At 22 °C, an excess amount of a generic metal hydroxide, M(OH)2, is mixed with pure water. The resulting equilibrium solution has a pH of 10.22. What is the Ksp of the salt at 22 °C?

2.29x10⁻¹² is Ksp of the salt

Explanation:

The Ksp of the metal hydroxide is:

M(OH)₂(s) ⇄ M²⁺ + 2OH⁻

Ksp = [M²⁺] [OH⁻]²

As you can see in the reaction, 2 moles of OH⁻ are produced per mole of M²⁺. It is possible to find [OH⁻] with pH, thus:

pOH = 14- pH

pOH = 14 - 10.22

pOH = 3.78

pOH = -log[OH⁻]

1.66x10⁻⁴ = [OH⁻]

And [M²⁺] is the half of [OH⁻], [M²⁺] = 8.30x10⁻⁵

Replacing in Ksp formula:

Ksp = [8.30x10⁻⁵] [1.66x10⁻⁴]²

Ksp = 2.29x10⁻¹² is Ksp of the salt

the answer is b. dilution

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