# Homework Solutions Problem 14 - 17

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```					                                        Homework Solutions

Problem 14 – 17

+
a) Arrhenius acid – any substance that produces H ions in water
+
b) Bronsted – Lowry acid – a proton (H ) donor

c) Lewis acid – an electron pair acceptor
Problem 14 – 29
acid         base                      conj. base           conj. acid
-                       +
a)   HF           H 2O                      F                    H 3O
-               +
b)   H2SO4        H 2O                      HSO4                 H 3O
4-                                     2-                  +
c)   HSO          H 2O                      SO4                  H 3O

Problem 14 – 31
a) weak      b) strong c) strong d) weak
Problem 14 – 33
+
HNO3 > HOCl > NH4 > H2O

Problem 14 – 35

a) HCl       b) HNO2           c) HCN

Problem 14 – 37
-        +
[OH ] = Kw/[H ]
-            -7
a) [OH ] = 1.0 x 10        M ; neutral
-            -11
b) [OH ] = 1.5 x 10            M ; acidic
-            -4
c) [OH ] = 5.3 x 10        M ; basic
-            -15
d) [OH ] = 4.3 x 10            M ; acidic
Problem 14 – 38
[H+] = Kw/[OH-]
+               -15
a) [H ] = 2.8 x 10         M ; basic
+               -6
b) [H ] = 1.0 x 10        M ; acidic
+               -12
c) [H ] = 4.5 x 10         M ; basic
+               -7
d) [H ] = 1.0 x 10        M ; neutral

Problem 14 – 41
+
pH = - log [H ] ; pOH = 14.00 – pH
-7
a) pH = - log (1.0 x 10 ) = 7.00 ; pOH = 14.00 – 7.00 = 7.00
-4
b) pH = - log (6.7 x 10 ) = 3.17 ; pOH = 14.00 – 3.17 = 10.83
-11
c) pH = - log (1.9 x 10           ) = 10.72 ; pOH = 14.00 – 10.72 = 3.28

d) pH = - log (2.3) = - 0.36 ; pOH = 14.00 – (- 0.36) = 14.36
Problem 14 – 42
-
pOH = - log [OH ] ; pH = 14.00 – pOH

a) pOH = - log (3.6) = - 0.56 ; pH = 14.00 – (- 0.56) – 14.56
-9
b) pOH = - log (9.7 x 10 ) = 8.01 ; pH = 14.00 – 8.01 = 5.99
-3
c) pOH = - log (2.2 x 10 ) = 2.66 ; pH = 14.00 – 2.66 = 11.34
-7
d) pOH = - log (1.0 x 10 ) = 7.00 ; pH = 14.00 – 7.00 = 7.00
Problem 14 – 43
-pH                                                  -pOH
pH = antilog (-pH) = 10                         ; pOH = antilog (-pOH) = 10

pH + pOH = 14.00
+         -7.40                   -8               -             -6.60                  -7
a) [H ] = 10                 = 4.0 x 10           M ; [OH ] = 10                   = 2.5 x 10          M ; basic
+         -15.3               -16                 -             1.3
b) [H ] = 10                 = 5 x 10             M ; [OH ] = 10                 = 20 M ; basic
+         1.1                       -          -15                    -15
c) [H ] = 10               = 10 M ; [OH ] = 10                   = 1 x 10              M ; acidic
+         -3.20                    -4                 -          -10.80                   -11
d) [H ] = 10                 = 6.3 x 10           M ; [OH ] = 10                       = 1.6 x 10          M ; acidic
-       -5.0              -5              +               -9.0              -9
e) [OH ] = 10                = 1 x 10        M ; [H ] = 10                     = 1 x 10        M; basic
-      -9.60                   -10                +              -4.40                -5
f) [OH ] = 10                 = 2.5 x 10              M ; [H ] = 10                    = 4.0 x 10        M; acidic

Problem 14 – 44

+         0.42                          -         -14.42                       -15
a) [H ] = 10                = 2.6 M ; [OH ] = 10                          = 3.8 x 10           M ; acidic
+         -3.42                    -4                 -          -10.58                   -11
b) [H ] = 10                 = 3.8 x 10           M ; [OH ] = 10                       = 2.6 x 10          M ; acidic
+         -10.67                    -11                   -           -3.33                -4
c) [H ] = 10                  = 2.1 x 10              M ; [OH ] = 10                    = 4.7x 10         M ; basic
-       -14.30                      -15               +           0.30
d) [OH ] = 10                  = 5.0 x 10               M ; [H ] = 10                   = 2.0 M; acidic
-       -9.67                   -10               +              -4.33                -5
e) [OH ] = 10                 = 2.1 x 10              M ; [H ] = 10                    = 4.7 x 10         M; acidic
-      -1.15                   -2             +             -12.85                    -13
f) [OH ] = 10                 = 7.1 x 10           M ; [H ] = 10                       = 1.4 x 10         M; basic

Problem 14 – 45
pOH = 14.00 – 6.77 = 7.23
+               -6.77                -7
[H ] = 10                 = 1.7 x 10        M
-            -7.23                -8
[OH ] = 10                 = 5.9 x 10         M

acidic
Problem 14 – 49
a) pH = - log (0.10) = 1.00
b) pH = - log (5.0) = - 0.70
-11                                                                          +   -7
c) pH = - log (1.0 x 10                           ) = 11.00 (not possible); the H2O has [H ] = 1.0 x 10                           M

pH = 7.00
Problem 14 – 51
+            -
HCl : 50.0 mL x 0.050 M = 2.5 mmol (H and Cl )
+                 -
HNO3 : 150.0 mL x 0.10 M = 15 mmol (H and NO3 )
+
[H ] = (2.5 mmol + 15 mmol)/(50.0 mL + 150.0 mL) = 0.088 M
-
[Cl ] = 2.5 mmol/(50.0 mL + 150.0 mL) = 0.013 M
-
[NO3 ] = 15 mmol/(50.0 mL + 150.0 mL) = 0.075 M
-                    +                       -14                             -13
[OH ] = Kw/[H ] = 1.0 x 10                                  /0.088 = 1.1 x 10             M

Problem 14 – 53
+            -2.50                           -3                                                          +
[H ] = 10                   = 3.2 x 10              M ; Since HCl is strong, the [H ] is also the acid

concentration.
Problem 14 – 57
+                      -
HOCl          +         H 2O                H 3O              +       OCl

I:                1.5 M                                            0                    0
C:                     -x                                             +x                    +x
E:            1.5 – x                                                  x                     x
+             -                                                                -8
Ka = [H3O ][OCl ]/[HOCl] = (x)(x)/(1.5 – x) = 3.5 x 10
+                     -                      -4                                     -4
x = [H3O ] = [OCl ] = 2.3 x 10                                   M ; pH = -log (2.3 x 10 ) = 3.64
-                        -14                    -4                -11
[OH ] = 1.0 x 10                         /2.3 x 10          = 4.3 x 10          M
-4
[HOCl] = 1.5 M – 2.3 x 10                              M = 1.5 M
Problem 14 – 59
+                      -
HF    +          H 2O                    H 3O           +         F

I:          0.020 M                                          0                        0
C:                -x                                       +x                      +x
E:           0.020 – x                                     x                           x
+        -                                                        -4
Ka = [H3O ][F ]/[HF] = (x)(x)/(0.020 – x) = 7.2 x 10
+                -                  -3                                      -3
x = [H3O ] = [F ] = 3.5 x 10                       M ; pH = - log (3.5 x 10 ) = 2.46
-                     -14                 -3                -12
[OH ] = 1.0 x 10                  /3.5 x 10        = 2.9 x 10         M
-3
[HF] = 0.020 M – 3.5 x 10                     M = 0.017 M

Problem 14 – 61
+                              -
HClO2          +      H 2O                 H 3O            +        ClO2

I:              0.50 M                                   0                        0
C:               -x                                       +x                      +x
E:           0.50 – x                                      x                       x
+                -                                                               -2
Ka = [H3O ][ClO2 ]/[HClO2] = (x)(x)/(0.50 – x) = 1.2 x 10
+
x = [H3O ] = 0.072 M

pH = -log (0.072) = 1.14
Problem 14 – 63
HCl is a strong acid and will dominate the pH. The pH of the solution can be
based on the HCl concentration.
pH = - log (0.10) = 1.00
Problem 14 – 67
-                        +
HC3H5O2               +       H 2O                                 C 3H 5O 2               +        H 3O

I:            0.10 M                                                                    0                            0
C:               -x                                                                    +x                           +x
E:              0.10 – x                                                                   x                            x
-          +                                                                              -5
Ka = [C3H5O2 ][H3O ]/[HC3H5O2] = (x)(x)/(0.10 – x) = 1.3 x 10
+                       -3                                    -3
x= [H3O ] = 1.1 x 10                       M ; pH = - log (1.1 x 10 ) = 2.96
+
% ionization = [H3O ]/[HC3H5O2]o x 100% = 1.1 %

Problem 14 – 69
-                           +
HA        +       H 2O                     A           +       H 3O

I: 0.15 M                                        0                       0
C:     -x                                         +x                      +x
E: 0.15 – x                                        x                          x
+                                              +
[H3O ]/[HA]o = 0.030 ; [H3O ] = 0.030 x [HA]o = 0.030 x 0.15 M = 0.0045 M = x
-                +
Ka = [A ][H3O ]/[HA] = (x)(x)/(0.15 – x) = (0.0045)(0.0045)/(0.15 – 0.0045)
-4
Ka = 1.4 x 10

Problem 14 – 71
-                            +
HBrO        +        H 2O                         BrO                 +           H 3O

I: 0.063 M                                                         0                            0
C:         -x                                                  +x                              +x
E: 0.063 – x                                                       x                            x
-             +
Ka = [BrO ][H3O ]/[HBrO] = (x)(x)/(0.063 – x)
+                     -                -4.95                 -5
[H3O ] = [BrO ] = x = 10                             = 1.1 x 10        M
-5                  -5                          -5                        -9
Ka = (1.1 x 10 )(1.1 x 10 )/(0.063 – 1.1 x 10 ) = 1.9 x 10
Problem 14 – 73
-                   +
HCOOH               +         H 2O                        COOH                 +       H 3O

I:           ?                                                             0                      0
C:            -x                                                        +x                       +x
E: ? – x                                                                    x                     x
-            +                                                                    -4
Ka = [COOH ][H3O ]/[HCOOH] = (x)(x)/(? – x) = 1.8 x 10
+             -2.70                          -3
x = [H3O ] = 10                     = 2.0 x 10             M
-3                     -3                             -3                -4
(2.0 x 10 )(2.0 x 10 )/(? – 2.0 x 10 ) = 1.8 x 10
-2
? = 2.4 x 10           M

Problem 14 – 75
+                -
a)   NH3 +             H 2O                                NH4            +   OH
+             -
Kb = [NH4 ][OH ]/[NH3]
+                        -
b)   C 5H 5N           +        H 2O                                    C5H5NH               +       OH
+          -
Kb = [C5H5NH ][OH ]/[C5H5N]

Problem 14 – 77
-
NH3 > C5H5N > H2O > NO3

Problem 14 – 79
-
a) NH3        b) NH3                     c) OH             d) CH3NH2
Problem 14 – 81
-
pOH = - log [OH ] ; pH = 14.00 – pOH

a) pOH = - log (0.10) = 1.00 ; pH = 14.00 – 1.00 = 13.00
-10
b) pOH = - log (1.0 x 10              ) = 10.00 ; pH = 14.00 – 10.00 = 2.00 (impossible)
-                 -7
Water is the most important contributor to [OH ] (1.0 x 10                               M) ; pH = 7.00

c) pOH = - log (2.0) = - 0.30 ; pH = 14.00 – (- 0.30) = 14.30
Problem 14 – 85
KOH is a strong base. Therefore, the concentration of the base will be the same
-
as the [OH ].
pH + pOH = 14.00 ; pOH = 14.00 – 10.50 = 3.50
-        -pOH          -3.50                -4                               -4
[OH ] = 10          = 10           = 3.2 x 10        M ; [KOH] = 3.2 x 10             M

Problem 14 – 87
+                      -
NH3        +    H 2O                 NH4             +        OH

I: 0.150 M                                    0                      0
C:     -x                                     +x                     +x
E: 0.150 – x                                   x                     x
+      -                                                   -5
Kb = [NH4 ][OH ]/[NH3] = (x)(x)/(0.150 – x) = 1.8 x 10
-               -3
x = [OH ] = 1.6 x 10         M
-3
pOH = - log (1.6 x 10 ) = 2.80 ; pH = 14.00 – 2.80 = 11.20
Problem 14 – 89
+                        -
a)            (C2H5)3N          +     H 2O                       (C2H5)3NH                +    OH

I:    0.20 M                                                   0                          0
C:     -x                                                  +x                             +x
E: 0.20 – x                                                    x                              x
-4
Kb = (x)(x)/(0.20 – x) = 4.0 x 10
-                -3
x = [OH ] = 8.7 x 10           M
+                -                 -14               -3                -12
[H3O ] = Kw/[OH ] = 1.0 x 10                  /8.7 x 10        = 1.1 x 10         M
-12
pH = - log (1.1 x 10            ) = 11.96
+                      -
b)            HONH2         +       H 2O                         HONH3                +       OH

I: 0.20 M                                                 0                         0
C:    -x                                                   +x                        +x
E: 0.20 – x                                                 x                         x
-8
Kb = (x)(x)/(0.20 – x) = 1.1 x 10
-                -5
x = [OH ] = 4.7 x 10           M
+                -14               -5                -10
[H3O ] = 1.0 x 10           /4.7 x 10        = 2.1 x 10         M
-10
pH = - log (2.1 x 10            ) = 9.68
Problem 14 – 91
+                  -
C2H5NH2             +   H 2O               C2H5NH3               +        OH

I:        0.20                                          0                      0
C:        -x                                            +x                     +x
E:   0.20 – x                                            x                      x
+         -                                                         -4
Kb = [C2H5NH3 ][OH ]/[C2H5NH2] = (x)(x)/(0.20 – x) = 5.6 x 10
-              -2                                  -2
x = [OH ] = 1.0 x 10        M ; pOH = - log (1.0 x 10 ) = 2.00

pH = 14.00 – 2.00 = 12.00
Problem 14 – 93
+                 -
NH3           +   H 2O               NH4          +       OH

a)                I: 0.10 M                                     0                0
C: - x                                    +x                   +x
E: 0.10 – x                                   x                 x
-5
Kb = (x)(x)/(0.10 – x) = 1.8 x 10
-              -3
x = [OH ] = 1.3 x 10        M
-                                   -3
% ionization = [OH ]/[NH3]o x 100% = 1.3 x 10                   M/0.10 M x 100% = 1.3 %
-5
b) Kb = (x)(x)/(0.010 – x) = 1.8 x 10
-              -4
x = [OH ] = 4.2 x 10        M
-4
% ionization = 4.2 x 10              M/0.010 M x 100 % = 4.2 %
Problem 14 – 103

HCl > NH4Cl > KCl > KCN > KOH

Problem 14 – 109
+                                                                            +
a)             CH3NH3                     +        H 2O                           CH3NH2          +    H 3O

I:        0.10 M                                                           0                   0
C:            -x                                                             +x                  +x
E: 0.10 – x                                                                    x                   x
+                     +
Ka = [CH3NH2][H3O ]/[CH3NH3 ] = Kw/Kb
-14                -4                      -11
Ka = (x)(x)/(0.10 – x) = 1 x 10                             /4.38 x 10        = 2.28 x 10
+                          -6
x = [H3O ] = 1.5 x 10                          M
-5
pH = - log (1.5 x 10 ) = 5.82

-                                                                        -
b)             CN            +       H 2O                             HCN +           OH

I: 0.050 M                                                        0                 0
C:        -x                                                      +x             +x
E: 0.050 – x                                                       x                x
-                -
Kb = [HCN][OH ]/[CN ] = Kw/Ka
-14               -10                  -5
Kb = (x)(x)/(0.050 – x) = 1 x 10                                /6.2 x 10         = 1.6 x 10
-4
x = [OH-] = 8.9 x 10                      M
-4
pOH = - log (8.9 x 10 ) = 3.05 ; pH = 14.00 – 3.05 = 10.95
Problem 14 – 110
-                                                                      -
a)          NO2                 +    H 2O                      HNO2          +         OH

I: 0.12 M                                                    0                      0
C: - x                                                    +x                     +x
E: 0.12 – x                                                    x                     x
-14               -4                -11
Kb = (x)(x)/(0.12 – x) = 1 x 10                         /4.0 x 10        = 2.5 x 10
-                        -6
x = [OH ] = 1.7 x 10                      M
-6
pOH = - log (1.7 x 10 ) = 5.77 ; pH = 14.00 – 5.77 = 8.23

-                                                                      -
b)          OCl             +       H 2O                       HOCl           +   OH

I: 0.45 M                                                      0                0
C: - x                                                         +x                +x
E: 0.45 – x                                                     x                 x
-14               -8                -7
Kb = (x)(x)/(0.45 – x) = 1 x 10                         /3.5 x 10        = 2.9 x 10
-                        -4
x = [OH ] = 3.6 x 10                      M
-4
pOH = - log (3.6 x 10 ) = 3.44 ; pH = 14.00 – 3.44 = 10.56

+                                                                   +
c)          NH4                 +         H 2O                 NH3          +     H 3O

I:       0.40 M                                          0                     0
C:        -x                                              +x                    +x
E: 0.40 – x                                                   x                  x
-14               -5                -10
Ka = (x)(x)/(0.40 – x) = 1 x 10                         /1.8 x 10        = 5.6 x 10
+                        -5
x = [H3O ] = 1.5 x 10                         M
-5
pH = - log (1.5 x 10 ) = 4.82
Problem 14 – 115
a) neutral
-                                                            -
b) basic ;    NO2           +       H 2O                  HNO2      +       OH
+                                                           +
c) acidic ;     C5H5NH                   +     H 2O               C 5H 5N            +   H 3O

d) neutral
-                                                    -
e) basic ;     OCl              +       H 2O              HOCl      +   OH

f) neutral
Problem 14 – 121

a) basic ;    CaO           +   H 2O                  Ca(OH)2

b) acidic ; SO2             +   H 2O                  H2SO3

c) acidic ;   Cl2O          +       H 2O              2HClO

Problem 14 – 122

a) basic ;    Li2O          +       H 2O              2LiOH

b) acidic ;   CO2           + H 2O                    H2CO3

c) basic ;     SrO          +       H 2O              Sr(OH)2

```
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