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Complete the following solubility constant expression for Mg(OH)2?

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Final answer:

The solubility product constant (Ksp) for Mg(OH)2 is found by applying the expression Ksp = [Mg^2+][OH^-]^2, yielding a Ksp value of 9.0 × 10^-12 when provided with the Mg^2+ concentration in a saturated solution.

Step-by-step explanation:

To complete the solubility constant expression for Mg(OH)2, we must consider its dissociation in water:

Mg(OH)2 (s) → Mg^2+ (aq) + 2OH^- (aq)

The solubility product constant, Ksp, expression for this dissociation is:

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

Given that the concentration of Mg^2+ is 1.31 × 10^-4 M in a saturated solution, and assuming the stoichiometry of the dissociation, the concentration of OH^- would be twice that of Mg^2+, which is 2 × (1.31 × 10^-4 M) = 2.62 × 10^-4 M.

Substituting these values into the Ksp expression gives us:

Ksp = (1.31 × 10^-4) × (2.62 × 10^-4)^2

Ksp = 1.31 × 10^-4 × 6.8644 × 10^-8

Ksp = 8.998644 × 10^-12

After calculating, we round off to two significant figures, to match the precision of the given concentration of Mg^2+:

Ksp = 9.0 × 10^-12

answered
User Jordi Xucla
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