Solubility Equilibria Review and Questions - KEY
Understand what is meant by molar solubility (S in molarity), solubility (S in other units), and
solubility product (Ksp). Be able to calculate S from Ksp or visa versa.
Understand how common ions affect solubility and be able to illustrate this effect by calculation.
Recognize the ion product (Qip) as the reaction quotient (Q) for solubility processes. Be able to
predict precipitation or dissolution based upon the value of the ion product vs. the value of Ksp.
Recognize the influence of pH and complex formation on solubility. Be able to illustrate through
calculations involving hydroxides, sulfides, or complexes.
Do not memorize facts about qualitative analysis, but do understand and be able to apply the
methods qualitative in theory.
Note: The values of Ksp (of all salts), Ka, Kb, Kf, etc. may all differ slightly from your textbook
values. This doesn’t alter the method by which the problem is solved!
1.
Calculate the molar solubility of Cu(OH)2 [Ksp = 8 x 10-16] in:
a. distilled water, b. pH = 13.00 NaOH(aq), and c. pH = 4.00 buffer
Also, calculate these solubilities in g solid/L solution.
1 (a)
Cu(OH)2(s)
ÅÆ
2+
At equil
2+
-
+
Cu (aq)
S
-
Ksp= 2.2x10-20
-
Ksp= 2.2x10-20
2 OH (aq)
2S
2
Ksp= 2.2x10-20 = [Cu ][ OH ]2 = (S)(2S)
giving the answer as
2+
S = [Cu ] = 1.8x10-7 M
-1
-1
and in g solid/L solution = 1.8x10-7 M *(97.57 g mole ) = 1.76x10-5 gL
1 (b)
In pH = 13.00 NaOH, [OH-] = 0.10 M
ÅÆ
Cu(OH)2(s)
2+
Initial
change
At equil
2+
-
+
Cu (aq)
0
+S
S
2
2 OH (aq)
0.10
+2S
0.10 + 2S
2
Ksp= 2.2x10-20 = [Cu ][ OH ]2 = (S)(0.10 + 2S) ≈ (S)(0.10)
giving the answer as
2+
S = [Cu ] = 2.2x10-18 M
-1
-1
and in g solid/L solution = 2.2x10-18 M *(97.57 g mole ) = 2.15x10-16 gL
NOTE: this basic solid is less soluble in a solution containing a source of common ion.
122
1 (c)
In pH = 4.00 buffer, pOH is 10.00 and [OH-] = 1.0x10-10 M
ÅÆ
Cu(OH)2(s)
2+
Initial
change
At equil
2+
-
-
+
Cu (aq)
0
+S
S
Ksp= 2.2x10-20
2 OH (aq)
1.0x10-10
+2S
~1.0x10-10
2
2
Ksp= 2.2x10-20 = [Cu ][ OH ]2 = (S)( 1.0x10-10 + 2S) ≈ (S)( 1.0x10-10)
Because the system is buffered at that pH . . .. . giving the answer as
2+
S = [Cu ] = 2.2 M . . . this means that 4.4 mole/L worth of hydroxide had to be neutralized by the acidic
buffering component
-1
-1
and in g solid/L solution = 2.2 M *(97.57 g mole ) = 2.15x10+2 gL
NOTE: this basic solid is more soluble in a solution that can remove hydroxide from solution as the solid
dissolves.
+
ALTERNATE METHOD: note: pH = 4.00 = 1.0x10-4 M H
ÅÆ
Cu(OH)2(s)
2
+
H (aq)
Cu(OH)2(s)
-------------
ÅÆ
-
+
2 OH (aq)
+
2 H (aq)
1.0x10-4
-2S
1.0x10-4
ÅÆ
+
Initial
change
equil
2+
Cu (aq)
-
+
2 OH
Ksp= 2.2x10-20
(aq)
2 H2O (l)
2 H2O (l)
-------------
2+
+
Cu (aq)
0
+S
S
K = (1/Kw)2 =
1.0x10+28
Kov = Ksp/Kw2
Kov = Ksp/Kw2 = 2.2x10+8 = [Cu2+]/[H+]2
and since [H+] is buffered to 1.0x10-4 M we find
[Cu2+] = [H+]2*Kov = 2.2 M = S still the same answer!
2.
Calculate the molar solubility of AgCl [Ksp = 1.8 x 10-10] in:
a. distilled water and b. 0.100 M NH3 [Kf = 1.6 x 10+7 for Ag(NH3)2+].
2 (a)
AgCl(s)
ÅÆ
At equil
+
-
+
+
Ag (aq)
S
Ksp= 1.8x10-10 = [Ag ][Cl ] = (S)(S) = (S)
giving the answer as
+
S = [Ag ] = [Cl ] = 1.3x10-5 M
2
123
-
Cl (aq)
S
Ksp= 1.8x10-10
2 (b)
2 NH3 (aq)
+
2 NH3 (aq)
0.20
-2S
0.20 – 2S
+
AgCl(s)
ÅÆ
+
ÅÆ
Ag (aq)
AgCl(s)
-------------
Initial
change
equil
ÅÆ
+
Ag (aq)
+
Ag(NH3)2 (aq)
+
Ag(NH3)2 (aq)
0
+S
S
-
+
Cl (aq)
+
Cl (aq)
0
+S
S
Ksp= 1.8x10-10
Kf = 1.6x10+7
-
Kov = KspKf
+
Kov = Ksp*Kf = 2.9x10-3 = [Ag(NH3)2 ][Cl-]/[NH3]2 = S2/(0.100-2s)2
We can simply take the square root of both sides of the equation to find
+
S = [Ag(NH3)2 ] = [Cl-] = 4.9x10-3 M higher solubility because the complex formation reaction
between silver ions and ammonia shifts the solubility equilibria in the forward direction
3.
Calculate the molar solubility of Ag2S [Ksp = 6 x 10-51] in:
a. distilled water and b. saturated solution of H2S in a pH 6.00 buffer.
3 (a)
Ag2S(s)
ÅÆ
+
At equil
+ 2
2-
2
2-
+
2 Ag (aq)
2S
Ksp= 6.3x10-50
S (aq)
S
3
Ksp= 6.3x10-50 = [Ag ] [S ] = (2S) (S) = 4S
giving the answer as
+
2S = ½ [Ag ] = [S ] = 2.5x10-17 M
but this neglects the hydrolysis of the basic sulfide ion!!
with hydrolysis the story can be made more complete . .. .
Ag2S(s)
H2O (l)
+
H2O (l)
+
2 H2O (l)
+
2-
S (aq)
1HS (aq)
Ag2S(s)
ÅÆ
ÅÆ
ÅÆ
ÅÆ
+
2 Ag (aq)
HS (aq)
+
H2S (aq)
2 Ag+(aq) + H2S (aq)
----Initial
----0
----change
----+2S
----equil
----2S
-14
-7
-19
note: Kw = 1.0x10 , Ka1 = 1.0x10 and Ka2 = 1.0x10
0
+S
S
2-
+
S (aq)
OH (aq)
Ksp= 6.3x10-50
Kb1 = Kw/Ka2
+
OH (aq)
Kb2 = Kw/Ka1
+
-
-
2 OH (aq)
~0
+2S
~0 + 2S
Kov =
Ksp*Kb1*Kb2
Kov =6.3x10-52
Kov = Ksp*Kb1*Kb2 = 6.3x10-52
This is still a very small K indicating that the forward reaction is not very important! As such, I will
assume that the reaction produces very little products, but there will be hydroxide present from the
autoprotolysis reaction! Air exposed water has a pH of about 5.6 (pOH = 8.40, [OH ]=4.0x10-9 M)
124
Now if we plug in that hydroxide concentration and solve we find:
+ 2
-
+ 2
= [Ag ] [H2S][OH ]2 ≈ [Ag ] [H2S][ 4.0x10-9 ]2 = 4S3(4.0x10-9)2
+
so ½ [Ag ] = 2.1x10-12 M
Higher solubility due to the reaction of some of the sulfide ion with water.
[H2S] = 0.10 M in a saturated solution at room temperature . . . .let’s solve this problem using a
bit of reasoning . . . let’s figure out how much sulfide will be present in our solution and see how
this relates to the solubility equilibria.
3 (b)
ÅÆ
H2S (aq)
+
H2O (l)
HS (aq)
H2S (aq)
+
H2O (l)
+
2 H2O (l)
1-
ÅÆ
2-
ÅÆ
-
HS (aq)
2S (aq)
2-
S (aq)
+
+
+
H3O (aq)
+
H3O (aq)
+
2 H3O (aq)
+
-7
Ka1 = 1.0x10
-19
Ka1 = 1.0x10
Kov = Ka1*Ka2
+ 2
Kov = Ka1*Ka2 = 1.0x10-26 = [S ][H3O ] /[H2S]
In a saturated solution of H2S [0.10 M] adjusted to pH 6.00
2-
2
1.0x10-26 = [S ][1.0x10-6] /[0.10]
giving
2-
[S ] = 1.0x10-15 M
This might seem low, but remember Ksp is really tiny for silver sulfide.
Ag2S(s)
ÅÆ
+
At equil
+ 2
2-
+
2 Ag (aq)
2S
2-
S (aq)
1.0x10-15 M
Ksp= 6.3x10-50
2
Ksp= 6.3x10-50 = [Ag ] [S ] ≈ (2S) (1.0x10-15 M)
giving the answer as
+
S = ½ [Ag ] = 4.0x10-18 M
so neglecting the additional sulfide ion is not a bad bet, but you could solve the equation exactly.
4.
Ammonium chloride solutions are slightly acidic. Find the solubility of Mg(OH)2 in
units of moles per liter in a solution that is 0.20M NH4Cl and compare it to the solubility
in water. HINT: find K for the following reaction:
Mg(OH)2 (s) + 2 NH4+ (s) → Mg2+ (aq) + 2 NH3 (aq) + 2 H2O (l)
This problem involves more of the same . .. find the overall reaction and solve the ICE table.
125
Mg(OH)2(s)
-
2 OH (aq)
Mg(OH)2(s)
+
2
+
2
-------------
Initial
change
equil
+
NH4 (aq)
+
NH4 (aq)
ÅÆ
ÅÆ
ÅÆ
0.20 M
-2S
0.20 – 2S
2+
-
+
2 H2O (l)
+
Mg (aq)
2 NH3 (aq)
+
2 H2O (l)
2+
2 NH3 (aq) + Mg (aq)
0
+2S
2S
2
2+
2 OH (aq)
+ 2
0
+S
S
2
~0
+2S
~0 + 2S
Ksp=6x10-12
2
K = (1/Kb )
Kov
=Ksp*(1/Kb)2
Kov =1.9x10-2
2
Kov = Ksp*(1/Kb)2= 1.9x10-2 = [Mg ][NH3] /[NH4 ] = [S][2S] /[0.20 – 2S]
This solves, with either “iterative solution” or one of those fancy calculators that will solve
equations for us to yield
S = [Mg2+] = 4.1 x10-2 M
so we would not have been safe in neglecting S when compared to 0.20 M
btw, this also yields [NH4+] = 0.12 M at equilibrium
126