Question

± Common-Ion Effect on Solubility for Lead Thiocyanate Lead thiocyanate, Pb(SCN ) 2 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 1.00 M KSCN . Express your answer with the appropriate units. , has a K sp 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.

Answer 1

PbSCN (s) < ----------> Pb²⁺ (aq.) + 2 SCN^- (aq.)

S                                S                2S    Mol/L of solubility

Ksp = [Pb²⁺][SCN^-]²

In pure water,

2.00 * 10^-5 = (S)(2S)²

2.00 * 10^-5 = 4S³

S = (2.00 *10^-5 / 4 ) ^1/3

S = Solubility of lead thiocyanate in pure water = 2.236 * 10^-3 M

In KSCN solution,

KSCN -------------> K⁺ (aq.) + SCN^- (aq.)

1.00                    1.00         1.00M

[SCN^-] = 1.00 M Since KSCN is completely soluble assume the [SCN^-] is approximately equals to conc. of SCN^- coming from KSCN.

therefore,

Ksp = [Pb²⁺]SCN^-]²

2.00 * 10^-5 = (S)(1.00)²

S = solubility of lead thiocyanate in 1.00 M KSCN solution = 2.00 * 10^-5 M