g/L
1L

(b)           Ag2CrO4(s)  2 Ag+(aq) + CrO42S(aq)
equil (M)                      2x           x
Ksp = [Ag+]2[CrO42S] = 1.1 x 10S12 = (2x)2(x) = 4x3
1.1 x 10 _12
molar solubility = x =   3                = 6.5 x 10S5 mol/L
4
Ag2CrO4, 331.73 amu
         _5        331.73 g 
 6.5 x 10 mol x             
solubility =                      1 mol 
= 0.022 g/L
1L
Ag2CrO4 has both the higher molar and gram solubility, despite its smaller value of Ksp.

16.24   Let the number of ions be proportional to its concentration.
For AgX, Ksp = [Ag+][XS]  (4)(4) = 16
For AgY, Ksp = [Ag+][YS]  (1)(9) = 9
For AgZ, Ksp = [Ag+][ZS]  (3)(6) = 18
(a) AgZ      (b) AgY

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16.25   [Mg2+]0 is from 0.10 M MgCl2.
MgF2(s)  Mg2+(aq) + 2 FS(aq)
initial (M)                 0.10             0
change (M)                  +x             +2x
equil (M)                  0.10 + x         2x
Ksp = 7.4 x 10 = [Mg ][F ] = (0.10 + x)(2x)2  (0.10)(4x2)
S11      2+  S 2

x = 1.4 x 10S5, molar solubility = x = 1.4 x 10S5 M

16.26   Compounds that contain basic anions are more soluble in acidic solution than in pure
water. AgCN, Al(OH)3, and ZnS all contain basic anions.

16.27  [Cu2+] = (5.0 x 10S3 mol)/(0.500 L) = 0.010 M
Cu2+(aq) + 4 NH3(aq)  Cu(NH3)42+(aq)
before reaction (M) 0.010              0.40              0
assume 100 % reaction (M) S0.010             S 4(0.010)        +0.010
after reaction (M)     0              0.36           0.010
assume small back reaction (M) +x                  +4x              Sx
equil (M)    x            0.36 + 4x        0.010 S x
[Cu( NH3) 2+]                    (0.010 _ x)       0.010
Kf =             4
4
= 5.6 x 1011 =                   
2+
[Cu ][ NH3]                                     4
(x)(0.36 + 4 x ) x(0.36) 4
Solve for x. x = [Cu2+] = 1.1 x 10S12 M

16.28                 AgBr(s)  Ag+(aq) + BrS(aq)                Ksp = 5.4 x 10S13
Ag+(aq) + 2 S2O32S  Ag(S2O3)23S(aq)              Kf = 4.7 x 1013
dissolution   AgBr(s) + 2 S2O32S(aq)  Ag(S2O3)23S(aq) + BrS(aq)
reaction
K = (Ksp)(Kf) = (5.4 x 10S13)(4.7 x 1013) = 25.4
AgBr(s) + 2 S2O32S(aq)  Ag(S2O3)23S(aq) + BrS(aq)
initial (M)                0.10              0           0
change (M)                 S2x               x           x
equil (M)                0.10 S 2x           x           x
3_    _                   2
[Ag(S2O3) 2 ][Br ]                x
K=             2_ 2
= 25.4 =
[S2O3 ]                (0.10 _ 2 x ) 2
Take the square root of both sides and solve for x.
2
x                          x
25.4 =                   ;   5.04 =              ;   x = molar solubility = 0.045 mol/L
(0.10 _ 2 x ) 2              0.10 _ 2 x

16.29   On mixing equal volumes of two solutions, the concentrations of both solutions are cut
in half.
For BaCO3, Ksp = 2.6 x 10S9
(a) IP = [Ba2+][CO32S] = (1.5 x 10S3)(1.0 x 10S3) = 1.5 x 10S6

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IP > Ksp; a precipitate of BaCO3 will form.
(b) IP = [Ba2+][CO32S] = (5.0 x 10S6)(2.0 x 10S5) = 1.0 x 10S10
IP < Ksp; no precipitate will form.

[base]               [ NH 3]
16.30   pH = pKa + log          = pK a + log
[acid]               [ NH + ]
4
For NH4+, Ka = 5.6 x 10S10, pKa = Slog Ka = Slog(5.6 x 10S10) = 9.25
(0.20)
pH = 9.25 + log           = 9.25; [H3O+] = 10SpH = 10S9.25 = 5.6 x 10S10 M
(0.20)
_14
K w = 1.0 x 10
[OHS] =                            = 1.8 x 10S5 M
[H 3O + ] 5.6 x 10 _10
(25 mL)(1.0 x 10 _ 3 M)
[Fe2+] = [Mn2+] =                                = 1.0 x 10S4 M
250 mL
For Mn(OH)2, Ksp = 2.1 x 10S13
IP = [Mn2+][OHS]2 = (1.0 x 10S4)(1.8 x 10S5)2 = 3.2 x 10S14
IP < Ksp; no precipitate will form.
For Fe(OH)2, Ksp = 4.9 x 10S17
IP = [Fe2+][OHS]2 = (1.0 x 10S4)(1.8 x 10S5)2 = 3.2 x 10S14
IP > Ksp; a precipitate of Fe(OH)2 will form.

16.31   MS(s) + 2 H3O+(aq)  M2+(aq) + H2S(aq) + 2 H2O(l)
[M 2+ ][H 2 S]
Kspa =
[ H 3O + ] 2
For ZnS, Kspa = 3 x 10S2;      for CdS, Kspa = 8 x 10S7
[Cd2+] = [Zn2+] = 0.005 M
Because the two cation concentrations are equal, Qc is the same for both.
[M 2+]t[H 2 S ]t (0.005)(0.10)
Qc =          + 2
=          2
= 6 x 10S3
[H3O ]t            (0.3)
Qc > Kspa for CdS; CdS will precipitate. Qc < Kspa for ZnS; Zn2+ will remain in
solution.

16.32   This protein has both acidic and basic sites. H3PO4-H2PO4S is an acidic buffer. It
protonates the basic sites in the protein making them positive and the protein migrates
towards the negative electrode. H3BO3-H2BO3S is a basic buffer. At basic pH's, the
acidic sites in the protein are dissociated making them negative and the protein migrates
towards the positive electrode.

16.33   To increase the rate at which the proteins migrates toward the negative electrode,
increase the number of basic sites that are protonated by lowering the pH. Decrease the
[HPO42S]/[H2PO4S] ratio (less HPO42S, more H2PO4S) to lower the pH.

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Understanding Key Concepts

16.34   A buffer solution contains a conjugate acid-base pair in about equal concentrations.
(a) (1), (3), and (4)
(b) (4) because it has the highest buffer concentration.

16.35   (a) (2) has the highest pH, [AS] > [HA]
(3) has the lowest pH, [HA] > [AS]

(b)                                                              (c)

16.36   (4); only AS and water should be present

16.37   (a) (1) corresponds to (iii); (2) to (i); (3) to (iv); and (4) to (ii)
(b) Solution (3) has the highest pH; solution (2) has the lowest pH.

16.38   (a) (i) (1), only B present        (ii) (4), equal amounts of B and BH+ present
+
(iii) (3), only BH present         (iv) (2), BH+ and H3O+ present
(b) The pH is less than 7 because BH+ is an acidic cation.

16.39   (2) is supersaturated; (3) is unsaturated; (4) is unsaturated

16.40   Let the number of ions be proportional to its concentration.
For Ag2CrO4, Ksp = [Ag+]2[CrO42S]  (4)2(2) = 32
For (2), IP = [Ag+]2[CrO42S]  (2)2(4) = 16
For (3), IP = [Ag+]2[CrO42S]  (6)2(2) = 72
For (4), IP = [Ag+]2[CrO42S]  (2)2(6) = 24
A precipitate will form when IP > Ksp. A precipitate will form only in (3).

16.41   (a) The lower curve represents the titration of a strong acid; the upper curve represents
the titration of a weak acid.
(b) pH = 7 for titration of the strong acid; pH = 10 for titration of the weak acid.
(c) Halfway to the equivalence point, the pH = pKa ~ 6.3.

Neutralization Reactions

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16.42   (a) HI(aq) + NaOH(aq)  H2O(l) + NaI(aq)
net ionic equation: H3O+(aq) + OHS(aq)  2 H2O(l)
The solution at neutralization contains a neutral salt (NaI); pH = 7.00.

(b) 2 HOCl(aq) + Ba(OH)2(aq)  2 H2O(l) + Ba(OCl)2(aq)
net ionic equation: HOCl(aq) + OHS(aq)  H2O(l) + OClS(aq)
The solution at neutralization contains a basic anion (OClS); pH > 7.00
(c) HNO3(aq) + C6H5NH2(aq)  C6H5NH3NO3(aq)
net ionic equation: H3O+(aq) + C6H5NH2(aq)  H2O(l) + C6H5NH3+(aq)
The solution at neutralization contains an acidic cation (C6H5NH3+); pH < 7.00.
(d) C6H5CO2H(aq) + KOH(aq)  H2O(l) + C6H5CO2K(aq)
net ionic equation: C6H5CO2H(aq) + OHS(aq)  H2O(l) + C6H5CO2S(aq)
The solution at neutralization contains a basic anion (C6H5CO2S); pH > 7.00.

16.43   (a) HNO2(aq) + CsOH(aq)  H2O(l) + CsNO2(aq)
net ionic equation: HNO2(aq) + OHS(aq)  H2O(l) + NO2S(aq)
The solution at neutralization contains a basic anion (NO2S); pH > 7.00
(b) HBr(aq) + NH3(aq)  NH4Br(aq)
net ionic equation: H3O+(aq) + NH3(aq)  H2O(l) + NH4+(aq)
The solution at neutralization contains an acidic cation (NH4+); pH < 7.00
(c) HClO4(aq) + KOH(aq)  H2O(l) + KClO4(aq)
net ionic equation: H3O+(aq) + OHS(aq)  2 H2O(l)
The solution at neutralization contains a neutral salt (KClO4); pH = 7.00
(d) HOBr(aq) + NH3(aq)  NH4OBr(aq)
net ionic equation: HOBr(aq) + NH3(aq)  NH4+(aq) + OBrS(aq)
The solution at neutralization contains the salt NH4OBr.
Ka(NH4+) = 5.6 x 10S10 and Kb(OBrS) = 5.0 x 10S5; Kb(OBrS) > Ka(NH4+); pH > 7.00

1            1
16.44   (a) Strong acid - strong base reaction          Kn =       =                = 1.0 x 1014
Kw            1.0 x 10 _14
_8
K a = 3.5 x 10
(b) Weak acid - strong base reaction Kn =               _14
= 3.5 x 106
K w 1.0 x 10
K     4.3 x 10 _10
(c) Strong acid - weak base reaction       Kn = b =                  = 4.3 x 104
Kw    1.0 x 10 _14
K     6.5 x 10 _ 5
(d) Weak acid - strong base reaction Kn = a =           _14
= 6.5 x 109
K w 1.0 x 10
(c) < (b) < (d) < (a)

_4
K a = 4.5 x 10
16.45   (d) Weak acid - strong base reaction Kn =                         = 4.5 x 1010
Kw    1.0 x 10 _14

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1.8 x 10 _ 5
(c) Strong acid - weak base reaction           Kn = K b =           _14
= 1.8 x 109
K w 1.0 x 10
1          1
(a) Strong acid - strong base reaction         Kn =     =               = 1.0 x 1014
Kw     1.0 x 10 _14
_9           _5
K a K b = (2.0 x 10 )(1.8 x 10 )
(d) Weak acid - weak base reaction             Kn =                                     =
Kw              1.0 x 10 _14
3.6
(d) < (b) < (a) < (c)

16.46   (a) After mixing, the solution contains the basic salt, NaF; pH > 7.00
(b) After mixing, the solution contains the neutral salt, NaCl; pH = 7.00
Solution (a) has the higher pH.

16.47   (a) After mixing, the solution contains the neutral salt, NaClO4; pH = 7.00
(b) After mixing, the solution contains the acidic salt, NH4ClO4; pH < 7.00
Solution (b) has the lower pH.

(1.3 x 10 _10)(1.8 x 10 _ 9 )
16.48   Weak acid - weak base reaction         Kn = K a K boverK w =                                 =
1.0 x 10 _14
2.3 x 10S5
Kn is small so the neutralization reaction does not proceed very far to completion.

_5        _10
K a K b = (8.0 x 10 )(4.3 x 10 )
16.49   Weak acid - weak base reaction         Kn =                                    = 3.4
Kw              1.0 x 10 _14
Because Kn is close to 1, there will be an appreciable amount of aniline present at
equilibrium.

The CommonSIon Effect

16.50   HNO2(aq) + H2O(l)  H3O+(aq) + NO2S(aq)
(a) NaNO2 is a source of NO2S (reaction product). The equilibrium shifts towards
reactants, and the percent dissociation of HNO2 decreases.
(c) HCl is a source of H3O+ (reaction product). The equilibrium shifts towards reactants,
and the percent dissociation of HNO2 decreases.
(d) Ba(NO2)2 is a source of NO2S (reaction product). The equilibrium shifts towards
reactants, and the percent dissociation of HNO2 decreases.

16.51   NH3(aq) + H2O(l)  NH4+(aq) + OHS(aq)
(a) KOH is a strong base, and it increases the [OHS]. The pH increases.
(b) NH4NO3 is a source of NH4+ (reaction product). The equilibrium shifts towards
reactants, and the [OHS] decreases. The pH decreases.
(c) NH4Br is a source of NH4+ (reaction product). The equilibrium shifts towards
reactants, and the [OHS] decreases. The pH decreases.

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(d) KBr does not affect the pH of the solution.

16.52   (a) HF(aq) + H2O(l)  H3O+(aq) + FS(aq)
LiF is a source of FS (reaction product). The equilibrium shifts toward reactants, and the
[H3O+] decreases. The pH increases.
(b) Because HI is a strong acid, addition of KI, a neutral salt, does not change the pH.
(c) NH3(aq) + H2O(l)  NH4+(aq) + OHS(aq)
NH4Cl is a source of NH4+ (reaction product). The equilibrium shifts toward reactants,
and the [OHS] decreases. The pH decreases.

16.53   (a) NH3(aq) + H2O(l)  NH4+(aq) + OHS(aq)
NH4Cl is a source of NH4+ (reaction product). The equilibrium shifts toward reactants,
and the [OHS] decreases. The pH decreases.
(b) HCO3S(aq) + H2O(l)  H3O+(aq) + CO32S(aq)
Na2CO3 is a source of CO32S (reaction product). The equilibrium shifts toward reactants,
and the [H3O+] decreases. The pH increases.
(c) Because NaOH is a strong base, addition of NaClO4, a neutral salt, does not change
the pH.

16.54   For 0.25 M HF and 0.10 M NaF
HF(aq) + H2O(l)  H3O+(aq) + FS(aq)
initial (M)         0.25                      ~0        0.10
change (M)            Sx                      +x        +x
equil (M)           0.25 S x                   x      0.10 + x
[H 3O +][F _ ]                  x(0.10 + x) x(0.10)
Ka =                  = 3.5 x 10 _ 4 =            
[HF]                         0.25 _ x    0.25
+             S4
Solve for x. x = [H3O ] = 8.8 x 10 M
pH = Slog[H3O+] = Slog(8.8 x 10S4) = 3.06

16.55   On mixing equal volumes of two solutions, both concentrations are cut in half.
[CH3NH2] = 0.10 M; [CH3NH3Cl] = 0.30 M
CH3NH2(aq) + H2O(l)  CH3NH3+(aq) + OHS(aq)
initial (M)      0.10                           0.30            ~0
change (M)       Sx                              +x             +x
equil (M)       0.10 S x                       0.30 + x          x
+      _
[CH 3 NH 3 ][OH ]                  (0.30 + x) x (0.30) x
Kb =                     = 3.7 x 10 _ 4 =              
[CH 3 NH 2]                       0.10 _ x     0.10
S            S4
Solve for x. x = [OH ] = 1.2 x 10 M
_14
K w = 1.0 x 10
[H3O+] =        _             _4
= 8.1 x 10S11 M
[OH ] 1.2 x 10
pH = Slog[H3O+] = Slog(8.1 x 10S11) = 10.09

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16.56   For 0.10 M HN3:
HN3(aq) + H2O(l)  H3O+(aq) + N3S(aq)
initial (M)     0.10                             ~0             0
change (M)       Sx                              +x            +x
equil (M)       0.10 S x                          x              x
+     _                           2         2
[H 3O ][ N 3 ]                       x
Ka =                  = 1.9 x 10 _ 5 =               x
[HN 3]                        0.10 _ x 0.10
S3
Solve for x. x = 1.4 x 10 M
[HN 3]diss               1.4 x 10 _ 3 M
% dissociation =                    x 100% =                    x 100% = 1.4%
[HN 3]initial                0.10 M
For 0.10 M HN3 in 0.10 M HCl:
HN3(aq) + H2O(l)  H3O+(aq) + N3S(aq)
initial (M)     0.10                            0.10            0
change (M)       Sx                              +x            +x
equil (M)       0.10 S x                       0.10 + x          x
+     _
[H 3O ][ N 3 ]                   (0.10 + x)(x) (0.10)(x)
Ka =                   = 1.9 x 10 _ 5 =                            =x
[HN 3]                           0.10 _ x           0.10
Solve for x. x = 1.9 x 10S5 M
[HN 3]diss               1.9 x 10 _ 5 M
% dissociation =                    x 100% =                    x 100% = 0.019%
[HN 3]initial                0.10 M
The % dissociation is less because of the common ion (H3O+) effect.

16.57                     NH3(aq) + H2O(l)  NH4+(aq) +          OHS(aq)
initial (M)       0.30                        0          ~0
change (M)         Sx                       +x           +x
equil (M)         0.30 S x                    x           x
+      _                        2        2
[ NH 4 ][OH ]                      x
Kb =                  = 1.8 x 10 _ 5 =            x
[ NH 3]                      0.30 _ x 0.30
S
Solve for x. x = [OH ] = 2.3 x 10S3 M
_14
K w = 1.0 x 10
[H3O+] =                            = 4.3 x 10S12 M
[OH _ ] 2.3 x 10 _ 3
pH = Slog[H3O+] = Slog(4.3 x 10S12) = 11.37
Add 4.0 g of NH4NO3.
          1 mol 
 4.0 g x 80.04 g 
NH4NO3, 80.04 amu;        [NH4+] = molarity of NH4NO3 =                   = 0.50 M
0.100 L

NH3(aq) + H2O(l)  NH4+(aq) + OHS(aq)
initial (M)     0.30               0.50       ~0
change (M)       Sx                 +x        +x
equil (M)       0.30 S x          0.50 + x     x

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[ NH +][OH _ ]                  (0.50 + x) x (0.50) x
Kb =       4
= 1.8 x 10 _ 5 =              
[ NH 3]                       0.30 _ x     0.30
S             S5
Solve for x. x = [OH ] = 1.1 x 10 M
_14
K w = 1.0 x 10
[H3O+] =                          = 9.1 x 10S10 M
[OH _ ] 1.1 x 10 _ 5
pH = Slog[H3O+] = Slog(9.1 x 10S10) = 9.04
The % dissociation decreases because of the common ion effect.

Buffer Solutions

16.58   Solutions (a), (c) and (d) are buffer solutions. Neutralization reactions for (c) and (d)
result in solutions with equal concentrations of HF and FS.

16.59   Solutions (b), (c) and (d) are buffer solutions. Neutralization reactions for (b) and (d)
result in solutions with equal concentrations of NH3 and NH4+.

16.60   Both solutions buffer at the same pH because in both cases the [NO2S]/[HNO2] = 1.
Solution (a), however, has a higher concentration of both HNO2 and NO2S, and therefore
it has the greater buffer capacity.

16.61   Both solutions buffer at the same pH because in both cases the [NH3]/[NH4+] = 1.5.
Solution (b), however, has a higher concentration of both NH3 and NH4+, and therefore it
has the greater buffer capacity.

16.62   When blood absorbs acid, the equilibrium shifts to the left, decreasing the pH, but not by
much because the [HCO3S]/[H2CO3] ratio remains nearly constant. When blood absorbs
base, the equilibrium shifts to the right, increasing the pH, but not by much because the
[HCO3S]/[H2CO3] ratio remains nearly constant.

16.63   H2PO4S(aq) + H2O(l)  H3O+(aq) + HPO42S(aq)
For H2PO4S, K a 2 = 6.2 x 10 _ 8, pK a 2 = _log K a 2 = 7.21
[HPO 2 _ ]
4                   [HPO 2 _ ]
4
pH = 7.4 = p K a 2 + log          _
= 7.21 + log            _
[H 2 PO 4 ]                [H 2 PO 4 ]
[HPO 2 _ ]
4                [HPO 2 _ ]
4
To maintain pH near 7.4, need log                _
= 0.19 and          _
= 100.19 = 1.5
[H 2 PO 4 ]             [H 2 PO 4 ]
The principal buffer reactions are:
H3O+(aq) + HPO42S(aq)  H2PO4S(aq) + H2O(l)
OHS(aq) + H2PO4S(aq)  HPO42S(aq) + H2O(l)

[base]              [CN _ ]
16.64   pH = pKa + log       = pK a + log
[acid]              [HCN]
S10
For HCN, Ka = 4.9 x 10 , pKa = Slog Ka = Slog(4.9 x 10S10) = 9.31

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 0.12 
pH = 9.31 + log        = 9.09
 0.20 
The pH of a buffer solution will not change on dilution because the acid and base
concentrations will change by the same amount and their ratio will remain the same.

16.65   NaHCO3, 84.01 amu; Na2CO3, 105.99 amu
            1 mol 
 4.2 g x 84.01 g 
[HCO3S] = molarity of NaHCO3 =                            = 0.25 M
0.20 L
              1 mol 
 5.3 g x 105.99 g 
[CO32S] = molarity of Na2CO3 =                           = 0.25 M
0.20 L
[base]                     2
[CO 3 _ ]
pH = pKa + log         = pKa + log           _
[acid]               [ HCO 3 ]
For HCO3S, K a 2 = 5.6 x 10S11, p K a 2 = _log K a 2 = Slog(5.6 x 10S11) = 10.25
[0.25]
pH = 10.25 + log         = 10.25
[0.25]
The pH of a buffer solution will not change on dilution because the acid and base
concentrations will change by the same amount and their ratio will remain the same.

[base]                [ NH 3]
16.66   pH = pKa + log           = pK a + log
[acid]                [ NH +]
4
S10
For NH4 , Ka = 5.6 x 10 , pKa = Slog Ka = Slog(5.6 x 10S10) = 9.25
+

(0.200)
For the buffer: pH = 9.25 + log                = 9.25
(0.200)
(a) add 0.0050 mol NaOH, [OHS] = 0.0050 mol/0.500 L = 0.010 M
NH4+(aq) + OHS(aq)  NH3(aq) + H2O(l)
before reaction (M)        0.200            0.010          0.200
change (M) S0.010                 S0.010        +0.010
after reaction (M) 0.200 S 0.010              0      0.200 + 0.010
[ NH 3]                 (0.200 + 0.010)
pH = 9.25 + log             = 9.25 + log                   = 9.29
[ NH +]
4                 (0.200 _ 0.010)
(b) add 0.020 mol HCl, [H3O+] = 0.020 mol/0.500 L = 0.040 M
NH3(aq) + H3O+(aq)  NH4+(aq) + H2O(l)
before reaction (M)        0.200             0.040           0.200
change (M) S0.040                  S0.040         +0.040
after reaction (M) 0.200 S 0.040              0        0.200 + 0.040
[ NH 3]                 (0.200 _ 0.040)
pH = 9.25 + log         +
= 9.25 + log                   = 9.07
[ NH 4 ]                (0.200 + 0.040)

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[base]                      2
[SO 3 _ ]
16.67   pH = pKa + log            = pK a + log         _
[acid]                [ HSO 3 ]
For HSO3S, Ka = 6.3 x 10S8, pKa = Slog Ka = Slog(6.3 x 10S8) = 7.20
(0.300)
For the buffer: pH = 7.20 + log                = 6.98
(0.500)
(a) add (0.0050 L)(0.20 mol/L) = 0.0010 mol HCl = 0.0010 mol H3O+
mol HSO3S = (0.300 L)(0.500 mol/L) = 0.150 mol
mol SO32S = (0.300 L)(0.300 mol/L) = 0.0900 mol
SO32S(aq) + H3O+(aq)  HSO3S(aq) + H2O(l)
before reaction (mol)            0.0900           0.0010       0.150
change (mol)            S0.0010          S0.0010      +0.0010
after reaction (mol)       0.0900 S 0.0010         0       0.150 + 0.0010
2
[SO 3 _ ]               (0.0900 _ 0.0010)
pH = 7.20 + log           _
= 7.20 + log                     = 6.97
[ HSO 3 ]                 (0.150 + 0.0010)
(b) add (0.0050 L)(0.10 mol/L) = 0.00050 mol NaOH = 0.00050 mol OHS
HSO3S(aq) + OHS(aq)  SO32S(aq) + H2O(l)
before reaction (mol)            0.150            0.00050    0.0900
change (mol)         S0.00050           S0.00050    +0.00050
after reaction (mol) 0.150 S 0.00050                0      0.0900 + 0.00050
2
[SO 3 _ ]               (0.0900 + 0.00050)
pH = 7.20 + log           _
= 7.20 + log                       = 6.98
[ HSO 3 ]                 (0.150 _ 0.00050)

16.68          Acid           Ka                      pKa = Slog Ka
S10
(a)    H3BO3          5.8 x 10                9.24
(b)    HCO2H          1.8 x 10S4              3.74
S8
(c)    HOCl           3.5 x 10                7.46
The stronger the acid (the larger the Ka), the smaller is the pKa.

16.69   (a) Ka = 10 _ pK a = 10S5.00 = 1.0 x 10S5             (b) Ka = 10 _ pK a = 10S8.70 = 2.0 x 10S9
(b) is the weaker acid

_
[base]               [ HCO 2 ]
16.70   pH = pKa + log        = pK a + log
[acid]              [ HCO 2 H]
S4
For HCO2H, Ka = 1.8 x 10 ; pKa = Slog Ka = Slog(1.8 x 10S4) = 3.74
(0.50)
pH = 3.74 + log        = 4.04
(0.25)

_
[base]              [ HCO 3 ]
16.71   pH = pKa + log          = pKa + log
[acid]              [ H 2CO 3]
S7
For H2CO3, Ka = 4.3 x 10 ; pKa = Slog Ka = Slog(4.3 x 10S7) = 6.37
_                       _
[ HCO 3 ]               [ HCO 3 ]
7.40 = 6.37 + log            ; 1.03 = log
[ H 2CO 3]              [ H 2CO 3]

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_
[ HCO 3 ]                         [H 2CO 3]
= 101.03 = 10.7;             _
= 0.093
[ H 2CO 3]                        [HCO 3 ]

[base]              [ NH 3]
16.72   pH = pKa + log          = pKa + log
[acid]              [ NH +]
4
S10
For NH4 , Ka = 5.6 x 10 ; pKa = Slog Ka = Slog(5.6 x 10S10) = 9.25
+

[ NH 3]                   [ NH 3]   [ NH 3]
9.80 = 9.25 + log      +
;  0.550 = log       +
;      +
= 100.55 = 3.5
[ NH 4 ]                 [ NH 4 ]   [ NH 4 ]
The volume of the 1.0 M NH3 solution should be 3.5 times the volume of the 1.0 M
NH4Cl solution so that the mixture will buffer at pH 9.80.

_
[base]               [CH 3CO 2 ]
16.73   pH = pKa + log          = pKa + log
[acid]              [CH 3CO 2 H]
S5
For CH3CO2H, Ka = 1.8 x 10 ; pKa = Slog Ka = Slog(1.8 x 10S5) = 4.74
_                              _
[CH 3CO 2 ]                     [CH 3CO 2 ]
4.44 = 4.74 + log               ;     S0.30 = log
[CH 3CO 2 H]                    [CH 3CO 2 H]
_
[CH 3CO 2 ]
= 10S0.30 = 0.50
[CH 3CO 2 H]
The solution should have 0.50 mol of CH3CO2S per mole of CH3CO2H. For example,
you could dissolve 41g of CH3CO2Na in 1.00 L of 1.00 M CH3CO2H.

16.74   H3PO4, K a1 = 7.5 x 10 _ 3; pK a1 = _log K a1 = 2.12
H2PO4S, K a 2 = 6.2 x 10 _ 8; pK a 2 = _log K a 2 = 7.21
HPO42S, K a 3 = 4.8 x 10 _13; pK a 3 = _log K a 3 = 12.32
The buffer system of choice for pH 7.00 is (b) H2PO4S S HPO42S because the pKa for
H2PO4S (7.21) is closest to 7.00.

16.75   HSO4S, Ka2 = 1.2 x 10S2; pKa2 = Slog Ka2 = 1.92
HOCl, Ka = 3.5 x 10S8; pKa = Slog Ka = 7.56
C6H5CO2H, Ka = 6.5 x 10S5; pKa = Slog Ka = 4.19
The buffer system of choice for pH = 4.50 is (c) C6H5CO2H - C6H5CO2S because the pKa
for C6H5CO2H (4.19) is closest to 4.50.

pH Titration Curves

16.76   (a) (0.060 L)(0.150 mol/L)(1000 mmol/mol) = 9.00 mmol HNO3
 1 mmol NaOH         1 mL NaOH       
(b) vol NaOH = (9.00 mmol HNO3)                                            = 20.0 mL NaOH
 1 mmol HNO3   0.450 mmol NaOH 
(c) At the equivalence point the solution contains the neutral salt NaNO3. The pH is 7.00.

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(d)

16.77

mmol NaOH = (50.0 mL)(1.0 mmol/mL) = 50 mmol
mmol HCl = mmol NaOH = 50 mmol
 1.0 mL 
vol HCl = (50 mmol)            = 50 mL
 1.0 mmol 
50 mL of 1.0 M HCl is needed to reach the equivalence point.

16.78   mmol OHS = (20.0 mL)(0.150 mmol/mL) = 3.00 mmol
mmol acid present = mmol OHS added = 3.00 mmol
3.00 mmol
[acid] =            = 0.0500 M
60.0 mL

16.79   mmol OHS = (60.0 mL)(0.240 mmol/mL) = 14.4 mmol
1 mmol acid
mmol acid present = 14.4 mmol OHS x             = 7.20 mmol acid
2 mmol OH _
7.20 mmol
[acid] =            = 0.288 M
25.0 mL

16.80   HBr(aq) + NaOH(aq)  Na+(aq) + BrS(aq) + H2O(l)

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(a) [H3O+] = 0.120 M; pH = Slog[H3O+] = Slog (0.120) = 0.92
(b) (50.0 mL)(0.120 mmol/mL) = 6.00 mmol HBr
(20.0 mL)(0.240 mmol/mL) = 4.80 mmol NaOH
6.00 mmol HBr S 4.80 mmol NaOH = 1.20 mmol HBr after neutralization
1.20 mmol
[H3O+] =                            = 0.0171 M
(50.0 mL + 20.0 mL)
pH = Slog[H3O+] = Slog(0.0171) = 1.77
(c) (24.9 mL)(0.240 mmol/mL) = 5.98 mmol NaOH
6.00 mmol HBr S 5.98 mmol NaOH = 0.02 mmol HBr after neutralization
0.02 mmol
[H3O+] =                            = 3 x 10S4 M
(50.0 mL + 24.9 mL)
pH = Slog[H3O+] = Slog(3 x 10S4) = 3.5
(d) The titration reaches the equivalence point when 25.0 mL of 0.240 M NaOH is
added. At the equivalence point the solution contains the neutral salt NaBr. The pH is
7.00.
(e) (25.1 mL)(0.240 mmol/mL) = 6.024 mmol NaOH
6.024 mmol NaOH S 6.00 mmol HBr = 0.024 mmol NaOH after neutralization
0.024 mmol
[OHS] =                            = 3.2 x 10S4 M
(50.0 mL + 25.1 mL)
_14
K w = 1.0 x 10
[H3O+] =                          = 3.1 x 10S11 M
[OH _ ] 3.2 x 10 _ 4
pH = Slog[H3O+] = Slog(3.1 x 10S11) = 10.5
(f) (40.0 mL)(0.240 mmol/mL) = 9.60 mmol NaOH
9.60 mmol NaOH S 6.00 mmol HBr = 3.60 mmol NaOH after neutralization
3.60 mmol
[OHS] =                            = 0.040 M
(50.0 mL + 40.0 mL)
_14
K w = 1.0 x 10
[H3O+] =         _
= 2.5 x 10S13 M
[OH ]        0.040
pH = Slog[H3O ] = Slog(2.5 x 10S13) = 12.60
+

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16.81   Ba(OH)2(aq) + 2 HNO3(aq)  Ba2+(aq) + 2 NO3S(aq) + 2 H2O(l)
(a) [OHS] = 2(0.150 M) = 0.300 M
1.0 x 10 _14
[H3O+] = K w_ =                    = 3.33 x 10S14 M
[OH ]          0.300
pH = Slog[H3O ] = Slog(3.33 x 10S14) = 13.48
+

(b) (40.0 mL)(0.150 mmol/mL) = 6.00 mmol Ba(OH)2
2 mmol OH _
6.00 mmol Ba(OH)2 x                            = 12.0 mmol OHS
1 mmol Ba(OH ) 2
(10.0 mL)(0.400 mmol/mL) = 4.00 mmol HNO3
12.0 mmol OHS S 4.00 mmol HNO3 = 8.00 mmol OHS after neutralization
8.00 mmol
[OHS] =                                = 0.160 M
(40.0 mL + 10.0 mL)
_14
K w = 1.0 x 10
[H3O+] =                           = 6.25 x 10S14 M
[OH _ ]        0.160
pH = Slog[H3O+] = Slog(6.25 x 10S14) = 13.20
(c) (20.0 mL)(0.400 mmol/mL) = 8.00 mmol HNO3
12.0 mmol OHS S 8.00 mmol HNO3 = 4.00 mmol OHS after neutralization
4.00 mmol
[OHS] =                                 = 0.0667 M
(40.0 mL + 20.0 mL)
_14
K w = 1.0 x 10
[H3O+] =        _
= 1.50 x 10S13 M
[OH ]         0.0667
pH = Slog[H3O ] = Slog(1.50 x 10S13) = 12.82
+

(d) The titration reaches the equivalence point when 30.0 mL of 0.400 M HNO3 is
added. At the equivalence point the solution contains the neutral salt Ba(NO3)2. The pH
is 7.00.
(e) (40.0 mL)(0.400 mmol/mL) = 16.0 mmol HNO3
16.0 mmol HNO3 S 12.0 mmol OHS = 4.00 mmol H3O+ after neutralization
4.00 mmol
[H3O+] =                                 = 0.0500 M
(40.0 mL + 40.0 mL)
pH = Slog[H3O+] = Slog(0.0500) = 1.30

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16.82   mmol HF = (40.0 mL)(0.250 mmol/mL) = 10.0 mmol
mmol NaOH required = mmol HF = 10.0 mmol
 1.00 mL 
mL NaOH required = (10.0 mmol)                      = 50.0 mL
 0.200 mmol 
50.0 mL of 0.200 M NaOH is required to reach the equivalence point.
For HF, Ka = 3.5 x 10S4; pKa = Slog Ka = Slog(3.5 x 10S4) = 3.46
(a) mmol HF = 10.0 mmol
mmol NaOH = (0.200 mmol/mL)(10.0 mL) = 2.00 mmol
Neutralization reaction:     HF(aq) + OHS(aq)  FS(aq) + H2O(l)
before reaction (mmol)       10.0         2.00           0
change (mmol)                S2.00       S2.00        +2.00
after reaction (mmol) 8.0         0            2.00
8.0 mmol                                     2.00 mmol
[HF] =                        = 0.16 M; [FS] =                            = 0.0400 M
(40.0 mL + 10.0 mL)                          (40.0 mL + 10.0 mL)
HF(aq) + H2O(l)  H3O+(aq) + FS(aq)
initial (M)     0.16                       ~0      0.0400
change (M) Sx                              +x      +x
equil (M)       0.16 S x                    x     0.0400 + x
+   _
[H 3O ][F ]                   x(0.0400 + x) x(0.0400)
Ka =                = 3.5 x 10 _ 4 =              
[HF]                         0.16 _ x      0.16
+              S3
Solve for x. x = [H3O ] = 1.4 x 10 M
pH = Slog[H3O+] = Slog(1.4 x 10S3) = 2.85
(b) Halfway to the equivalence point,
pH = pKa = Slog Ka = Slog(3.5 x 10S4) = 3.46
(c) At the equivalence point only the salt NaF is in solution.
10.0 mmol
[FS] =                              = 0.111 M
(40.0 mL + 50.0 mL)
FS(aq) + H2O(l)  HF(aq) + OHS(aq)
initial (M)    0.111          0         ~0
change (M)      Sx                 +x           +x
equil (M)      0.111 S x             x           x
_14
Kw        1.0 x 10
For FS, Kb =             =          _4
= 2.9 x 10S11
K a for HF 3.5 x 10

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[HF][OH _ ]                       x
2        2
Kb =              = 2.9 x 10 _11 =             x
[F _ ]                    0.111 _ x 0.111
S
Solve for x. x = [OH ] = 1.8 x 10S6 M
_14
K w = 1.0 x 10
[H3O+] =                         = 5.6 x 10S9 M
[OH _ ] 1.8 x 10 _ 6
pH = Slog[H3O+] = Slog(5.6 x 10S9) = 8.25
(d) mmol HF = 10.0 mmol
mol NaOH = (0.200 mmol/mL)(80.0 mL) = 16.0 mmol

Neutralization reaction:      HF(aq) + OHS(aq)  FS(aq) + H2O(l)
before reaction (mmol)        10.0       16.0           0
change (mmol)               S10.0       S10.0         +10.0
after reaction (mmol) 0           6.0         10.0
After the equivalence point, the pH of the solution is determined by the [OHS].
6.0 mmol
[OHS] =                          = 5.0 x 10S2 M
(40.0 mL + 80.0 mL)
_14
K w = 1.0 x 10
[H3O+] =                        = 2.0 x 10S13 M
[OH _ ] 5.0 x 10 _ 2
pH = Slog[H3O+] = Slog(2.0 x 10S13) = 12.70

16.83   mmol CH3NH2 = (100.0 mL)(0.100 mmol/mL) = 10.0 mmol
mmol HNO3 required = mmol CH3NH2 = 10.0 mmol
 1.00 mL 
vol HNO3 required = (10.0 mmol)                     = 40.0 mL
 0.250 mmol 
40.0 mL of 0.250 M HNO3 are required to reach the equivalence point.
(a)             CH3NH2(aq) + H2O(l)  CH3NH3+(aq) + OHS(aq)
initial (M)     0.100                           0             ~0
change (M)       Sx                            +x             +x
equil (M)       0.100 S x                         x            x
+     _                         2
[CH 3 NH 3 ][OH ]                      x
Kb =                     = 3.7 x 10 _ 4 =
[CH 3 NH 2]                     0.100 _ x
2             S4              S5
x + (3.7 x 10 )x S (3.7 x 10 ) = 0
Use the quadratic formula to solve for x.
_ (3.7 x 10 _ 4)    (3.7 x 10 _ 4) 2 _ (4)(_ 3.7 x 10 _ 5) _ 3.7 x 10 _ 4  0.0122
x=                                                              =
2(1)                                         2
x = 0.0059 and S0.0063
Of the two solutions for x, only the positive value of x has physical meaning because x is
the [OHS].
[OHS] = x = 0.0059 M
_14
K w = 1.0 x 10
[H3O+] =      _               _3
= 1.7 x 10S12 M
[OH ] 5.9 x 10

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Chapter 16 S Applications of Aqueous Equilibria
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pH = Slog[H3O+] = Slog(1.7 x 10S12) = 11.77
(b) 20.0 mL of HNO3 is halfway to the equivalence point.
Kw            1.0 x 10 _14
For CH3NH3+, Ka =                      =              = 2.7 x 10S11
Kb  for CH 3 NH 2 3.7 x 10 _ 4
pH = pKa = Slog(2.7 x 10S11) = 10.57
(c) At the equivalence point only the salt CH3NH3NO3 is in solution.
mmol CH3NH3NO3 = (0.100 mmol/mL)(100.0 mL) = 10.0 mmol
10.0 mmol
[CH3NH3+] =                            = 0.0714 M
(100.0 mL + 40.0 mL)
CH3NH3+(aq) + H2O(l)  H3O+(aq) + CH3NH2(aq)
initial (M)     0.0714                ~0              0
change (M)        Sx                          +x              +x
equil (M)       0.0714 S x                     x              x
+                                    2           2
[H 3O ][CH 3 NH 2]                      x
Ka =                +
= 2.7 x 10 _11 =               x
[CH 3 NH 3 ]                    0.0714 _ x 0.0714
+            S6
Solve for x. x = [H3O ] = 1.4 x 10 M
pH = Slog[H3O+] = Slog(1.4 x 10S6) = 5.85
(d) mmol CH3NH2 = (0.100 mmol/mL)(100.0 mL) = 10.0 mmol
mmol HNO3 = (0.250 mmol/mL)(60.0 mL) = 15.0 mmol
Neutralization reaction:       CH3NH2(aq) + H3O+(aq)  CH3NH3+(aq) + H2O(l)
before reaction (mmol)          10.0             15.0             0
change (mmol)                  S10.0            S10.0           +10.0
after reaction (mmol) 0                   5.0            10.0
After the equivalence point the pH of the solution is determined by the [H3O+].
5.0 mmol
[H3O+] =                             = 3.1 x 10S2 M
(100.0 mL + 60.0 mL)
pH = Slog[H3O+] = Slog(3.1 x 10S2) = 1.51

16.84   For H2A+, Ka1 = 4.6 x 10S3 and Ka2 = 2.0 x 10S10
(a) (10.0 mL)(0.100 mmol/mL) = 1.00 mmol NaOH added = 1.00 mmol HA produced.
(50.0 mL)(0.100 mmol/mL) = 5.00 mmol H2A+
5.00 mmol H2A+ S 1.00 mmol NaOH = 4.00 mmol H2A+ after neutralization
4.00 mmol
[H2A+] =                         = 6.67 x 10S2 M
(50.0 mL + 10.0 mL)
1.00 mmol
[HA] =                         = 1.67 x 10S2 M
(50.0 mL + 10.0 mL)
[HA]                               1.67 x 10 _ 2 
pH = pKa1 + log            = Slog(4.6 x 10S3) + log            _2 
= 1.74
[H 2A +]                           6.67 x 10 
(b) Halfway to the first equivalence point, pH = pKa1 = 2.34
pK a1 + pK a 2
(c) At the first equivalence point, pH =                = 6.02
2
(d) Halfway between the first and second equivalence points, pH = pKa2 = 9.70

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(e) At the second equivalence point only the basic salt, NaA, is in solution.
Kw              1.0 x 10 _14
Kb =               = Kw =             _10
= 5.0 x 10S5
K a for HA K a 2 2.0 x 10
mmol AS = (50.0 mL)(0.100 mmol/mL) = 5.00 mmol
5.0 mmol
[AS] =                             = 3.3 x 10S2 M
(50.0 mL + 100.0 mL)
AS(aq) + H2O(l)  HA(aq) + OHS(aq)
initial (M)      0.033                   0       ~0
change (M) Sx                            +x      +x
equil (M)        0.033 S x                x        x
_                                2
[HA][OH ]                        (x)(x)
Kb =                = 5.0 x 10 _ 5 =            x
[A _ ]                     0.033 _ x 0.033
Solve for x.
x = [OHS] = (5.0 x 10_5)(0.033) = 1.3 x 10S3 M
_14
K w = 1.0 x 10
[H3O+] =                      = 7.7 x 10S12 M
[OH _ ] 1.3 x 10 _ 3
pH = Slog[H3O+] = Slog(7.7 x 10S12) = 11.11

16.85   For H2CO3, Ka1 = 4.3 x 10S7 and Ka2 = 5.6 x 10S11
(a) (25.0 mL)(0.0200 mmol/mL) = 0.500 mmol H2CO3
(10.0 mL)(0.0250 mmol/mL) = 0.250 mmol KOH added
0.500 mmol H2CO3 S 0.250 mmol KOH = 0.250 mmol HCO3S produced
This is halfway to the first equivalence point where pH = pKa1 = Slog(4.3 x 10S7) = 6.37
pK a1 + pK a 2
(b) At the first equivalence point, pH =                 = 8.31
2
(c) Halfway between the first and second equivalence points, pH = pKa2 = 10.25
(d) At the second equivalence point only the basic salt, K2CO3, is in solution.
Kw                  1.0 x 10 _14
Kb =                  _
= Kw =             _11
= 1.8 x 10S4
K a for HCO 3 K a 2 5.6 x 10
mmol CO32S = mmol H2CO3 = 0.500 mmol
0.500 mmol
[CO32S] =                              = 0.00769 M
(25.0 mL + 40.0 mL)
CO32S(aq) + H2O(l)  HCO3S(aq) + OHS(aq)
initial (M)    0.00769                      0      ~0
change (M)        Sx                        +x     +x
equil (M)      0.00769 S x                   x       x
_       _
[HCO 3 ][OH ]                       (x)(x)
Kb =           2_
= 1.8 x 10 _ 4 =
[CO 3 ]                      0.00769 _ x
Use the quadratic formula to solve for x.
x=

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Chapter 16 S Applications of Aqueous Equilibria
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_ (1.8 x 10 _ 4)   (1.8 x 10 _ 4) 2 _ (4)(1)(_ 1.4 x 10 _ 6) (_1.8 x 10 _ 4)  (2.37 x 10 _ 3)
=
2(1)                                                2
S3                  S3
x = S1.27 x 10 and 1.09 x 10
Of the two solutions for x, only the positive value of x has physical meaning because x is
the [OHS].
[OHS] = x = 1.09 x 10S3 M
_14
K w = 1.0 x 10
[H3O+] =                            = 9.2 x 10S12 M
[OH _ ] 1.09 x 10 _ 3
pH = Slog[H3O+] = Slog(9.2 x 10S12) = 11.04
(e) excess KOH
(50.0 mL S 40.0 mL)(0.025 mmol/mL) = 0.250 mmol KOH = 0.250 mmol OHS
0.250 mmol
[OHS] =                              = 3.33 x 10S3 M
(25.0 mL + 50.0 mL)
_14
K w = 1.0 x 10
+
[H3O ] =       _                   _3
= 3.0 x 10S12 M
[OH ] 3.33 x 10
pH = Slog[H3O+] = Slog(3.0 x 10S12) = 11.52

16.86   When equal volumes of acid and base react, all concentrations are cut in half.
(a) At the equivalence point only the salt NaNO2 is in solution.
[NO2S] = 0.050 M
Kw         1.0 x 10 _14
For NO2S, Kb =                      =               = 2.2 x 10S11
Ka  for HNO 2 4.5 x 10 _ 4
NO2S(aq) + H2O(l)  HNO2(aq) + OHS(aq)
Initial (M)         0.050                 0           ~0
change (M)           Sx                         +x            +x
equil (M)           0.050 S x                     x             x
_                                    2
[HNO 2][OH ]                          (x)(x)
Kb =              _
= 2.2 x 10 _11 =              x
[ NO 2 ]                       0.050 _ x 0.050
S             S6
Solve for x. x = [OH ] = 1.1 x 10 M
_14
K w = 1.0 x 10
[H3O+] =           _             _6
= 9.1 x 10S9 M
[OH ] 1.1 x 10
pH = Slog[H3O+] = Slog(9.1 x 10S9) = 8.04
Phenol red would be a suitable indicator. (see Figure 15.4)
(b) The pH is 7.00 at the equivalence point for the titration of a strong acid (HI) with a
strong base (NaOH).
Bromthymol blue or phenol red would be suitable indicators. (Any indicator that
changes color in the pH range 4 S 10 is satisfactory for a strong acid S strong base
titration.)
(c) At the equivalence point only the salt CH3NH3Cl is in solution.
[CH3NH3+] = 0.050 M
Kw             1.0 x 10 _14
For CH3NH3+, Ka =                            =              = 2.7 x 10S11
Kb  for CH 3 NH 2 3.7 x 10 _ 4

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Chapter 16 S Applications of Aqueous Equilibria
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CH3NH3+(aq) + H2O(l)  H3O+(aq) + CH3NH2(aq)
initial (M)     0.050                          ~0          0
change (M) Sx                                 +x          +x
equil (M)       0.050 S x                      x           x
+                                          2
[H 3O ][CH 3 NH 2]                    (x)(x)
Ka =                +
= 2.7 x 10 _11 =           x
[CH 3 NH 3 ]                    0.050 _ x 0.050
+           S6
Solve for x. x = [H3O ] = 1.2 x 10 M
pH = Slog[H3O+] = Slog(1.2 x 10S6) = 5.92
Chlorphenol red would be a suitable indicator.

16.87   When equal volumes of acid and base react, all concentrations are cut in half.
(a) At the equivalence point only the salt C5H11NHNO3 is in solution.
[C5H11NH+] = 0.10 M
Kw            1.0 x 10 _14
For C5H11NH+, Ka =                           =            _3
= 7.7 x 10S12
K b for C5H11 N 1.3 x 10
C5H11NH+(aq) + H2O(l)  H3O+(aq) + C5H11N(aq)
initial (M)      0.10                               ~0                0
change (M) Sx                                      +x                +x
equil (M)        0.10 S x                            x                x
+                                                  2
[H 3O ][C5H11 N]                         (x)(x)
Ka =                  +
= 7.7 x 10 _12 =               x
[C5H11NH ]                           0.10 _ x 0.10
+            S7
Solve for x. x = [H3O ] = 8.8 x 10 M
pH = Slog[H3O+] = Slog(8.8 x 10S7) = 6.06
Alizarin would be a suitable indicator (see Figure 15.4)
(b) At the equivalence point only the salt Na2SO3 is in solution.
[SO32S] = 0.10 M
Kw         1.0 x 10 _14
For SO32S, Kb =                    _
=            _8
= 1.6 x 10S7
K a for HSO 3 6.3 x 10
SO32S(aq) + H2O(l)  HSO3S(aq) + OHS(aq)
Initial (M)        0.10                           0               ~0
change (M)          Sx                            +x              +x
equil (M)          0.10 S x                        x                x
_       _                                      2
[HSO 3 ][OH ]                         (x)(x)
Kb =            2_
= 1.6 x 10 _ 7 =               x
[SO 3 ]                        0.10 _ x 0.10
S              S4
Solve for x. x = [OH ] = 1.26 x 10 M
_14
K w = 1.0 x 10
[H3O+] =                             = 7.9 x 10S11 M
[OH _ ] 1.26 x 10 _ 4
pH = Slog[H3O+] = Slog(7.9 x 10S11) = 10.10
Thymolphthalein would be a suitable indicator.
(c) The pH is 7.00 at the equivalence point for the titration of a strong acid (HBr) with a
strong base (Ba(OH)2).
Alizarin, bromthymol blue, or phenol red would be suitable indicators. (Any indicator

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that changes color in the pH range 4 - 10 is satisfactory for a strong acid - strong base
titration.)
Solubility Equilibria

16.88   (a)   Ag2CO3(s)  2 Ag+(aq) + CO32S(aq)                Ksp = [Ag+]2[CO32S]
(b)   PbCrO4(s)  Pb2+(aq) + CrO42S(aq)                Ksp = [Pb2+][CrO42S]
(c)   Al(OH)3(s)  Al3+(aq) + 3 OHS(aq)                Ksp = [Al3+][OHS]3
(d)   Hg2Cl2(s)  Hg22+(aq) + 2 ClS(aq)                Ksp = [Hg22+][ClS]2

16.89   (a) Ksp = [Ca2+][OHS]2                   (b) Ksp = [Ag+]3[PO43S]
(c) Ksp = [Ba2+][CO32S]                  (d) Ksp = [Ca2+]5[PO43S]3[OHS]

16.90   (a) Ksp = [Pb2+][IS]2 = (5.0 x 10S3)(1.3 x 10S3)2 = 8.4 x 10S9
K sp       (8.4 x 10 _ 9
(b) [IS] =          =                   = 5.8 x 10S3 M
[Pb 2+]     (2.5 x 10 _ 4)
K sp   (8.4 x 10 _ 9)
(c) [Pb2+] = _ 2 =                    = 0.13 M
[I ] (2.5 x 10 _ 4) 2

16.91   (a) Ksp = [Ca2+]3[PO42S]2 = (2.9 x 10S7)3(2.9 x 10S7)2 = 2.1 x 10S33
K sp      2.1 x 10 _ 33
(b) [Ca2+] =    3
2_ 2
=3           2
= 2.8 x 10S10 M
[PO 4 ]       (0.010)
K sp     2.1 x 10 _ 33
(c) [PO42S] =                 =               = 4.6 x 10S14 M
[Ca 2+]3     (0.010) 3

16.92                 Ag2CO3(s)  2 Ag+(aq) + CO32S(aq)
equil (M)                      2x            x
[Ag+] = 2x = 2.56 x 10S4 M; [CO32S] = x = (2.56 x 10S4 M)/2 = 1.28 x 10S4 M
Ksp = [Ag+]2[CO32S] = (2.56 x 10S4)2(1.28 x 10S4) = 8.39 x 10S12

16.93   (a) [Cd2+] = [CO32S] = 2.5 x 10S6 M
Ksp = [Cd2+][CO32S] = (2.5 x 10S6)2 = 6.2 x 10S12
(b) [Ca2+] = 1.06 x 10S2 M
[OHS] = 2[Ca2+] = 2(1.06 x 10S2 M) = 2.12 x 10S2 M
Ksp = [Ca2+][OHS]2 = (1.06 x 10S2)(2.12 x 10S2)2 = 4.76 x 10S6
(c) PbBr2, 367.01 amu
            1 mol 
 4.34 g x 367.01 g 
[Pb2+] = molarity of PbBr2 =                        = 1.18 x 10S2 M
1L
S                          S2
[Br ] = 2[Pb ] = 2(1.18 x 10 M) = 2.36 x 10S2 M
2+

Ksp = [Pb2+][BrS]2 = (1.18 x 10S2)(2.36 x 10S2)2 = 6.57 x 10S6
(d) BaCrO4, 253.32 amu

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         -3    1 mol 
 2.8 x 10 g x 253.32 g 
[Ba2+] = [CrO42S] = molarity of BaCrO4 =                            = 1.1 x 10S5 M
1L
Ksp = [Ba2+][CrO42S] = (1.1 x 10S5)2 = 1.2 x 10S10

16.94   (a)                  BaCrO4(s)  Ba2+(aq) + CrO42S(aq)
equil (M)                      x      x
Ksp = [Ba2+][CrO42S] = 1.2 x 10S10 = (x)(x)
molar solubility = x = 1.2 x 10_10 = 1.1 x 10S5 M
(b)                   Mg(OH)2(s)  Mg2+(aq) + 2 OHS(aq)
equil (M)                     x            2x
2+      S 2        S12        2     3
Ksp = [Mg ][OH ] = 5.6 x 10 = x(2x) = 4x
5.6 x 10 _12
molar solubility = x =   3          = 1.1 x 10S4 M
4
(c)                  Ag2SO3(s)  2 Ag+(aq) + SO32S(aq)
equil (M)                     2x          x
Ksp = [Ag+]2[SO32S] = 1.5 x 10S14 = (2x)2x = 4x3
1.5 x 10 _14
molar solubility = x =   3                = 1.6 x 10S5 M
4

16.95   (a)                  Ag2CO3(s)  2 Ag+(aq) + CO32S(aq)
equil (M)                      2x           x
Ksp = [Ag+]2[CO32S] = 8.4 x 10S12 = (2x)2(x) = 4x3
8.4 x 10 _12
molar solubility = x =   3                = 1.3 x 10S4 M
4
Ag2CO3, 275.75 amu
solubility = (1.3 x 10S4 mol/L)(275.75 g/mol) = 0.036 g/L
(b)                    CuBr(s)  Cu+(aq) + BrS(aq)
equil (M)                   x x
+     S           S9
Ksp = [Cu ][Br ] = 6.3 x 10 = (x)(x)
molar solubility = x = 6.3 x 10-9 = 7.9 x 10S5 M
CuBr, 143.45 amu
solubility = (7.9 x 10S5 mol/L)(143.45 g/mol) = 0.011 g/L
(c)                     Cu3(PO4)2(s)  3 Cu2+(aq) + 2 PO43S(aq)
equil (M)                         3x           2x
S37
2+ 3     3S 2                   3    2
Ksp = [Cu ] [PO4 ] = 1.4 x 10 = (3x) (2x) = 108x5
1.4 x 10 _ 37
molar solubility = x =   5                 = 1.7 x 10S8
108
Cu3(PO4)2 , 380.58 amu
solubility = (1.7 x 10S8 mol/L)(380.58 g/mol) = 6.5 x 10S6 g/L

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Factors That Affect Solubility

16.96   Ag2CO3(s)  2 Ag+(aq) + CO32S(aq)
(a) AgNO3, source of Ag+; equilibrium shifts left
(b) HNO3, source of H3O+, removes CO32S; equilibrium shifts right
(c) Na2CO3, source of CO32S; equilibrium shifts left
(d) NH3, forms Ag(NH3)2+; removes Ag+; equilibrium shifts right

16.97   BaF2(s)  Ba2+(aq) + 2 FS(aq)
(a) H+ from HCl reacts with FS forming the weak acid HF. The equilibrium shifts to the
right increasing the solubility of BaF2.
(b) KF, source of FS; equilibrium shifts left, solubility of BaF2 decreases.
(c) No change in solubility.
(d) Ba(NO3)2, source of Ba2+; equilibrium shifts left, solubility of BaF2 decreases.

16.98   (a)                  PbCrO4(s)  Pb2+(aq) + CrO42S(aq)
equil (M)                      x      x
Ksp = [Pb2+][CrO42S] = 2.8 x 10S13 = (x)(x)
molar solubility = x =  2.8 x 10_13 = 5.3 x 10S7 M
(b)                   PbCrO4(s)  Pb2+(aq) + CrO42S(aq)
initial(M)                     0          1.0 x 10S3
equil (M)                      x        1.0 x 10S3 + x
Ksp = [Pb2+][CrO42S] = 1.2 x 10S10 = (x)(1.0 x 10S3 + x)  (x)(1.0 x 10S3)
2.8 x 10 _13
molar solubility = x =              = 2.8 x 10S10 M
1 x 10 _ 3

16.99   (a)                    SrF2(s)  Sr2+(aq) + 2 FS(aq)
initial (M)                0.010        0
equil (M)                  0.010 + x    2x
Ksp = [Sr2+][FS]2 = 4.3 x 10S9 = (0.010 + x)(2x)2  (0.010)(2x)2 = 0.040 x2
4.3 x 10 _ 9
molar solubility = x =                  = 3.3 x 10S4 M
0.040
(b)                    SrF2(s)  Sr2+(aq) + 2 FS(aq)
initial (M)                 0             0.010
equil (M)                   x             0.010 + 2x
Ksp = [Sr2+][FS]2 = 4.3 x 10S9 = (x)(0.010 + 2x)2  (x)(0.010)2 = x(0.00010)
4.3 x 10 _ 9
molar solubility = x =                = 4.3 x 10S5 M
0.00010

16.100 (b), (c), and (d) are more soluble in acidic solution.
(a) AgBr(s)  Ag+(aq) + BrS(aq)
(b) CaCO3(s) + H3O+(aq)  Ca2+(aq) + HCO3S(aq) + H2O(l)

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(c) Ni(OH)2(s) + 2 H3O+(aq)  Ni2+(aq) + 4 H2O(l)
(d) Ca3(PO4)2(s) + 2 H3O+(aq)  3 Ca2+(aq) + 2 HPO42S(aq) + 2 H2O(l)

16.101 (a), (b), and (d) are more soluble in acidic solution.
(a) MnS(s) + 2 H3O+(aq)  Mn2+(aq) + H2S(aq) + 2 H2O(l)
(b) Fe(OH)3(s) + 3 H3O+(aq)  Fe3+(aq) + 6 H2O(l)
(c) AgCl(s)  Ag+(aq) + ClS(aq)
(d) BaCO3(s) + H3O+(aq)  Ba2+(aq) + HCO3S(aq) + H2O(l)

16.102 On mixing equal volumes of two solutions, the concentrations of both solutions are cut in half.
Ag+(aq) + 2 CNS(aq)  Ag(CN)2S(aq)
before reaction (M)            0.0010          0.10               0
assume 100% reaction           S0.0010        S2(0.0010)        0.0010
after reaction (M)               0            0.098            0.0010
assume small back rxn          +x             +2x                Sx
equil (M)        x         0.098 + 2x        0.0010 S x
_
[Ag(CN ) 2 ]     (0.0010 _ x)     0.0010
Kf = 3.0 x 1020 =                  =                
+                          2
[Ag ][CN _ ] x(0.098 + 2 x ) x(0.098 ) 2
2

Solve for x. x = [Ag+] = 3.5 x 10S22 M

16.103                                   Cr3+(aq) + 4 OHS(aq)  Cr(OH)4S(aq)
before reaction (M)          0.0050            1.0               0
assume 100% reaction S0.0050                S(4)(0.0050)       +0.0050
after reaction(M)               0                0.98             0.0050
assume small back rxn           +x               +4x                Sx
equil (M)         x            0.98 + 4x         0.0050 S x
_
[Cr(OH ) 4 ]                 (0.0050 _ x)        (0.0050)
Kf =              _ 4
= 8 x 1029 =                    
3+
[Cr ][OH ]                                    4
(x)(0.98 + 4 x ) (x)(0.98) 4
Solve for x. x = [Cr3+] = 6.8 x 10S33 M = 7 x 10S33 M
[Cr 3+]      7 x 10 _ 33 M
fraction uncomplexed Cr3+ =                _
=                = 1.4 x 10S30 = 1 x 10S30
[Cr(OH ) 4 ]     0.0050 M

16.104 (a)            AgI(s)  Ag+(aq) + IS(aq)                                 Ksp = 8.5 x 10S17
Ag+(aq) + 2 CNS(aq)  Ag(CN)2S(aq)                        Kf = 3.0 x 1020
dissolution rxn AgI(s) + 2 CNS(aq)  Ag(CN)2S(aq) + IS(aq)
K = (Ksp)(Kf) = (8.5 x 10S17)(3.0 x 1020) = 2.6 x 104
(b)            Al(OH)3(s)  Al3+(aq) + 3 OHS(aq)                        Ksp = 1.9 x 10S33
Al3+(aq) + 4 OHS(aq)  Al(OH)4S(aq)                      Kf = 3 x 1033
dissolution rxn Al(OH)3(s) + OHS(aq)  Al(OH)4S(aq)
K = (Ksp)(Kf) = (1.9 x 10S33)(3 x 1033) = 6

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(c)          Zn(OH)2(s)  Zn2+(aq) + 2 OHS(aq)              Ksp = 4.1 x 10S17
Zn2+(aq) + 4 NH3(aq)  Zn(NH3)42+(aq)        Kf = 7.8 x 108
dissolution rxn Zn(OH)2(s) + 4 NH3(aq)  Zn(NH3)42+ + 2 OHS(aq)
K = (Ksp)(Kf) = (4.1 x 10S17)(7.8 x 108) = 3.2 x 10S8

16.105 (a)            Zn(OH)2(s)  Zn2+(aq) + 2 OHS(aq)             Ksp = 4.1 x 10S17
Zn2+(aq) + 4 OHS(aq)  Zn(OH)42S(aq)          Kf = 3 x 1015
dissolution rxn Zn(OH)2(s) + 2 OHS(aq)  Zn(OH)42S(aq)
K = (Ksp)(Kf) = (4.1 x 10S17)(3 x 1015) = 0.1
(b)           Cu(OH)2(s)  Cu2+(aq) + 2 OHS(aq)             Ksp = 1.6 x 10S19
Cu2+(aq) + 4 NH3(aq)  Cu(NH3)42+(aq)       Kf = 5.6 x 1011
dissolution rxn Cu(OH)2(s) + 4 NH3(aq)  Cu(NH3)42+ + 2 OHS(aq)
K = (Ksp)(Kf) = (1.6 x 10S19)(5.6 x 1011) = 9.0 x 10S8
(c)           AgBr(s)  Ag+(aq) + BrS(aq)                   Ksp = 5.4 x 10S13
Ag+(aq) + 2 NH3(aq)  Ag(NH3)2+(aq)           Kf = 1.7 x 107
dissolution rxn AgBr(s) + 2 NH3(aq)  Ag(NH3)2+(aq) + BrS(aq)
K = (Ksp)(Kf) = (5.4 x 10S13)(1.7 x 107) = 9.2 x 10S6

16.106 (a)                     AgI(s)  Ag+(aq) + IS(aq)
equil (M)               x        x
+ S             S17
Ksp = [Ag ][I ] = 8.5 x 10 = (x)(x)
molar solubility = x =   8.5 x 10_17 = 9.2 x 10S9 M
(b)                   AgI(s) + 2 CNS(aq)  Ag(CN)2S(aq)    + IS(aq)
initial (M)                     0.10             0    0
change (M)                      S2x             +x   +x
equil (M)                  0.10 S 2x             x    x
S17             20
K = (Ksp)(Kf) = (8.5 x 10 )(3.0 x 10 ) = 2.6 x 104
_
4   [Ag(CN ) 2 ][I _ ]          x
2
K = 2.6 x 10 =                      =
[CN _ ]2         (0.10 _ 2 x ) 2
Take the square root of both sides and solve for x.
molar solubility = x = 0.050 M

16.107                          Cr(OH)3(s) + OHS(aq)  Cr(OH)4S(aq)
initial (M)                      0.50      0
change (M)                       Sx       +x
equil (M)                     0.50 S x     x
K = (Ksp)(Kf) = (6.7 x 10S31)(8 x 1029) = 0.54
_
[Cr(OH ) 4 ]       x
K = 0.54 =         _
=
[OH ]       0.50 _ x
0.27 S 0.54x = x

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0.27 = 1.54x
0.27
molar solubility = x =        = 0.2 M
1.54

Precipitation; Qualitative Analysis

16.108 For BaSO4, Ksp = 1.1 x 10S10
Total volume = 300 mL + 100 mL = 400 mL
(4.0 x 10 _ 3 M)(100 mL)
[Ba2+] =                             = 1.0 x 10S3 M
(400 mL)
(6.0 x 10 _ 4 M)(300 mL)
[SO42S] =                             = 4.5 x 10S4 M
(400 mL)
IP = [Ba ]t[SO4 ]t = (1.0 x 10S3)(4.5 x 10S4) = 4.5 x 10S7
2+      2S

IP > Ksp; BaSO4(s) will precipitate.

16.109 On mixing equal volumes of two solutions, the concentrations of both solutions are cut in
half.
For PbCl2, Ksp = 1.2 x 10S5 = [Pb2+][ClS]2
IP = (0.0050)(0.0050)2 = 1.2 x 10S7
IP < Ksp; no precipitate will form.
K sp       1.2 x 10 _ 5
[ClS] =           =                = 0.049 M
[Pb 2+]     5.0 x 10 _ 3
A [ClS] just greater than 0.049 M will result in precipitation.

16.110 BaSO4, Ksp = 1.1 x 10S10; Fe(OH)3, Ksp = 2.6 x 10S39
Total volume = 80 mL + 20 mL = 100 mL
(1.0 x 10 _ 5 M)(80 mL)
[Ba2+] =                            = 8.0 x 10S6 M
(100 mL)
S
[OH ] = 2[Ba ] = 2(8.0 x 10S6) = 1.6 x 10S5 M
2+

2(1.0 x 10 _ 5 M)(20 mL)
[Fe3+] =                             = 4.0 x 10S6 M
(100 mL)
3(1.0 x 10 _ 5 M)(20 mL)
[SO42S] =                              = 6.0 x 10S6 M
(100 mL)
For BaSO4, IP = [Ba2+]t[SO42S]t = (8.0 x 10S6)(6.0 x 10S6) = 4.8 x 10S11
IP < Ksp; BaSO4 will not precipitate.
For Fe(OH)3, IP = [Fe3+]t[OHS]t3 = (4.0 x 10S6)(1.6 x 10S5)3 = 1.6 x 10S20
IP > Ksp; Fe(OH)3(s) will precipitate.

(2.0 x 10 _ 3 M)(0.10 mL)
16.111 (a)     [CO32S] =                              = 8.0 x 10S7 M
(250 mL)
S9      2+
Ksp = 5.0 x 10 = [Ca ][CO32S]
IP = [Ca2+][CO32S] = (8.0 x 10S4)(8.0 x 10S7) = 6.4 x 10S10

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IP < Ksp; no precipitate will form.
(b) Na2CO3, 106 amu; 10 mg = 0.010 g
             1 mol 
 0.010 g x 106 g 
[CO32S] =                        = 3.8 x 10S4 M
0.250 L
IP = [Ca2+][CO32S] = (8.0 x 10S4)(3.8 x 10S4) = 3.0 x 10S7
IP > Ksp; CaCO3(s) will precipitate.

16.112 pH = 10.80; [H3O+] = 10SpH = 10S10.80 = 1.6 x 10S11 M
_14
K w = 1.0 x 10
[OHS] =                         = 6.2 x 10S4 M
[H 3O +] 1.6 x 10 _11
For Mg(OH)2, Ksp = 5.6 x 10S12
IP = [Mg2+]t[OHS]t2 = (2.5 x 10S4)(6.2 x 10S4)2 = 9.6 x 10S11
IP > Ksp; Mg(OH)2(s) will precipitate

16.113 Mg(OH)2, Ksp = 5.6 x 10S12; Al(OH)3, Ksp = 1.9 x 10S33
pH = 8; [H3O+] = 10SpH = 10S8 = 1 x 10S8 M
_14
K w = 1.0 x 10
[OHS] =                          = 1 x 10S6 M
[ H 3O + ] 1 x 10 _ 8
For Mg(OH)2, IP = [Mg2+][OHS]2 = (0.01)(1 x 10S6)2 = 1 x 10S14
IP < Ksp; no Mg(OH)2 will precipitate.
For Al(OH)3, IP = [Al3+][OHS]3 = (0.01)(1 x 10S6)3 = 1 x 10S20
IP > Ksp; Al(OH)3 will precipitate.

[ M 2+ ][ H 2 S]
16.114 Kspa =            + 2
; FeS, Kspa = 6 x 102; SnS, Kspa = 1 x 10S5
[ H 3O ]
2+
Fe and Sn2+ can be separated by bubbling H2S through an acidic solution containing the
two cations because their Kspa values are so different.
(0.01)(0.10)
For FeS and SnS, Qc =                     = 1.1 x 10S2
(0.3) 2
For FeS, Qc < Kspa, and no FeS will precipitate.
For SnS, Qc > Kspa, and SnS will precipitate.

[Co 2+][ H 2 S]
16.115 CoS, Kspa =                     =3
[ H 3O + ]2
(i) In 0.5 M HCl, [H3O+] = 0.5 M
[Co 2+]t[H 2 S ]t (0.10)(0.10)
Qc =                   =              = 0.04; Qc < Kspa; CoS will not precipitate
[H3O+]2 t            (0.5) 2

(ii) pH = 8;   [H3O+] = 10SpH = 10S8 = 1 x 10S8 M

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[Co 2+]t[H 2 S ]t (0.10)(0.10)
Qc =            + 2
=         _8 2
= 1 x 1014; Qc > Kspa; CoS(s) will precipitate
[H 3O ]t         (1 x 10 )

16.116 (a) add ClS to precipitate AgCl
(b) add CO32S to precipitate CaCO3
(c) add H2S to precipitate MnS
(d) add NH3 and NH4Cl to precipitate Cr(OH)3
(Need buffer to control [OHS]; excess OHS produces the soluble Cr(OH)4S.)

16.117 (a) add ClS to precipitate Hg2Cl2
(b) add (NH4)2HPO4 to precipitate MgNH4PO4
(c) add HCl and H2S to precipitate HgS
(d) add ClS to precipitate PbCl2

General Problems

16.118 Prepare aqueous solutions of the three salts. Add a solution of (NH4)2HPO4. If a white
precipitate forms, the solution contains Mg2+. Perform flame test on the other two
solutions. A yellow flame test indicates Na+. A violet flame test indicates K+.

16.119 (a), solution contains HCN and CNS
(c), solution can contain HCN and CNS
(e), solution can contain HCN and CNS

16.120 (a), solution contains H2CO3 and HCO3S
(b), solution contains HCO3S and CO32S
(d), solution contains HCO3S and CO32S

16.121

(a) The pH for the weak acid is higher.
(b) Initially, the pH rises more quickly for the weak acid, but then the curve becomes
more level in the region halfway to the equivalence point.
(c) The pH is higher at the equivalence point for the weak acid.

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(d) Both curves are identical beyond the equivalence point because the pH is determined
by the excess [OHS].
(e) If the acid concentrations are the same, the volume of base needed to reach the
equilavence point is the same.

16.122 (a)

(b) mol NaOH required =
 0.010 mol HA                  1 mol NaOH 
                 (0.0500 L)                   = 0.000 50 mol
        L                      1 mol HA 
    1L     
vol NaOH required = (0.000 50 mol)                  = 0.050 L = 50 mL
 0.010 mol 
(c) A basic salt is present at the equivalence point; pH > 7.00
(d) Halfway to the equivalence point, the pH = pKa = 4.00

16.123 (a)    AgBr(s)  Ag+(aq) + BrS(aq)
(i) HBr is a source of BrS (reaction product). The solubility of AgBr is decreased.
(ii) unaffected
(iii) AgNO3 is a source of Ag+ (reaction product). The solubility of AgBr is decreased.
(iv) NH3 forms a complex with Ag+, removing it from solution. The solubility of AgBr is
increased.
(b)    BaCO3(s)  Ba2+(aq) + CO32S(aq)
(i) HNO3 reacts with CO32S, removing it from the solution. The solubility of BaCO3 is
increased.
(ii) Ba(NO3)2 is a source of Ba2+ (reaction product). The solubility of BaCO3 is
decreased.
(iii) Na2CO3 is a source of CO32S (reaction product). The solubility of BaCO3 is
decreased.
(iv) CH3CO2H reacts with CO32S, removing it from the solution. The solubility of BaCO3
is increased.

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_14
Kw            1.0 x 10
16.124 For NH4+, Ka =                      =               = 5.6 x 10S10
Kb  for NH 3 1.8 x 10 _ 5
pKa = Slog Ka = Slog(5.6 x 10S10) = 9.25
[ NH 3]                              [ NH 3]
pH = pKa + log        +
;       9.40 = 9.25 + log
[ NH 4 ]                             [ NH +]
4

[ NH 3]                                [ NH 3]
log         = 9.40 S 9.25 = 0.15;                  = 100.15 = 1.41
[ NH +]
4                                 [ NH +]
4

mol NH 3
Because the volume is the same for both NH3 and NH4+,                     = 1.41.
mol NH +
4
mol NH3 = (0.20 mol/L)(0.250 L) = 0.050 mol NH3
mol NH 3 0.050
mol NH4+ =                  =         = 0.035 mol NH4+
1.41          1.41
 1L 
vol NH4+ = (0.035 mol)                  = 0.012 L = 12 mL
 3.0 mol 
12 mL of 3.0 M NH4Cl must be added to 250 mL of 0.20 M NH3 to obtain a buffer
solution having a pH = 9.40.

16.125 H2PO4S(aq) + H2O(l)  H3O+(aq) + HPO42S(aq)
(a) Na2HPO4, source of HPO42S, equilibrium shifts left, pH increases.
(b) Addition of the strong acid, HBr, decreases the pH.
(c) Addition of the strong base, KOH, increases the pH.
(d) There is no change in the pH with the addition of the neutral salt KI.
(e) H3PO4, source of H2PO4S, equilibrium shifts right, pH decreases.
(f) Na3PO4, source of PO43S, decreases [H3O+] by forming HPO42S, pH increases.

16.126 pH = 10.35; [H3O+] = 10SpH = 10S10.35 = 4.5 x 10S11 M
_14
K w = 1.0 x 10
[OHS] =                         = 2.2 x 10S4 M
[H 3O +] 4.5 x 10 _11
[OH _ ] 2.2 x 10 _ 4
2+
[Mg ] =         =              = 1.1 x 10S4 M
2         2
Ksp = [Mg2+][OHS]2 = (1.1 x 10S4)(2.2 x 10S4)2 = 5.3 x 10S12

16.127 mmol Hg22+ = (0.010 mmol/mL)(1.0 mL) = 0.010 mmol
mmol ClS = (6 mmol/mL)( 0.05 mL) = 0.3 mmol
Assume complete reaction.
Hg22+(aq) + 2 ClS(aq)  Hg2Cl2(s)
before reaction (mmol)       0.010         0.3
change (mmol)               S0.010      S2(0.010)
after reaction (mmol)          0          0.28
0.28 mmol
[ClS] =             = 0.27 M
1.05 mL
Allow Hg2Cl2 to establish a new equilibrium.

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Hg2Cl2(s)  Hg22+(aq) + 2 ClS(aq)
initial (M)                     0             0.27
equil (M)                      x           0.27 + 2x
Ksp = [Hg22+][ClS]2 = 1.4 x 10S18 = x(0.27 + 2x)2  x(0.27)2
1.4 x 10 _18
x = [Hg22+] =                = 2 x 10S17 mol/L
(0.27) 2
Hg22+, 401.18 amu
Hg22+ concentration = (2 x 10S17 mol/L)(401.18 g/mol) = 8 x 10S15 g/L

1 mol
16.128 NaOH, 40.0 amu; 20 g x                   = 0.50 mol NaOH
40.0 g
(0.500 L)(1.5 mol/L) = 0.75 mol NH4Cl
NH4+(aq) + OHS(aq)  NH3(aq) + H2O(l)
before reaction (mol)            0.75             0.50             0
change (mol)             S0.50            S0.50          +0.50
after reaction (mol)             0.25               0             0.50
This reaction produces a buffer solution.
[NH4+] = 0.25 mol/0.500 L = 0.50 M;             [NH3] = 0.50 mol/0.500 L = 1.0 M
[base]                [ NH 3]
pH = pKa + log            = pK a + log
[acid]                [ NH +]
4

Kw          1.0 x 10 _14
For NH4+, Ka =                    =            _5
= 5.6 x 10S10; pKa = Slog Ka = 9.25
K b for NH 3 1.8 x 10
 1.0 
pH = 9.25 + log         = 9.55
 0.5 

16.129 (a) AgCl, Ksp = [Ag+][ClS] = 1.8 x 10S10
K sp     1.8 x 10 _10
[ClS] =          =              = 6.0 x 10S9 M
[ Ag + ]      0.030
(b) Hg2Cl2, Ksp = [Hg22+][ClS]2 = 1.4 x 10S18
K sp     1.4 x 10 _18
[ClS] =           =              = 6.8 x 10S9 M
[Hg 2+]
2
0.030
(c) PbCl2, Ksp = [Pb2+][ClS]2 = 1.2 x 10S5
S      K sp      1.2 x 10 _ 5
[Cl ] =           =                = 0.020 M
[Pb 2+]         0.030
AgCl(s) will begin to precipitate when the [ClS] just exceeds 6.0 x 10S9 M. At this ClS
concentration, IP < Ksp for PbCl2 so all of the Pb2+ will remain in solution.
Kw          1.0 x 10 _14
16.130 For NH4+, Ka =                  =           _5
= 5.6 x 10S10; pKa = Slog Ka = 9.25
K b for NH 3 1.8 x 10
[ NH 3]               (0.50)
pH = pKa + log       +
= 9.25 + log         = 9.47
[ NH 4 ]              (0.30)

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[H3O+] = 10SpH = 10S9.47 = 3.4 x 10S10 M
[ Mn 2+ ][ H 2 S]
For MnS, Kspa =            + 2
= 3 x 1010
[ H 3O ]
+ 2                      2
K spa[H3O ] (3 x 1010)(3.4 x 10 _10)
2+
molar solubility = [Mn ] =                =                        = 3.5 x 10S8 M
[H 2 S]           (0.10)
MnS, 87.00 amu;                             S8
solubility = (3.5 x 10 mol/L)(87.00 g/mol) = 3 x 10S6 g/L

16.131 pH = 9.00;  [H3O+] = 10SpH = 10S9.00 = 1.0 x 10S9 M
1.0 x 10 _14
[OHS] = K w + =              = 1.0 x 10S5 M
[H 3O ] 1.0 x 10 _ 9
Mg(OH)2(s)  Mg2+(aq) + 2 OHS(aq)
equil (M)                           x      1.0 x 10S5 (fixed by buffer)
Ksp = [Mg2+][OHS]2 = 5.6 x 10S12 = x(1.0 x 10S5)2
5.6 x 10 _12
molar solubility = x =                  = 0.056 M
(1.0 x 10 _ 5) 2

16.132 60.0 mL = 0.0600 L
1.00 mol H3PO4
mol H3PO4 = 0.0600 L x                = 0.0600 mol H3PO4
1.00 L
0.100 mol LiOH
mol LiOH = 1.00 L x                = 0.100 mol LiOH
1.00 L

H3PO4(aq) + OHS(aq)  H2PO4S(aq) + H2O(l)
before reaction (mol)         0.0600 0.100     0
change (mol)                 S0.0600       S0.0600   +0.0600
after reaction (mol)             0          0.040     0.0600

H2PO4S(aq) + OHS(aq)  HPO42S(aq) + H2O(l)
before reaction (mol)           0.0600 0.040      0
change (mol)                   S0.040          S0.040 +0.040
after reaction (mol)             0.020            0           0.040
The resulting solution is a buffer because it contains the conjugate acid-base pair,
H2PO4S and HPO42S, at acceptable buffer concentrations.
For H2PO4S, Ka2 = 6.2 x 10S8 and pKa2 = S log Ka2 = S log (6.2 x 10S8) = 7.21
[HPO 2 _ ]
4                 (0.040 mol / 1.06 L)
pH = pKa2 + log            _
= 7.21 + log
[H 2 PO 4 ]              (0.020 mol / 1.06 L)
(0.040)
pH = 7.21 + log           = 7.21 + 0.30 = 7.51
(0.020)

16.133 (a) The mixture of 0.100 mol H3PO4 and 0.150 mol NaOH is a buffer and contains
mainly H2PO4S and HPO42S from the reactions:

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H3PO4(aq) + OHS(aq)  H2PO4S(aq) + H2O(l)
before (mol) 0.100       0.150         0
change (mol) S0.100      S0.100    +0.100
after (mol)    0         0.050      0.100

H2PO4S(aq) + OHS(aq)  HPO42S(aq) + H2O(l)
before (mol) 0.100              0.050    0
change (mol) S0.050 S0.050            +0.050
after (mol) 0.050                 0            0.050
If water were used to dilute the solution instead of HCl, the pH would be equal to pKa2
because [H2PO4S] = [HPO42S] = 0.050 mol/1.00 L = 0.050 M
H2PO4S(aq) + H2O(l)  H3O+(aq) + HPO42S(aq)                        Ka2 = 6.2 x 10S8
S8
pKa2 = Slog K2a = Slog(6.2 x 10 ) = 7.21
[HPO 2 _ ]
4
pH = pKa2 + log            _
= pKa2 + log(1) = pKa2 = 7.21
[H 2 PO 4 ]

The pH is lower (6.73) because the added HCl converts some HPO42S to H2PO4S.
HPO42S(aq) + H3O+(aq)  H2PO4S(aq) + H2O(l)
before (M) 0.050                      x              0.050
change (M)       Sx                 Sx                +x
after (M)     0.050 S x              0            0.050 + x
2S             S
[HPO4 ] + [H2PO4 ] = (0.050 S x) + (0.050 + x) = 0.100 M
[HPO 2 _ ]
4
pH = pKa2 + log             _
[H 2 PO 4 ]
[HPO42S] = 0.100 S [H2PO4S]
_
(0.100 _ [H 2 PO 4 ])
6.73 = 7.21 + log                   _
[H 2 PO 4 ]
_
(0.100 _ [H 2 PO 4 ])
6.73 S 7.21 = S0.48 = log                   _
[H 2 PO 4 ]
_
(0.100 _ [H 2 PO 4 ])
10S0.48 = 0.331 =                 _
[H 2 PO 4 ]
(0.331)[H2PO4S] = 0.100 S [H2PO4S]
(1.331)[H2PO4S] = 0.100
[H2PO4S] = 0.100/1.331 = 0.075 M
[HPO42S] = 0.100 S [H2PO4S] = 0.100 S 0.075 = 0.025 M

H3PO4(aq) + H2O(l)  H3O+(aq) + H2PO4S(aq)                 Ka1 = 7.5 x 10S3
_
[H 3O + ][H 2 PO 4 ]
Ka1 =
[ H 3PO 4]
_
[H 3O + ][ H 2 PO 4 ]
[H3PO4] =
K a1
SpH
[H3O ] = 10 = 10S6.73 = 1.86 x 10S7 M
+

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(1.86 x 10 _ 7 )(0.075)
[H3PO4] =                           = 1.9 x 10S6 M
7.5 x 10 _ 3

(b)    If distilled water were used and not HCl, the mole amounts of both H2PO4S and HPO42S
would be 0.050 mol. The HCl converted some HPO42S to H2PO4S.
HPO42S(aq) + H3O+(aq)  H2PO4S(aq) + H2O(l)
before (mol) 0.050               x             0.050
change (mol) Sx                  Sx              +x
after (mol) 0.050 S x            0            0.050 + x
2S
From part (a), [HPO4 ] = 0.025 M
mol HPO42S = (0.025 mol/L)(1.00 L) = 0.025 mol = 0.050 S x
x = mol H3O+ = mol HCl inadvertently added = 0.050 S 0.025 = 0.025 mol HCl

16.134 For CH3CO2H, Ka = 1.8 x 10S5 and pKa = Slog Ka = Slog(1.8 x 10S5) = 4.74
The mixture will be a buffer solution containing the conjugate acid-base pair, CH3CO2H
and CH3CO2S, having a pH near the pKa of CH3CO2H.
_
[CH 3CO 2 ]
pH = pKa + log
[CH 3CO 2 H]
_                                   _
[CH 3CO 2 ]                          [CH 3CO 2 ]
4.85 = 4.74 + log                ; 4.85 S 4.74 = log
[CH 3CO 2 H]                         [CH 3CO 2 H]
_                  _
[CH 3CO 2 ]         [CH 3CO 2 ]
0.11 = log              ;                   = 100.11 = 1.3
[CH 3CO 2 H]        [CH 3CO 2 H]
In the Henderson-Hasselbalch equation, moles can be used in place of concentrations
because both components are in the same volume so the volume terms cancel.
20.0 mL = 0.0200 L

Let X equal the volume of 0.10 M CH3CO2H and Y equal the volume of 0.15 M
CH3CO2S. Therefore, X + Y = 0.0200 L and
_
Y x [CH 3CO 2 ]     Y (0.15 mol/L)
=                = 1.3
X x [CH 3CO 2 H] X (0.10 mol/L)
X = 0.0200 S Y
Y (0.15 mol/L)
= 1.3
(0.020 _ Y)(0.10 mol/L)
0.15 Y
= 1.3
0.0020 _ 0.10 Y
0.15Y = 1.3(0.0020 S 0.10Y)
0.15Y = 0.0026 S 0.13Y
0.15Y + 0.13Y = 0.0026
0.28Y = 0.0026
Y = 0.0026/0.28 = 0.0093 L
X = 0.0200 S Y = 0.0200 S 0.0093 = 0.0107 L
X = 0.0107 L = 10.7 mL and Y = 0.0093 L = 9.3 mL
You need to mix together 10.7 mL of 0.10 M CH3CO2H and 9.3 mL of 0.15 M

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Chapter 16 S Applications of Aqueous Equilibria
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NaCH3CO2 to prepare 20.0 mL of a solution with a pH of 4.85.

16.135 [H3O+] = 10SpH = 10S2.37 = 0.004 27 M
H3Cit(aq) + H2O(l)  H3O+(aq) + H2CitS(aq)
[ O + ][ H 2Cit _ ]
Ka1 = 7.1 x 10S4 = H 3
[H 3 Cit]
S4
(7.1 x 10 )[H3Cit] = (0.004 27)[H2CitS]
[H3Cit] = (0.004 27)[H2CitS]/(7.1 x 10S4) = (6.01)[H2CitS]

H2CitS(aq) + H2O(l)  H3O+(aq) + HCit2S(aq)
[H 3O +][HCit 2 _ ]
Ka2 = 1.7 x 10S5 =
[H 2Cit _ ]
(1.7 x 10S5)[H2CitS] = (0.004 27)[HCit2S]
[HCit2S] = (1.7 x 10S5)[H2CitS]/(0.004 27) = (0.003 98)[H2CitS]
[H3Cit] + [H2CitS] + [HCit2S] + [Cit3S] = 0.350 M
Now assume [Cit3S]  0, so [H3Cit] + [H2CitS] + [HCit2S] = 0.350 M and then by
substitution:
(6.01)[H2CitS] + [H2CitS] + (0.003 98)[H2CitS] = 0.350 M
(7.01)[H2CitS] = 0.350 M
[H2CitS] = 0.350 M/7.01 = 0.050 M
[H3Cit] = (6.01)[H2CitS] = (6.01)(0.050 M) = 0.30 M
[HCit2S] = (0.003 98)[H2CitS] = (0.003 98)(0.050 M) = 2.0 x 10S4 M

HCit2S(aq) + H2O(l)  H3O+(aq) + Cit3S(aq)
[H 3O +][Cit 3_ ]
Ka3 = 4.1 x 10S7 =
[HCit 2 _ ]
(K a 3)[HCit 2 _ ]   (4.1 x 10 _ 7)(2.0 x 10 _ 4)
[Cit3S] =           +
=                              = 1.9 x 10S8 M
[H 3O ]                 (0.004 27)

16.136 (a) HCl is a strong acid. HCN is a weak acid with Ka = 4.9 x 10S10. Before the titration,
the [H3O+] = 0.100 M. The HCN contributes an insignificant amount of additional
H3O+, so the pH = Slog[H3O+] = Slog(0.100) = 1.00
(b) 100.0 mL = 0.1000 L
0.100 mol HCl
mol H3O+ = 0.1000 L x                  = 0.0100 mol H3O+
1.00 L
add 75.0 mL of 0.100 M NaOH; 75.0 mL = 0.0750 L
0.100 mol NaOH
mol OHS = 0.0750 L x                     = 0.00750 mol OHS
1.00 L

H3O+(aq) + OHS(aq)  2 H2O(l)
before reaction (mol)          0.0100 0.0075

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change (mol)                 S0.0075         S0.0075
after reaction (mol)         0.0025      0
0.0025 mol H 3O +
[H3O+] =                         = 0.0143 M
0.1000 L + 0.0750 L
pH = Slog[H3O+] = Slog(0.0143) = 1.84
(c) 100.0 mL of 0.100 M NaOH will completely neutralize all of the H3O+ from 100.0
mL of 0.100 M HCl. Only NaCl and HCN remain in the solution. NaCl is a neutral salt
and does not affect the pH of the solution. [HCN] changes because of dilution. Because
the solution volume is doubled, [HCN] is cut in half.
[HCN] = 0.100 M/2 = 0.0500 M

HCN(aq) + H2O(l)  H3O+(aq) + CNS(aq)
initial (M)   0.0500               ~0         0
change (M)       Sx                      +x       +x
equil (M)     0.0500 S x                    x       x
+    _                         2       2
[ O ][CN ]                        x
Ka = H 3            = 4.9 x 10S10 =             x
HCN                       0.0500 _ x 0.0500
[H3O+] = x = (0.0500)(4.9 x 10_10) = 4.95 x 10S6 M
pH = Slog[H3O+] = Slog(4.95 x 10S6) = 5.31
(d) Add an additional 25.0 mL of 0.100 M NaOH.
25.0 mL = 0.0250 L
0.100 mol NaOH
additional mol OHS = 0.0250 L x                         = 0.00250 mol OHS
1.00 L
0.0500 mol HCN
mol HCN = 0.200 L x                          = 0.0100 mol HCN
1.00 L
HCN(aq) + OHS(aq)  CNS(aq) + H2O(l)
before reaction (mol)          0.0100        0.00250          0
change (mol)                  S0.00250       S0.00250     +0.00250
after reaction (mol)            0.0075          0           0.00250
The resulting solution is a buffer because it contains the conjugate acid-base pair, HCN
and CNS, at acceptable buffer concentrations.
For HCN, Ka = 4.9 x 10S10 and pKa = S log Ka = S log (4.9 x 10S10) = 9.31
[CN _ ]              (0.00250 mol / 0.2250 L)
pH = pKa + log            = 9.31 + log
[HCN]                 (0.0075 mol / 0.2250 L)
(0.00250)
pH = 9.31 + log             = 9.31 _ 0.48 = 8.83
(0.0075)

16.137 (a)          Cd(OH)2(s)  Cd2+(aq) + 2 OHS(aq)
initial (M)                 0          ~0
equil (M)                   x          2x
S 2         S15
2+                         2
Ksp = [Cd ][OH ] = 5.3 x 10 = (x)(2x) = 4x3

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Chapter 16 S Applications of Aqueous Equilibria
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5.3 x 10 _15
molar solubility = x =   3            = 1.1 x 10S5 M
4
S                   S5
[OH ] = 2x = 2(1.1 x 10 M) = 2.2 x 10S5 M
1.0 x 10 _14
[H3O+] =            _5
= 4.5 x 10S10 M
2.2 x 10
pH = Slog[H3O+] = Slog(4.5 x 10S10) = 9.35
(b) 90.0 mL = 0.0900 L
mol HNO3 = (0.100 mol/L)(0.0900 L) = 0.009 00 mol HNO3
The addition of HNO3 dissolves some Cd(OH)2(s).
Cd(OH)2(s) + 2 HNO3(aq)  Cd2+(aq) + 2 H2O(l)
before (mol) 0.100 0.009 00              1.1 x 10S5
change (mol) S0.0045          S2(0.0045) 1.1 x 10S5 + 0.0045
after (mol)     0.0955            0             ~0.0045
total volume = 100.0 mL + 90.0 mL = 190.0 mL = 0.1900 L
[Cd2+] = 0.0045 mol/0.1900 L = 0.024 M
Ksp = 5.3 x 10S15 = [Cd2+][OHS]2 = (0.024)[OHS]2
5.3 x 10 _15
[OHS] =                = 4.7 x 10S7 M
0.024
1.0 x 10 _14
[H3O+] =           _7
= 2.1 x 10S8 M
4.7 x 10
pH = Slog[H3O+] = Slog(2.1 x 10S8) = 7.68
2 mol HNO 3        1.00 L    1000 mL
(c) volume HNO3 = 0.0100 mol Cd(OH)2 x                      x           x         = 200
1 mol Cd(OH ) 2   0.100 mol    1.00 L
mL
16.138 (a)          Zn(OH)2(s)  Zn2+(aq) + 2 OHS(aq)
initial (M)                 0          ~0
equil (M)                   x          2x
S 2         S17
2+                         2
Ksp = [Zn ][OH ] = 4.1 x 10 = (x)(2x) = 4x3
4.1 x 10 _17
molar solubility = x =   3              = 2.2 x 10S6 M
4
(b) [OHS] = 2x = 2(2.2 x 10S6 M) = 4.4 x 10S6 M
1.0 x 10 _14
[H3O+] =                 = 2.3 x 10S9 M
4.4 x 10 _ 6
pH = Slog[H3O+] = Slog(2.3 x 10S9) = 8.64
(c)             Zn(OH)2(s)  Zn2+(aq) + 2 OHS(aq)      Ksp = 4.1 x 10S17
Zn2+(aq) + 4 OHS(aq)  Zn(OH)42S(aq)   Kf = 3 x 1015
Zn(OH)2(s) + 2 OHS(aq)  Zn(OH)42S(aq) K = KspKf = 0.123
initial (M)                     0.10             0
change (M)                      S2x              +x
equil (M)                     0.10 S 2x          x

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Chapter 16 S Applications of Aqueous Equilibria
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[Zn(OH ) 2 _ ]
4                    x
K=                  = 0.123 =
[OH _ ]2
(0.10 _ 2 x ) 2
0.492x2 S 1.0492x + 0.00123 = 0
Use the quadratic formula to solve for x.
_ (_ 1.0492)    (_ 1.0492) 2 _ (4)(0.492)(0.00123) 1.0492  1.0480
x=                                                      =
2(0.492)                           0.984
x = 2.1 and 1.2 x 10S3
Of the two solutions for x, only 1.2 x 10S3 has physical meaning because the other
solution leads to a negative [OHS].
molar solubility of Zn(OH)42S in 0.10 M NaOH = x = 1.2 x 10S3 M

16.139 (a) Fe(OH)3(s)  Fe3+(aq) + 3 OHS(aq)                               Ksp = 2.6 x 10S39
H3Cit(aq) + H2O(l)  H3O+(aq) + H2CitS(aq)                          Ka1 = 7.1 x 10S4
H2CitS(aq) + H2O(l)  H3O+(aq) + HCit2S(aq)                         Ka2 = 1.7 x 10S5
HCit2S(aq) + H2O(l)  H3O+(aq) + Cit3S(aq)                          Ka3 = 4.1 x 10S7
Fe3+(aq) + Cit3S(aq)  Fe(Cit)(aq)                                  Kf = 6.3 x 1011
3 [H3O+(aq) + OHS(aq)  2 H2O(l)]                                  (1/Kw)3 = 1.0 x 1042
Fe(OH)3(s) + H3Cit(aq)  Fe(Cit)(aq) + 3 H2O(l)

K = Ksp Ka1 Ka2 Ka3Kf(1/Kw)3 = 8.1
(b)           Fe(OH)3(s) + H3Cit(aq)  Fe(Cit)(aq) + 3 H2O(l)
initial (M)                   0.500     0
change (M)                      Sx      +x
equil (M)                   0.500 S x    x

[Fe(Cit)]              x
K=              = 8.1 =
[H 3 Cit]         0.500 _ x
8.1(0.500 S x) = x
4.05 S 8.1x = x
4.05 = 9.1x
x = molar solubility = 4.05/9.1 = 0.45 M

Multi-Concept Problems

16.140 (a) HAS(aq) + H2O(l)  H3O+(aq) + A2S(aq)                 Ka2 = 10S10
K
HAS(aq) + H2O(l)  H2A(aq) + OHS(aq)                       Kb = w = 10S10
K a1
Ka2
2 HAS(aq)  H2A(aq) + A2S(aq)                              K=         = 10S6
Ka 1
2 H2O(l)  H3O+(aq) + OHS(aq)                              Kw = 1.0 x 10S14

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Chapter 16 S Applications of Aqueous Equilibria
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The principal reaction of the four is the one with the largest K, and that is the third
reaction.
[H 3O +][ HA _ ]             [H 3O +][A 2 _ ]
(b) Ka1 =                      and Ka2 =
[H 2 A]                     [HA _ ]
_
1 [ A]                      K 2 [ HA ]
[H3O+] = K a H_2           and [H3O+] = a 2 _
[HA ]                          [A ]
_
K a1 [H 2 A] x K a 2 [ HA ] = [H O+]2;          K a1 K a 2 [H 2 A] = [H O+]2
_               2_          3                                   3
[HA ]             [A ]                                [A 2 _ ]
Because the principal reaction is 2 HAS(aq)  H2A(aq) + A2S(aq), [H2A] = [A2S].
Ka1 Ka2 = [H3O+]2
log Ka1 + log Ka2 = 2 log [H3O+]
log K a1 + log K a 2                   _ log K a1 + (_ log K a 2)
= log [H 3O +] ;                            = _ log [H 3O +]
2                                       2
p K a1 + p K a 2
= pH
2
(c)                     2 HAS(aq)  H2A(aq) + A2S(aq)
initial (M)       1.0             0           0
change (M)        S2x            +x           +x
equil (M)      1.0 S 2x x              x

[ H 2 A][ A 2 _ ]                   x
2
K=                     = 1 x 10S6 =
[ HA _ ]2                   (1.0 _ 2 x ) 2
Take the square root of both sides and solve for x.
x = [A2S] = 1 x 10S3 M
mol A2S = (1 x 10S3 mol/L)(0.0500 L) = 5 x 10S5 mol A2S
number of A2S ions = (5 x 10S5 mol A2S)(6.022 x 1023 ions/mol) = 3 x 1019 A2S ions

16.141 (a) (i)                en(aq) + H2O(l)  enH+(aq) + OHS(aq)
initial (M)     0.100                    0  ~0
change (M)      Sx                       +x +x
equil (M)       0.100 S x                 x  x
+      _
[enH ][OH ]                        (x)(x)
Kb =                = 5.2 x 10 _ 4 =
[en]                       0.100 _ x
2             S4               S5
x + (5.2 x 10 )x S (5.2 x 10 ) = 0
Use the quadratic formula to solve for x.
_ (5.2 x 10 _ 4)    (5.2 x 10 _ 4) 2 _ 4(1)(_ 5.2 x 10 _ 5) _ 5.2 x 10 _ 4  0.01443
x=                                                               =
2(1)                                          2
x = S0.0075 and 0.0070
Of the two solutions for x, only the positive value of x has physical meaning because x is
the [OHS].
[OHS] = x = 0.0070 M

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Chapter 16 S Applications of Aqueous Equilibria
______________________________________________________________________________
_14
K w = 1.0 x 10
[H3O+] =                     = 1.43 x 10S12 M
[OH _ ]   0.0070
pH = Slog[H3O+] = Slog(1.43 x 10S12) = 11.84

(ii) (30.0 mL)(0.100 mmol/mL) = 3.00 mmol en
(15.0 mL)(0.100 mmol/mL) = 1.50 mmol HCl
Halfway to the first equivalence point, [OHS] = Kb1
_14
K w = 1.0 x 10
[H3O+] =                         = 1.92 x 10S11 M
[OH _ ] 5.2 x 10 _ 4
pH = Slog[H3O+] = Slog(1.92 x 10S11) = 10.72

p K a1 + p K a 2
(iii) At the first equivalence point pH =                = 9.14
2
(iv) Halfway between the first and second equivalence points, [OHS] = Kb2 = 3.7 x 10S7
M
_14
K w = 1.0 x 10
[H3O+] =     _             _7
= 2.70 x 10S8 M
[OH ] 3.7 x 10
pH = Slog[H3O+] = Slog(2.70 x 10S8) = 7.57

(v) At the second equivalence point only the acidic enH2Cl2 is in solution.
Kw              1.0 x 10 _14
For enH22+, Ka =               = Kw =                = 2.70 x 10S8
Kb  for enH + K b 2 3.7 x 10 _ 7
3.00 mmol
[enH22+] =                          = 0.0333 M
(30.0 mL + 60.0 mL)
enH22+(aq) + H2O(l)  H3O+(aq) + enH+(aq)
initial (M)     0.0333                     ~0        0
change (M)         Sx                      +x       +x
equil (M)       0.0333 S x                  x        x
+      +                                2
[H 3O ][enH ]                     (x)(x)
Ka =            2+
= 2.70 x 10S8 =            x
[enH 2 ]                   0.0333 _ x 0.0333
Solve for x. x = [H3O+] = (2.70 x 10_8)(0.0333) = 3.00 x 10S5 M
pH = Slog[H3O+] = Slog(3.00 x 10S5) = 4.52

(vi) excess HCl
(75.0 mL S 60.0 mL)(0.100 mmol/mL) = 1.50 mmol HCl = 1.50 mmol H3O+
1.50 mmol
[H3O+] =                         = 0.0143 M
(30.0 mL + 75.0 mL)
pH = Slog[H3O+] = Slog(0.0143) = 1.84

481
Chapter 16 S Applications of Aqueous Equilibria
______________________________________________________________________________

(b)
Each of the two nitrogens in ethylenediamine can
accept a proton.

(c) Each nitrogen is sp3 hybridized.

16.142 (a) The first equivalence point is reached when all the H3O+ from the HCl and the H3O+
form the first ionization of H3PO4 is consumed.
pK a1 + pK a 2
At the first equivalence point pH =                    = 4.66
2
[H3O+] = 10SpH = 10(S4.66) = 2.2 x 10S5 M
(88.0 mL)(0.100 mmol/mL) = 8.80 mmol NaOH are used to get to the first equivalence
point
(b) mmol (HCl + H3PO4) = mmol NaOH = 8.8 mmol
mmol H3PO4 = (126.4 mL S 88.0 mL)(0.100 mmol/mL) = 3.84 mmol
mmol HCl = (8.8 S 3.84) = 4.96 mmol
4.96 mmol                                   3.84 mmol
[HCl] =                 = 0.124 M; [H3PO4] =                     = 0.0960 M
40.0 mL                                     40.0 mL
(c) 100% of the HCl is neutralized at the first equivalence point.
(d)                     H3PO4(aq) + H2O(l)  H3O+(aq) + H2PO4S(aq)
initial (M)      0.0960                          0.124         0
change (M)        Sx                              +x         +x
equil (M)        0.0960 S x                     0.124 + x      x
+        _
[H 3O ][H 2 PO 4 ]                 (0.124 + x)(x)
Ka1 =                     = 7.5 x 10S3 =
[H 3PO 4]                       0.0960 _ x
2                          S4
x + 0.132x S (7.2 x 10 ) = 0
Use the quadratic formula to solve for x.

482
Chapter 16 S Applications of Aqueous Equilibria
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_ (0.132)     (0.132) 2 _ 4(1)(_ 7.2 x 10 _ 4) _ 0.132  0.142
x=                                                  =
2(1)                                 2
x = S0.137 and 0.005
Of the two solutions for x, only the positive value of x has physical meaning because the
other solution would give a negative [H3O+].
[H3O+] = 0.124 + x = 0.124 + 0.005 = 0.129 M
pH = Slog[H3O+] = Slog(0.129) = 0.89
(e)

(f) Bromcresol green or methyl orange are suitable indicators for the first equivalence
point. Thymolphthalein is a suitable indicator for the second equivalence point.

16.143 (a) PV = nRT; 25oC = 298 K
                  1.00 atm 
 732 mm Hg x 760 mm Hg  (1.000 L)
PV                                   
n HCl =     =                                           = 0.0394 mol HCl
RT                     L  atm 
 0.082 06            (298 K)
           K  mol 
Na2CO3, 105.99 amu
1 mol Na 2CO 3
mol Na2CO3 = 6.954 g Na2CO3 x                          = 0.0656 mol Na2CO3
105.99 g Na 2CO 3
CO32S(aq) + H3O+(aq)  HCO3S(aq) + H2O(l)
before reaction (mol)           0.0656         0.0394          0
change (mol)          S0.0394        S0.0394       +0.0394
after reaction (mol)     0.0656 S 0.0394        0          0.0394

mol CO32S = 0.0656 S 0.0394 = 0.0262 mol and mol HCO3S = 0.0394 mol
Therefore, we have an HCO3S/CO32S buffer solution.
2
[CO 3 _ ]                            0.0262 mol/V
pH = pKa2 + log         _
= S log(5.6 x 10S11) + log
[ HCO 3 ]                             0.0394 mol/V
pH = 10.25 S 1.77 = 10.08

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Chapter 16 S Applications of Aqueous Equilibria
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(b) mol Na+ = 2(0.0656 mol) = 0.1312 mol
mol CO32S = 0.0262 mol
mol HCO3S = 0.0394 mol
mol ClS = 0.0394 mol
total ion moles = 0.2362 mol
      o
C  kg  0.2362 mol 
ΔTf = Kf  m,            ΔTf = 1.86                                 o
 0.2500 kg  = 1.76 C
        mol                
o
Solution freezing point = 0 C S ΔTf = S1.76oC
(c) H2O, 18.02 amu
1 mol H 2 O
mol H2O = 250.0 g x                      = 13.87 mol H2O
18.02 g H 2 O
mol H 2 O                   13.87 mol
Xsolv =                           =                           = 0.9833
mol H 2 O + mol ions       13.87 mol + 0.2362 mol
Psoln = Psolv  Xsolv = (23.76 mm Hg)(0.9833) = 23.36 mm Hg

16.144 25oC = 298 K
                    1.00 atm 
 74.4 mm Hg x 760 mm Hg 

Π = 2MRT; M =           =                                 = 0.00200 M
2 RT                 L  atm 
(2)  0.082 06           (298 K)
          K mol 
[M+] = [XS] = 0.00200 M
Ksp = [M+][XS] = (0.00200)2 = 4.00 x 10S6

16.145 (a) HCO3S(aq) + OHS(aq)  CO32S(aq) + H2O(l)
(b) mol HCO3S = (0.560 mol/L)(0.0500 L) = 0.0280 mol HCO3S
mol OHS = (0.400 mol/L)(0.0500 L) = 0.0200 mol OHS
HCO3S(aq) + OHS(aq)  CO32S(aq) + H2O(l)
before reaction (mol)     0.0280        0.0200          0
change (mol)     S0.0200        S0.0200      +0.0200
after reaction (mol) 0.0280 S 0.0200     0          0.0200

mol HCO3S = 0.0280 S 0.0200 = 0.0080 mol
0.0080 mol                                   0.0200 mol
[HCO3S] =                   = 0.080 M            [CO32S] =             = 0.200 M
0.1000 L                                     0.1000 L
HCO3S(aq) + H2O(l)  H3O+(aq) + CO32S(aq)
initial (M)      0.080                         ~0          0.200
change (M)        Sx                            +x          +x
equil (M)        0.080 S x                      x          0.200 + x
+      2_
[H 3O ][CO 3 ]                    x(0.200 + x) x(0.200)
Ka =              _
= 5.6 x 10 _11 =               
[HCO 3 ]                         0.080 _ x       0.080
+               S11
Solve for x. x = [H3O ] = 2.24 x 10 M

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Chapter 16 S Applications of Aqueous Equilibria
______________________________________________________________________________

pH = Slog[H3O+] = Slog(2.24 x 10S11) = 10.65
Because this solution contains both a weak acid (HCO3S) and its conjugate base, the
solution is a buffer.
(c) HCO3S(aq) + OHS(aq)  CO32S(aq) + H2O(l)
ΔHorxn = [ΔHof(CO32S) + ΔHof(H2O)] S [ΔHof(HCO3S) + ΔHof(OHS)]
ΔHorxn = [(1 mol)(S677.1 kJ/mol) + (1 mol)(S285.8 kJ/mol)]
S [(1 mol)(S692.0 kJ/mol) + (1 mol)(S230 kJ/mol)]
o
ΔH rxn = S 40.9 kJ
0.0200 moles each of HCO3S and OHS reacted.
heat produced = q = (0.0200 mol)(40.9 kJ/mol) = 0.818 kJ = 818 J
(d) q = m x specific heat x ΔT
q                      818 J
ΔT =                        =                            = 2.0oC
m x specific heat      (100.0 g)[4.18 J/(g o C)]
Final temperature = 25oC + 2.0oC = 27oC

16.146 (a) species present initially:
NH4+           CO32S            H2O
acid           base         acid or base
2H2O(l)  H3O+(aq) + OHS(aq)
NH4+(aq) + H2O(l)  NH3(aq) + H3O+(aq)
CO32S(aq) + H2O(l)  HCO3S(aq) + OHS(aq)

NH3, Kb = 1.8 x 10S5
NH4+, Ka = 5.6 x 10S10
CO32S, Kb = 1.8 x 10S4
HCO3S, Ka = 5.6 x 10S11

In the mixture, proton transfer takes place from the stronger acid to the stronger base, so
the principal reaction is NH4+(aq) + CO32S(aq)  HCO3S(aq) + NH3(aq)

(b)             NH4+(aq) + OHS(aq)  NH3(aq) + H2O(l)              K1 = 1/Kb(NH3)
CO3 (aq) + H2O(l)  HCO3 (aq) + OH (aq)
2S                            S            S
K2 = Kb(CO32S)
NH4+(aq) + CO32S(aq)  HCO3S(aq) + NH3(aq) K = K1K2
initial (M)     0.16           0.080             0            0.16
change (M)       Sx             Sx              +x             +x
equil (M)       0.16 S x     0.080 S x x             0.16 + x
_
[HCO 3 ][ NH 3] 1.8 x 10 _ 4                x(0.16 + x)
K=         +      2_
=          _5
= 10 =
[ NH 4 ][CO 3 ]    1.8 x 10           (0.16 _ x)(0.080 _ x)
2
9x S 2.56x + 0.128 = 0
Use the quadratic formula to solve for x.

485
Chapter 16 S Applications of Aqueous Equilibria
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_ (_ 2.56)         (_ 2.56) 2 _ (4)(9)(0.128) 2.56  1.395
x=                                                     =
2(9)                             18
x = 0.220 and 0.0647
Of the two solutions for x, only 0.00647 has physical meaning because 0.220 leads to
negative concentrations.
[NH4+] = 0.16 S x = 0.16 S 0.0647 = 0.0953 M = 0.095 M
[NH3] = 0.16 + x = 0.16 + 0.0647 = 0.225 M = 0.22 M
[CO32S] = 0.080 S x = 0.080 S 0.0647 = 0.0153 M = 0.015 M
[HCO3S] = x = 0.0647 M = 0.065 M
The solution is a buffer containing two different sets of conjugate acid-base pairs. Either
pair can be used to calculate the pH.
For NH4+, Ka = 5.6 x 10S10 and pKa = 9.25
[ NH 3]                    (0.225)
pH = pKa + log              +
= 9.25 + log              = 9.62
[ NH 4 ]                  (0.0953)
[H3O+] = 10SpH = 10S9.62 = 2.4 x 10S10 M
1.0 x 10 _14
[OHS] =                    = 4.2 x 10S5 M
2.4 x 10 _10
_
[HCO3 ][H 3O +]          (0.647)(2.4 x 10 _10)
[H2CO3] =                         =                  _7
= 3.6 x 10S4 M
Ka                  (4.3 x 10 )
(c) For MCO3, IP = [M ][CO3 ] = (0.010)(0.0153) = 1.5 x 10S4
2+      2S

Ksp(CaCO3) = 5.0 x 10S9, 103 Ksp = 5.0 x 10S6
Ksp(BaCO3) = 2.6 x 10S9, 103 Ksp = 2.6 x 10S6
Ksp(MgCO3) = 6.8 x 10S6, 103 Ksp = 6.8 x 10S3
IP > 103 Ksp for CaCO3 and BaCO3, but IP < 103 Ksp for MgCO3 so the [CO32S] is large
enough to give observable precipitation of CaCO3 and BaCO3, but not MgCO3.
(d) For M(OH)2, IP = [M2+][OHS]2 = (0.010)(4.17 x 10S5)2 = 1.7 x 10S11
Ksp(Ca(OH)2) = 4.7 x 10S6, 103 Ksp = 4.7 x 10S3
Ksp(Ba(OH)2) = 5.0 x 10S3, 103 Ksp = 5.0
Ksp(Mg(OH)2) = 5.6 x 10S12, 103 Ksp = 5.6 x 10S9
IP < 103 Ksp for all three M(OH)2. None precipitate.
(e)             CO32S(aq) + H2O(l)  HCO3S(aq) + OHS(aq)
initial (M)     0.08                               0             ~0
change (M)       Sx                               +x             +x
equil (M)       0.08 S x                             x             x
_         _                          2
[HCO 3 ][OH ]                            x
Kb =                       = 1.8 x 10S4 =
2
[CO 3 _ ]                        (0.08 _ x)
x2 + (1.8 x 10S4)x S (1.44 x 10S5) = 0
Use the quadratic formula to solve for x.
x=
_ (1.8 x 10 _ 4)  (1.8 x 10 _ 4) 2 _ (4)(1)(_ 1.44 x 10 _ 5) _ (1.8 x 10 _ 4)  7.59 x 10 _ 3
=
2(1)                                                2
x = 0.0037 and S0.0039

486
Chapter 16 S Applications of Aqueous Equilibria
______________________________________________________________________________

Of the two solutions for x, only 0.0037 has physical meaning because S0.0039 leads to
negative concentrations.
[OHS] = x = 3.7 x 10S3 M
For MCO3, IP = [M2+][CO32S] = (0.010)(0.08) = 8.0 x 10S4
For M(OH)2, IP = [M2+][OHS]2 = (0.010)(3.7 x 10S3)2 = 1.4 x 10S7
Comparing IP=s here and 103 Ksp=s in (c) and (d) above, Ca2+ and Ba2+ cannot be
separated from Mg2+ using 0.08 M Na2CO3. Na2CO3 is more basic than (NH4)2CO3 and
Mg(OH)2 would precipitate along with CaCO3 and BaCO3.

16.147 (a) H2SO4, 98.09 amu
Assume 1.00 L = 1000 mL of solution.
mass of solution = (1000 mL)(1.836 g/mL) = 1836 g
mass H2SO4 = (0.980)(1836 g) = 1799 g H2SO4
1 mol H 2SO 4
mol H2SO4 = 1799 g H2SO4 x                      = 18.3 mol H2SO4
98.09 g H 2SO 4
[H2SO4] = 18.3 mol/ 1.00 L = 18.3 M
(b) Na2CO3, 105.99 amu; 1 kg = 1000 g = 2.2046 lb
H2SO4(aq) + Na2CO3(s)  Na2SO4(aq) + H2O(l) + CO2(g)
2000 lb      1000 g
mass H2SO4 = (0.980)(36 tons) x             x             = 3.20 x 107 g H2SO4
1 ton      2.2046 lb
1 mol H 2SO 4
mol H2SO4 = 3.20 x 107 g H2SO4 x                      = 3.26 x 105 mol H2SO4
98.09 g H 2SO 4
1 mol Na 2CO 3 105.99 g Na 2CO 3
mass Na2CO3 = 3.26 x 105 mol H2SO4 x                       x                    x
1 mol H 2SO 4        1 mol Na 2CO 3
1 kg
= 3.5 x 104 kg Na2CO3
1000 g
1 mol CO 2
(c) mol CO2 = 3.26 x 105 mol H2SO4 x                    = 3.26 x 105 mol CO2
1 mol H 2SO 4
o
18 C = 18 + 273 = 291 K
PV = nRT
            L  atm 
(3.26 x 105 mol)  0.082 06             (291 K)
nRT                                K  mol 
V=        =                                                   = 7.9 x 106 L
P                               1.00 atm 
 745 mm Hg x 760 mm Hg 
                              

16.148 Pb(CH3CO2)2, 325.29 amu; PbS, 239.27 amu
1 mol Pb(CH 3CO 2) 2
(a) mass PbS = (2 mL)(1 g/mL)(0.003) x                         x
325.29 g Pb(CH 3CO 2) 2
1 mol PbS         239.27 g PbS
x              x (30 /100) = 0.0013 g
1 mol Pb(CH 3CO 2) 2    1 mol PbS
= 1.3 mg PbS per dye application

487
Chapter 16 S Applications of Aqueous Equilibria
______________________________________________________________________________

(b) [H3O+] = 10SpH = 10S5.50 = 3.16 x 10S6 M
PbS(s) + 2 H3O+(aq)  Pb2+(aq) + H2S(aq) + 2 H2O(l)
initial (M)                  3.16 x 10S6        0              0
change (M)                       S2x          +x              +x
equil (M)                3.16 x 10S6 S 2x        x              x
2+                       2                     2
[Pb ][H 2 S]               x                     x
Kspa =           + 2
=               _6     2
              _6 2
= 3 x 10S7
[ H 3O ]       (3.16 x 10 _ 2 x )       (3.16 x 10 )
2               S6 2       S7           S18
x = (3.16 x 10 ) (3 x 10 ) = 3.0 x 10
x = 1.7 x 10S9 M = [Pb2+] for a saturated solution.
mass of PbS dissolved per washing =
239.27 g PbS
(3 gal)(3.7854 L/1 gal)(1.7 x 10S9 mol/L) x                      = 4.7 x 10S6 g PbS/washing
1 mol PbS
Number of washings required to remove 50% of the PbS from one application =
(0.0013 g PbS)(50 /100)
_6
= 1.4 x 102
(4.7 x 10 g PbS/washing)
washings
(c) The number of washings does not look reasonable. It seems too high considering
that frequent dye application is recommended. If the PbS is located mainly on the
surface of the hair, as is believed to be the case, solid particles of PbS can be lost by
abrasion during shampooing.

488

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