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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 = 1.05×10−18. (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.)What is the solubility of M(OH)2 in pure water?

User JHBonarius
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Answer:

S = 6.40 × 10⁻⁷ M

Step-by-step explanation:

In order to calculate the solubility (S) of M(OH)₂ in pure water we will use an ICE Chart. We recognize 3 stages: Initial, Change and Equilibrium, and we complete each row with the concentration or change in concentration.

M(OH)₂(s) ⇄ M²⁺(aq) + 2 OH⁻(aq)

I 0 0

C +S +2S

E S 2S

The solubility product (Kps) is:

Kps = 1.05 × 10⁻¹⁸ = [M²⁺].[OH⁻]²=S.(2S)²

1.05 × 10⁻¹⁸ = 4S³

S = 6.40 × 10⁻⁷ M

User Adrian Coman
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