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?

Question

Chemistry

Ryder

28 August, 15:27

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?

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Answers (1)

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Presley Powell · Answer 1

S = 6.40 * 10⁻⁷ M

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