1. Study pp. 765 – 768 (up to section 16.2) pp. 774 – 779
Aim # 15: What are the factors that affect the solubility of slightly soluble salts?
H.W. # 15
Study pp. 765 – 768 (up to section 16.2)
pp. 774 – 779
Ans. ques. p. 782 # 37, 41, 45, 47, 63, 73

2. I The common Ion Effect – the solubility of a slightly soluble salt is decreased by the presence of a second solute that supplies a common ion.
e.g. 1. NaOH(aq) or BaCl2(aq) decreases the solubility of Ba(OH)2.
2. Na2SO4(aq) or CaCl2(aq) will decrease the solubility of CaSO4.
CaSO4(s) Ca2+(aq) + SO42-(aq) Adding Ca2+(aq) or SO42-(aq) ion shifts the equilibrium to the left.
Problem: What is the molar solubility of PbI2 in 0.10 M KI(aq)?

3. Ans: PbI2(s) Pb2+(aq) + 2I-(aq)
init. Conc M M
change x M x M
final conc x M ( x) M
Ksp = [Pb2+][I-]2= (x)( x)2 = 7.1 x 10-9
assume x << 0.10 M
then (x)(.10)2 ≈ 7.1 x x = 7.1 x 10-7 M
[Pb2+] = [PbI2] = 7.1 x 10-7 M
Since 7.1 x 10-7 << 0.10, our assumption was justified.

4. Without KI, Ksp = (x)(2x)2 = 7.1 x 10-9
x = [PbI2] = 1.2 x 10-3 M
II The effect of pH on solubility – the solubility of slightly soluble salts containing basic anions increases as [H+] increases (pH is lowered).
Examples of basic anions are OH-, CO32-, PO43-, CN-, S2-
Salts of such ions would include Ca(OH)2, MgCO3, Ba3(PO4)2, AgCN, and NiS2.

5. For (1) Ca(OH)2(s) Ca2+(aq) + 2OH-(aq) Ksp = 5.5 x 10-6
Adding H+ ions drives (2) H+(aq) + OH-(aq) H2O(ℓ) to the right,
AND drives equation (1) to the right as well.
III The effect of complex ion formation on solubility
The solubility of slightly soluble salts increases in the presence of suitable Lewis bases such as NH3, CN-, or OH- if the metal forms a complex with the base.
e.g. iron (II) carbonate, FeCO3, has a Ksp = 3.2 x 10-11
(1) FeCO3(s) Fe2+(aq) + CO32-(aq)

6. Adding KCN to the solution leads to complex ion formation.
(2) Fe2+(aq) + 6CN-(aq) Fe(CN)64-(aq)
Kf = [Fe(CN)64-] = 1.0 x [Fe2+][CN-] where Kf is the formation constant
How does reaction (2) affect reaction (1) and the solubility of FeCO3?
Problem: Calculate the molar solubility of AgI in 3.0 M NH The overall formation constant for Ag(NH3)2+ is x Ksp for AgI in pure water is 1.5 x

7. Ans: AgI(s) Ag+(aq) + I-(aq) Ksp = 1.5 x 10-16
AgI(s) + 2NH3(aq) Ag(NH3)2+(aq) + I-(aq)
Kf = 1.7 x ͘
AgI(s) + 2NH3(aq) Ag(NH3)2+(aq) + I-(aq) K = Ksp x Kf = 2.6 x 10-9
Let x mol/L = the amount of AgI that dissolves to reach equlibrium.
Initial M
Equil – 2x x x
K = 2.6 x 10-9 = [Ag(NH3)2+][I-] [NH3]2

8. 2.6 x 10-9 = (x)(x) ≈ x2 ͘ (3.0 – 2x) (3.0)2
x2 = 2.3 x 10-8
[AgI] = x = 1.5 x 10-4 M
How does this compare to the solubility of AgI in pure water?
In pure water,
Ksp = 1.5 x = [Ag+][I-] = (x)(x) x2 = 1.5 x x = [AgI] = 1.2 x 10-8 M
The solubility increases in 3.0 M NH3.