Question

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 = 8.45×10−12. (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. View Available Hint(s) nothing nothing 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. View Available Hint(s) nothing nothing

Answer 1

Answer:

A. 0.000128 M is the solubility of M(OH)2 in pure water.

B. $3.23\times 10^{-6} M$ is the solubility of $M(OH)_2$ in a 0.202 M solution of $M(NO_3)_2$.

Explanation:

A

Solubility product of generic metal hydroxide = $K_{sp}=8.45\times 10^{-12}$

$M(OH)_2\rightleftharpoons M^{2+}+2OH^-$

                      S         2S

The expression of a solubility product is given by :

$K_{sp}=[M^{2+}][OH^-]^2$

$K_{sp}=S\times (2S)^2=4S^3$

$8.45\times 10^{-12}=4S^3$

Solving for S:

$S=0.000128 M$

0.000128 M is the solubility of M(OH)2 in pure water

B

Concentration of $M(NO_3)_2$ = 0.202 M

Solubility product of generic metal hydroxide = $K_{sp}=8.45\times 10^{-12}$

$M(OH)_2\rightleftharpoons M^{2+}+2OH^-$

                   S          2S

So, $[M^{2+}]=0.202 M+S$

The expression of a solubility product is given by :

$K_{sp}=[M^{2+}][OH^-]^2$

$8.45\times 10^{-12}=(0.202 M+S)(2S)^2$

Solving for S:

$S=3.23\times 10^{-6} M$

$3.23\times 10^{-6} M$ is the solubility of $M(OH)_2$ in a 0.202 M solution of $M(NO_3)_2$.