Question

± Common-Ion Effect on Solubility for Lead Thiocyanate

Lead thiocyanate, Pb(SCN)2, has a Ksp value of 2.00×10−5.

Part A

Calculate the molar solubility of lead thiocyanate in pure water. The molar solubility is the maximum amount of lead thiocyanate the solution can hold.

Express your answer with the appropriate units.

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Common-Ion Effect

Consider the dissolution of AB(s) :

AB(s)⇌A+(aq)+B−(aq)

Le Châtelier's principle tells us that an increase in either [A+] or [B−] will shift this equilibrium to the left, reducing the solubility of AB. In other words, AB is more soluble in pure water than in a solution that already contains A+ or B− ions. This is an example of the common-ion effect.

Part B

Calculate the molar solubility of lead thiocyanate in 0.800 M KSCN.

Express your answer with the appropriate units.

Answer 1

A)

At equilibrium:

Pb(SCN)2 <----> Pb2+ + 2 SCN-

   s 2s

Ksp = [Pb2+][SCN-]^2

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

2*10^-5= 4(s)^3

s = 1.71*10^-2 M

Answer: 1.71*10^-2 M

B)

KSCN here is Strong electrolyte

It will dissociate completely to give [SCN-] = 0.8 M

At equilibrium:

Pb(SCN)2 <----> Pb2+ + 2 SCN-

   s 0.8 + 2s

Ksp = [Pb2+][SCN-]^2

2*10^-5=(s)*(0.8+ 2s)^2

Since Ksp is small, s can be ignored as compared to 0.8

Above expression thus becomes:

2*10^-5=(s)*(0.8)^2

2*10^-5= (s) * 0.64

s = 3.125*10^-5 M

Answer: 3.13*10^-5 M