Question

What is the Ksp for the following and arrange these in order of their increasing molar solubility. Show all work.

a. PbCrO₄
b. Zn(OH)₂
c. Ag₂CO3
d. Ag₃PO₄

_{Thank you! It will be greatly appreciated}

Answer 1

(a)K_sp of PbCrO₄ = 2.3 * 10^-13

Suppose solubility of PbCrO4 = S mole /L

Consider an ionization reaction of PbCrO₄ is

PbCrO_{4 (aq)} ------> Pb²⁺_(aq) + CrO₄^2-_(aq)

1.0mole of PbCrO4 produces 1.0 mole of Pb²⁺ (aq) and 1.0 mole of CrO4^2-(aq), hence solubility of Pb²⁺ (aq) = solubilty of CrO4^2-(aq) = Solubility of PbCrO₄(aq) = S

Ksp = [Pb2+] [CrO42-] = (S) (S)

S² = 2.3 * 10^-13 = 23 * 10^-14

S = √ (23*10^-14) = 4.796 * 10^-7

Solubility of PbCrO₄ = 4.796 * 10^-7 mol/L

(b) Ksp of Zn (OH) 2 = 4.5 x 10^-17

Similarly,

Consider an ionization reaction of Zn(OH)2 is        Zn(OH)2(aq) -----à Zn2+(aq) + 2 OH- (aq)

Ksp = [Zn²⁺] [OH^-]² = (S)(2S)² = 4S³

4S3 = Ksp = 4.5 * 10-17 = 45 * 10-18

S3 = (¼) 45 * 10-18 = 11.25 * 10-18

S = Cubic root (11.25 * 10-18) = 2.24 * 10-6

Solubility of Zn(OH)2 = 2.24 * 10-6 mole /L