Question

a. What is the molar solublity of lead(II) chloride in water?

b. Should we be concerned about the chloride ion acting as a base? Why or why not?

Answer 1

a. Molar solubility of Lead (II) Chloride in water:

In water, Lead (II) Chloride dissociates as:

PbCl2 (s) -----------> Pb^2+ (aq) + 2Cl^- (aq)

s. 2s

If solubility of Pb^2+ = s and therefore solubility of Cl^- = 2s (due to 2Cl^- ions).

From the definition of Ksp

Ksp = [Pb^2+] [Cl^-]^2

Ksp of PbCl2 = 1.6* 10^-5, by putting the values we get,

1.6*10^-5 = (s)(2s)^2

=> 1.6* 10^-5 = 4s^3

=> s = 0.0159M

Therefore, the molar solubility of Lead (II) Chloride in water is 0.0159M

b. The Chloride ion acts as base, it will abstract the H^+ ions from the water as,

Cl^- (aq) + H^+ (aq) -------> HCl (aq)

So, the acid HCl may be formed, but being the strong acid, it again dissociates completely in water to form the H^+ and Cl^- ions as

HCl(aq) ----------> H^+ (aq) + Cl^- (aq) Ka (very high)

so, we get back the Cl^- ions in water.

Therefore, we should not concerned about the Cl^- ion acting as a base.