School of Chemistry, University of KwaZulu-Natal, Westville Campus, Durban
CHEM 120R
Tutorial 2: Solubility
Solubility
(Please refer to a table for the solubility product constants)
1.
Calculate the molar solubility of silver bromide in water at 20 ˚C. The solubility product
at this temperature is 5.0 × 10-13.
Silver bromide dissolves and dissociates according to the reaction
AgBr(s) ⇌ Ag+ + BrFor which
Ksp = [Ag+] [Br-]
The solubility, or concentration of dissolved silver bromide, is equal to the concentration of free
silver ion:
Solubility = [AgBr]dissolved = [Ag+]
According to the reaction stoichiometry, silver ion and bromide ion are formed in equal
quantities. Since there is no other source of these ions and since they do not undergo reactions
with any other substances, their concentrations are equal.
[Br-] = [Ag+]
Substituting in the equilibrium expression gives
Ksp = [Ag+]2
[Ag+] = solubility =
K sp  5.0 x 10 13  7.1 x 10 7 M
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2.
The solubility of lead (II) chloride (PbCl 2 ) is 1.6 102 M . What is the K
sp
of
PbCl2 ?
The dissolution-precipitation equilibrium is given by
PbCl2(s) ⇌ Pb2+ + 2Cl ˉ
For which
Ksp = [Pb2+ ] [Cl ˉ]2
From the reaction stoichiometry,
[Cl ˉ ] = 2[Pb2+]
Substituting in the equilibrium expression gives
Ksp = [Pb2+] (2[Pb2+])2 = 4[Pb2+]3
= 4 × (0.016)3
= 1.6 × 105
3.
What must be the concentration of Ag+ to just start precipitation of AgCl in a 1.0x10-3 M
solution.
AgCl (Ksp = 1.8  10-7) just starts to precipitate when the ion product just exceeds Ksp
Ksp = [Ag+][Cl-]
1.8 x10-10 = [Ag+] × 1.0 × 10-3
[Ag+] = 1.8 × 10-7 M
4.
Calculate the solubility, in g/100 mL, of calcium iodate in water at 20 ˚C.
The Ksp for Ca(IO3)2 is 7.1 × 10-7.
Calcium iodate dissolves and dissociates according to the reaction
Ca(IO3)2(s) ⇌ Ca2+ + 2IO3ˉ
Solubility Equilibria for which
Ksp = [Ca2] [IO3ˉ]2
2
The solubility, or concentration of dissolved calcium iodate, is equal to the concentration of
calcium ion:
Solubility = [Ca(IO3)2]dissolved = [Ca2+]
According to the reaction stoichiometry, twice as much iodate as calcium is formed on
dissolution.
Thus
[IO3ˉ] = 2[Ca2+]
Substituting in the equilibrium expression gives
Ksp = [Ca2+](2[Ca2+])2 = 4[Ca2+]3
[Ca2+] = solubility =
3
K sp
4

3
7.1 x 107
 5.6 x 103 M
4
Converting to the desired concentration units, we obtain
Solubility = (5.6 × 10-3 mol/L)(389.9g/mol) = 2.2g/L
= 0.22g/100mL
5.
Calculate the molar solubility of lead iodide in (a) water and (b) in 0.200 M sodium
iodide solution. The Ksp for PbI2 is 7.9 × 10-9.
(a)
The dissolution-precipitation equilibrium is given by
PbI2(s) ⇌ Pb2+ + 2I ˉ
For which
Ksp = [Pb2+ ] [I ˉ]2
The solubility, or concentration of dissolved PbI2, is equal to the concentration of Pb2+ or
one-half the concentration of I ˉ ):
solubility = [PbI2]dissolved = [Pb2+]
From the reaction stoichiometry,
[I ˉ ] = 2[Pb2+]
Substituting in the equilibrium expression gives
Ksp = [Pb2+] (2[Pb2+])2 = 4[Pb2+]3
3
[Pb2+] = solubility =
(b)
3
K sp
4

3
7.9 x 10 9
 1.3  10 3 M
4
The same equilibrium expression holds:
Ksp = [Pb2+] [I ˉ ]2
There are now two sources of iodide: the NaI and the PbI2. The amount of iodide
coming from PbI2 is small compared to that from the NaI. Thus
[I ˉ] = Cnal + 2[Pb2+] ≈ Cnal = 0.200 M
Then
K sp 7.9 x 10 9
 2.0  10 7 M
[Pb2+] = solubility =  2 
2
[I ]
(0.200)
Note that the solubility has decreased markedly (four orders of magnitude) upon the addition of
an excess of Iˉ.
6.
The molar solubility of silver sulfate is 1.5 × 10-2 M. Calculate the solubility product of
this salt.
Ag2SO4(s) ⇌ 2Ag+ + SO42Ksp = [Ag+]2 [SO42- ]
[Ag+] = 2 × 1.5 × 10-2 M = 3.0 × 10-2 M
SO42- = 1.5 × 10-2 M
Substituting in the equilibrium expression gives
Ksp = [3.0 × 10-2]2 (1.5 × 10-2)= 1.4 × 10-5
7.
Calculate the solubility of Zn(OH)2 (Ksp = 3.0  10-16) in pure water.
[Zn2+] [OH-]2 = 3.0  10-16
as [OH-] = 2 [Zn2+] substituting in above equation we get
[Zn2+] (2 [Zn2+])2 = 3.0  10-16
[Zn2+]3 = 3.0  10-16/ 4 = 7.5  10-17
[Zn(OH)2] = [Zn2+] = (7.5  10-17)1/3
= 4.2  10-6 mol dm-3
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8. At 25 ºC the molar solubility of Mg(OH)2 in pure water is 1.4 × 10-4 M . Calculate
its molar solubility in a buffer medium whose pH is 12.00.
Mg(OH)2(s) ⇌
pH
Mg2+(aq) + 2 OH−(aq)
=
12.00
pOH =
14.00 - 12.00 = 2.00
[OH−] =
1.0 x 10-2M
Therefore
Ksp = [Mg2+][OH−]2 = 1.6 x 10-12
[Mg2] =
1.6 x 10-12
(1.0 x 10-2)2
= 1.6 x 10-8 M
High pH- means that there is more base around in the solution. Therefore we have a
common ion in solution and the solubility of Mg(OH)2 is decreased.
9.
The solubility product for Mg(OH)2 is 7.1 × 10-12. If you had a 0.0500 M Mg2+ solution
and adjusted the pH to 10 would you expect Mg(OH)2 to precipitate ? Justify your
answer.
Mg(OH)2 ⇌ Mg2+ + 2OHKsp ⇌ [Mg2+] [OH-]2 = 7.1 × 10-12
Q=0.0500 × (1.0 × 10-4)2 = 7.1 × 10-12
Q=0.0500 × (1.0 × 10-4)2 = 5.0 × 10-10
Q> Ksp Mg(OH)2 will precipitate.
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10.
Calculate the molar solubility of PbCl2 given the relevant Ksp value is 1.7  10-5.
Answer:
We are being asked to calculate [PbCl2].
PbCl2(s) ⇌ Pb2+(aq) + 2Cl-
Therefore
Ksp = [Pb2+] [Cl-]2
[Cl-] = 2 [Pb2+], substitute into the Ksp equation
Ksp = [Pb2+](2 [Pb2+])2
= 4 [Pb2+]3
 4 [Pb2+]3 = 1.7  10-5
[Pb2+] = 3 4.25  10 6
= 1.6  10-3 mol dm-3
 [PbCl2] = 1.6  10-3 mol dm-3
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