Question

± Common-Ion Effect on Solubility for a Metal Hydroxide 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. The generic metal hydroxide M(OH)2 has Ksp = 4.85×10−12. (NOTE: In this particular problem, because of the magnitude of the Ksp and the stoichiometry of the compound, the contribution of OH− from water can be ignored. However, this may not always be the case.) Part A What is the solubility of M(OH)2 in pure water? Express your answer with the appropriate units. Hints 1.135•10−6 M SubmitMy AnswersGive Up Incorrect; Try Again; 4 attempts remaining Part B What is the solubility of M(OH)2 in a 0.202 M solution of M(NO3)2? Express your answer with the appropriate units. Hints 1.13•10−9 M SubmitMy AnswersGive Up Incorrect; Try Again; 5 attempts remaining

Answer 1

Ksp = 4.85×10^−12

part A)

M(OH)2 ------------------------> Mg+2 + 2OH-

                                          S            2 S

Ksp = [Mg+2] [OH-]^2

Ksp = (S)(2S)^2 = 4 S^3

4.85 x 10^-12 = 4 S^3

S = 1.07 x 10^-4 M

solubility = 1.07 x 10^-4 M

part B)

M+2 = 0.202 M

4.85 x 10^-12 = (0.202) (2S)^2

S = 2.45 x 10^-6 M

solubility = 2.45 x 10^-6 M