Question

The molar solubility of PbCl2 in 0.20 M Pb(NO3)2 solution is:
(a) 1.7 x 10-4 M
(b) 9.2 x 10-3 M
(c) 1.7 x 10-5 M
(d) 4.6 x 10-3 M
(e) 8.5 x 10-5 M

Please show all work

Answer 1

Sol :-

Let molar solubility PbCl₂ in 0.20 M Pb(NO₃)₂ = S mol/L

Partial dissociation of PbCl₂ is :

PbCl₂ (s) <------------------> Pb²⁺ (aq).................+.......................2Cl^- (aq)

..............................................S mol/L................................................2S mol/L

Complete dissociation of Pb(NO₃)₂ is :

Pb(NO₃)₂ (aq) <-------------------> Pb²⁺ (aq) ..................+...................2NO₃^- (aq)

0.20 M.............................................0.20 M..............................................2 x 0.20 M

Because Pb²⁺ is a common ion therefore it total concentration in solution = [Pb²⁺]_Total = (S+0.20) mol/L

Solubility product (K_sp) is equal to the product of the molar concentration of products raise to the power of stoichiometric coefficient at equilibrium stage of the reaction.

Expression of K_sp is :

K_sp = [Pb²⁺].[Cl^-]²_Total

1.7 x 10^-5 = (S).(2S+0.20)²

1.7 x 10^-5 = S (S² + 0.8 S + 0.04 )

S³ + 0.8S² + 0.04S - 1.7 x 10^-5 = 0

On solving this cubic equation :

S = 4.6 x 10^-3 M

Hence, option (d) is the correct answer.