Question

Consider the dissolution of AB(s)AB(s):

AB(s)⇌A+(aq)+B−(aq)AB(s)⇌A+(aq)+B−(aq)

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

The generic metal hydroxide M(OH)2M(OH)2 has KspKspK_sp = 8.45×10⁻¹². (NOTE: In this particular problem, because of the magnitude of the KspKsp and the stoichiometry of the compound, the contribution of OH−OH− from water can be ignored. However, this may not always be the case.)

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

What is the solubility of M(OH)2M(OH)2 in a 0.202 MM solution of M(NO3)2M(NO3)2?

Answer 1

Solution to the given question is shown in below image

Therefore solubility of M(OH)2 in pure water is s=1.2831*10-4 and solubility of M(OH)2 in 0.202M solution of M(NO3)2 is s=3.234*10-6.