# Solubility Products, Ksp, and Ion Products Qsp

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```					Solubility Products, Ksp,
and Ion Products Qsp
Ch 15, Part 4
The Other Half of the
Prank
Take-Home Test Posted
Due Wed Night
heterogeneous equilibrium
 solubility products, Ksp:
 Equilibrium Expression for Ksp and
Qsp
Equilibrium                              Expression for Ksp and Qsp

AgCl(s) = Ag+(aq) + Cl-(aq)                [Ag+] [Cl-]

CaCO3(s) = Ca2+(aq) + CO32-(aq)                 [Ca2+] [CO3-]

Li2CO3(s) = 2 Li+(aq) + CO32-(aq)              [Li+]2 [CO3-]
Ksp and Qsp
 Ksp is the equilibrium expression used
when the solution is saturated
 When the solution is NOT saturated, Qsp is
usually used to describe the equilibrium
– Qsp is called the ION PRODUCT QUOTIENT
– Equilibrium expression is the same
– If solution is not saturated, precipitate will not
form.
Comparing Qsp and Ksp
Qsp < Ksp   Unsaturated solution

Qsp = Ksp   Saturate solution

Qsp > Ksp   Oversaturate solution
Molar Solubilities and Solubility
Products
   Solubility products, Ksp, of salts are indirect indication of their
solubilities expressed in mol/L (called molar solubility).
– However, the solubility products are more useful than molar solubility.
– The molar solubilities are affected when there are common ions
present in the solution.
– We need to employ the solubility products to estimate the molar
solubilities in these cases.
– When a salt is dissolved in pure water, solubility products and molar
solubilities are related. This is illustrated using calcium carbonate. If x
is the concentration of Ca2+ (= [CO32-]) in the saturated solution, then
   Ksp = x2
– Because Ksp = [Ca2+] [CO32-]
Example 1

   The Ksp for AgCl is 1.8e-10 M2.
– What is the molar solubility of AgCl in
pure water?
   Solution
– Let x be the molar solubility, then
– AgCl = Ag+ + Cl-
x     x
– x = (1.8e-10 M2)1/2 = 1.3e-5 M
Molar Solubility
 The solubility product, Ksp is a better indicator
than the usual solubility specification of g per
100 mL of solvent or moles per unit volume of
solvent.
 For the AgCl case, when the cation
concentration is not the same as the anion
concentration ([Ag+] =/= [Cl-]) solubility of
AgCl can not be defined in terms of moles per L.
Example 1
– In this case, the system can be
divided into three zones. The
condition [Ag+] [Cl-] = Ksp, is
represented by a line which divides
the plane into two zones.
– When [Ag+] [Cl-] < Ksp, no
precipitate will be formed.
– When [Ag+] [Cl-] > Ksp, a
precipitate will be formed.
– When AgCl and NaCl dissolve in a
solution, both salts give Cl- ions.
The effect of [Cl-] on the solubility
of AgCl is called the Common ion
effect
Example 2
   The Ksp for Ag2CrO4 is 9e-12 M3.
– What is the molar solubility of Ag2CrO4 in pure water?
   Solution
Let x be the molar solubility of Ag2CrO4, then
– Ag2CrO4 = 2 Ag+ + CrO42-
2x       x
– (2 x)2(x) = Ksp
– x = {(9e-12)/4}(1/3) = 1.3e-4 M
– [Ag+] = 2.6e-4 M and the molar solubility is 1.3e-4 M.
   A diagram similar to AgCl can be drawn but shape
of the curve is different
Example 3
   What is the pH in a saturated solution of
Ca(OH)2?
Ksp = 5.5e-6 M3 for Ca(OH)2.
   Let [Ca2+] = x, then [OH-] = 2 x. The equilibrium
and concentration are represented below:
   Ca(OH)2 = Ca2+ + 2 OH-
x       2x
   x = {5.5e-6/4}(1/3) [OH-] = 2 x.
Note the stoichiometry of equilibrium.
Example 4
   At 760◦C, Kc = 33.3 for the reaction
– PCl5(g) ⇀↽ PCl3(g) + Cl2(g)
– If a mixture that consists of 0.54 mol PCl3 and
0.85 mol Cl2 is placed in a 9 L reaction vessel
and heated to 760◦C, what is the equilibrium
composition of PCl5?
   Initially,
– [PCl3] = 0.54 mol/9 L = 0.06 mol/L
– [Cl2] = 0.85 mol/9 L = 0.0944444 mol/L
Example 4
Since 33.4543
is larger than
the initial
con-
centrations, it
can be
33.3 x = x2 − 0.154444 x + 0.00566667
0 = x2 − 33.4544 + 0.00566667
Thus [PCl5] =
0.000169385
mol/L
AP Question #1
    Propanoic acid, HC3H5O2, ionizes in water
according to the equation below.
– HC3H5O2(aq) ↔ C2H5O2-(aq) + H+(aq)
– Ka = 1.34 x 10-5
A.   Write the equilibrium expression
AP Question #1
    Propanoic acid, HC3H5O2, ionizes in water
according to the equation below.
– HC3H5O2(aq) ↔ C2H5O2-(aq) + H+(aq)
– Ka = 1.34 x 10-5
A.   Write the equilibrium expression

    
[C3 H 5O ][H ]
Ka          2
[ HC3 H 5O2 ]
AP Example #1b
HC3H5O2(aq)   ↔     C2H5O2-(aq)    +      H+(aq)
0.265                 0                  0
-x                  +x                 +x
0.265-x                X                  x


[C3 H 5O2 ][H  ]                     x2
Ka                     1.34 x10 5 
[ HC3 H 5O2 ]                   (0.265  x)

x  [C3 H 5O2 ]  [ H  ]  0.00188M
pH   log[ H  ]   log(0.00188)  2.726
AP Example #1
B.       Calculate the pH of a 0.265 M solution of
propanoic acid
        A 0.496 g sample of sodium propanate,
NaC3H5O2, is added to 50.0 mL sample of a
0.265 M solution of propanoic acid. Assuming
no change in volume of the solution occurs,
calculate each of the following:
–      The concentration of the propanate ion, C3H5O2-
(aq), in the solution.
–     The concentration of the H+ (aq) in the solution.
AP Example #1C
0.496gNaC3 H 5O2         1molNaC3 H 5O2 1000mmol                  
x                    x       5.17mmolC3 H 5O2
96.0 gNaC3 H 5O2   1mol

     5.17mmolC3 H 5O2
[C3 H 5O ] 
2                        0.103M
50.0mL
0.265molHC3 H 5O2 50.0mL
MV  moles ::                     x       13.25mmolHC3 H 5O2
1L            1
HC3H5O2(aq)          ↔      C2H5O2-(aq)    +       H+(aq)
13.25                     5.17                  0

-x                       +x                  +x

13.25-x                  5.17+x                 x
AP Example #1c
pKa   log( Ka)  4.873
 [C3 H 5O2 ] 

 5.17 
pH  pKa  log      [ HC H O  ]   4.873  log  13.25   4.464

     3 5 2                          
10 pH  [ H  ]  3.43x10 5 M
AP Example #2
The methanote ion, HCO2- (aq) reacts with water to form methanoic acid
and hydroxide ion, as shown in the following equation.
HCO2- (aq) + H2O (l)  HCO2H (aq) + OH- (aq)
1) Given that [OH-] is 4.18 x 10-6 M in a 0.309 M solution of sodium
methanoate, calculate each of the following
I)The value of Kb for the methanoate ion, HCO2- (aq)
II)The value of Ka for methanoic acid, HCO2H

[ HCO2 H ]  [OH  ]  4.18x106

[ HCO2 ]  0.309M
[ HCO2 H ][OH  ] (4.18x10 6 )(4.18x10 6 )
Kb             
                            5.65x1011
[ HCO2 ]              0.309
K a xKb  K w  1.0 x1014  ( K a )(5.65x1011 )
K a  1.77 x10 4
AP Example #3 (2003)
   1. C6H5NH2(aq) + H2O(l)  C6H5NH3+(aq) + OH–
(aq) Aniline, a weak base, reacts with water
according to the reaction represented above.
(a) Write the equilibrium constant expression, Kb, for the
reaction represented above.
(b) A sample of aniline is dissolved in water to produce
25.0 mL of 0.10 M solution. The pH of the solution is
8.82. Calculate the equilibrium constant, Kb, for this
reaction.
(c) The solution prepared in part (b) is titrated with 0.10
M HCl. Calculate the pH of the solution when 5.0 mL
of the acid has been titrated.
(d) Calculate the pH at the equivalence point of the
titration in part (c).
(e) The pKa values for several indicators are given below.
Which of the indicators listed is most suitable for this
AP Example #3 (2003)
(e) The pKa values for several indicators are given below.
Which of the indicators listed is most suitable for this

Indicator                pKa
Erythrosine              3
Litmus                   7
Thymolphthalei 10
n
AP Example #3 Answers
         3 
C 6H 5NH   OH 
(a)      Kb =

C6H 5NH 2 
(b)      pOH = 14 – pH = 14 – 8.82 = 5.18
-log[OH–] = 5.18; [OH–] = 6.61´10–6 M
[OH–] = [C6H5NH3+]


Kb =
6.6110 
6 2
= 4.4*10–10
0.10 – 6.61106


(c)
AP Example #3 (Answers)
25 mL = 2.5 mmol C H NH
´                 6 5    2
5 mL ´ = 0.5 mmol     H+ added

2.0 mmol base remains in 30.0 mL solution
X = 1.80´10–9 = [OH–]
   0.50 mmol
X X  30.0 mL                [H+] =
4.4*10–10 =
20.0 mmol                    11014
9
 30.0 mL                     1.810
= 5.6*10–6; pH = 5.26
(d)       when neutralized, there are 2.5 mmol of C6H5NH3+ in 50.0 mL of solution, giving
a [C6H5NH3+] = 0.050 M his cation will partially ionize according to the following
equilibrium:
C6H5NH3+(aq)  C6H5NH2(aq) + H+(aq)
at equilibrium, [C6H5NH2] = [H+] = X
[C6H5NH3+] = (0.050–X)
2            = Ka = 2.3´10-5
X                      X = 1.06´10–3 = [H+]
(0.050  X )
pH = –log[H+] = 2.98
AP Example #3 (Answers)
   (e) erythrosine; the indicator will change
color when the pH is near its pKa, since
the equivalence point is near pH 3, the
indicator must have a pKa near this value.
AP Example #4 (2004)
1. Answer the following questions relating to the solubilities
of two silver compounds, Ag2CrO4 and Ag3PO4.
Silver chromate dissociates in water according to the
equation shown below.
Ag2CrO4(s)  2 Ag+(aq) + CrO42–(aq)
Ksp = 2.6 × 10–12 at 25°C
a)   Write the equilibrium-constant expression for the
dissolving of Ag2CrO4(s).
b)   Calculate the concentration, in mol L-1, of Ag+(aq) in a
saturated solution of Ag2CrO4 at 25°C.
c)   Calculate the maximum mass, in grams, of Ag2CrO4
that can dissolve in 100. mL of water at 25°C.
AP Example #4 (2004)
1. Answer the following questions relating to the solubilities
of two silver compounds, Ag2CrO4 and Ag3PO4.
Silver chromate dissociates in water according to the
equation shown below.
Ag2CrO4(s)  2 Ag+(aq) + CrO42–(aq)
Ksp = 2.6 × 10–12 at 25°C
d) A 0.100 mol sample of solid AgNO3 is added to a 1.00 L
saturated solution of Ag2CrO4 . Assuming no volume change,
does [CrO42–] increase, decrease, or remain the same? Justify
In a saturated solution of Ag3PO4 at 25°C, the
concentration of Ag+(aq) is 5.3 × 10–5 M. The
equilibrium-constant expression for the dissolving of
Ag3PO4(s) in water is shown below.
Ksp = [Ag+]3 [PO43-]
d) Write the balanced equation for the dissolving of Ag3PO4 in
water. Calculate the value of Ksp for Ag3PO4 at 25°C.
e) A 1.00 L sample of saturated Ag3PO4 solution is allowed to
evaporate at 25°C to a final volume of 500. mL. What is [Ag+]
AP Example #4 (Answers)
a) Write the equilibrium constant
expression for the dissolving of Ag2CrO4.
Ksp = [Ag2+]2[CrO42-]
b) Calculate the concentration in mol
L-1, of Ag+ in a saturated solution of
Ag2CrO4 at 25˚C. Ksp  [Ag ]2 [CrO 2 ]
4

2.6x1012  [2s]2[s]  4s3
s  8.7x105
                 4
[Ag ]  2s  1.7x10 M
AP Example #4 (Answers)
   c) Calculate the maximum mass in grams
of Ag2CrO4 that can dissolve in 100. mL of
water at 25˚C.
8.7 x10 5 molAg 2 CrO 4 0.100 L 332 g
x       x       0.0029 g
1L               1     1mol
d) A 0.100 mol sample of solid AgNO3 is added to 1.00 L saturated solution of
Ag2CrO4. Assuming no volume change, does [CrO42-] increase, decrease, or
Common ion effect. Adding AgNO3 is adding Ag+ to the
product side. This will shift the equilibrium to the left (LeChatliers
Principle). Thus the concentration of the chromate will go down. You
could justify this with math but since they don’t ask it then don’t.
AP Example #5 (2001)
   1. Answer the following questions relating to
the solubility of the chlorides of silver and
(a)      At 10C, 8.9  10-5 g of AgCl(s) will
dissolve in 100. mL of water.
(i)     Write the equation for the dissociation of AgCl(s)
in water.
(ii) Calculate the solubility, in mol L–1, of AgCl(s) in
water at 10C.
(iii) Calculate the value of the solubility-product
constant, Ksp for AgCl(s) at 10C.
AP Example #5 (2001)
(b) At 25C, the value of Ksp for PbCl2(s) is 1.6  10-5
and the value of Ksp for AgCl(s) is 1.8  10-10.
(i)      If 60.0 mL of 0.0400 M NaCl(aq) is added to 60.0 mL of
0.0300 M Pb(NO3)2(aq), will a precipitate form? Assume that
volumes are additive. Show calculations to support your
(ii)     Calculate the equilibrium value of [Pb2+(aq)] in 1.00 L
of saturated PbCl2 solution to which 0.250 mole of NaCl(s)
has been added. Assume that no volume change occurs.
(iii)    If 0.100 M NaCl(aq) is added slowly to a beaker
containing both 0.120 M AgNO3(aq) and 0.150 M
Pb(NO3)2(aq) at 25C, which will precipitate first, AgCl(s) or
PbCl2(s)? Show calculations to support your answer.
AP Example #5 (2001)
(a) At 10° C, 8.9 x10-5 g of AgCl(s) will dissolve
in 100 mL of water.
(i) Write the equation for the dissociation of AgCl(s) in
water.
AgCl(s) <===> Ag+ + Cl-
(ii) Calculate the solubility, in mol/L, of AgCl(s) in water
at 10° C.
(8.9 x10-5 g AgCl)(mol AgCl/169 g AgCl) = 5.27 x10-7
mol
5.27 x10-7 mol/0.1 L = 5.27 x10-6 M
(iii) Calculate the value of the solubility product
constant, Ksp, for AgCl(s) at 10° C.
Ksp = [Ag +][Cl -] = [5.27 x10-6]2 = 2.78 x10-11
AP Example #5 (2001)
 (b) At 25° C, the value of Ksp for PbCl2(s) is 1.6
x10-5 and the value for AgCl(s) is 1.8 x10-10.
 (i) If 60.0 mL of 0.0400 M NaCl(aq) is added to
60.0 mL of 0.0300 M Pb(NO3)2(aq), will a
precipitate form? Assume that volumes are
additive. Show calculations to support your
– (0.06 L)(0.04 M)   = 2.4 x10-3 mol Cl-/0.120 L - =
0.02 mol M Cl-
– (0.06 L)(0.03 M)   = 1.8 x10-3 mol Pb2+/0.120 L =
0.015 M Pb2+
– [Pb2+][Cl-]2 = Q   = [0.015][0.02]2 = 6.0 x10-6
– No ppt 1.6 x10-5   > 6.0 x10-6
AP Example #5 (2001)
(ii) Calculate the equilibrium value of [Pb2(aq)] in
1.00 L of saturated PbCl2 solution to which
0.2050 mole of NaCl(s) has been added. Assume
that no volume change occurs.
1.6 x10-5 = [Pb2+][0.25]2 = 2.56 x10-4
(iii) If 0.100 M is added slowly to a beaker
containing both 0.120 M AgNO3(aq) and 0.150 M
Pb(NO3)2 at 25° C, which will precipitate first,
AgCl(s) or PbCl2(s)? Show calculations to support
QAgCl = [0.12][0.1]2 = 1.2 x10-3
QPbCl2 = [0.15][0.1]2 = 1.5 x10-3
AP Example #6

5. Answer the questions below that relate to the five aqueous solutions
at 25C shown above.
(a) Which solution has the highest boiling point? Explain.
(b) Which solution has the highest pH? Explain.
(c) Identify a pair of the solutions that would produce a precipitate
when mixed together. Write the formula of the precipitate.
(d) Which solution could be used to oxidize the Cl–(aq) ion?
Identify the product of the oxidation.
(e) Which solution would be the least effective conductor of
electricity? Explain.
AP Example #6
(a) Which solution has the highest boiling point?
Explain.
Pb(NO3)2: greatest number of moles of particles
(b)   Which solution has the highest pH? Explain
C2H3O2 - + H2O <==> HC2H3O2 + OH -
(c) Identify a pair of solutions that would
produce a precipitate when mixed together.
Write the formula of the precipitate.
Pb2+ + Cl - ===> PbCl2
(d) Which solution could be used to oxidize the
Cl -(aq) ion? Identify the product of the
solution.
Cl - + MnO4 ===> Cl2 + Mn 2+
(e) Which solution would be the least effective
conductor of electricity? Explain.
C2H5OH:    stays in solution as a molecular species

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