Question

Suppose that two hydroxides, MOH and
M′(OH)2, both have a Ksp of 2.15 × 10−12
and that initially both cations are present in
a solution at concentrations of 0.001 mol/L.
At what pH will the hydroxide precipitate
when solid NaOH is added to the solution?

Answer 1

This problem is providing us with the solubility product of two hydroxides, which can be used to calculate the pH at which they will precipitate. At the end, the answer turns out to be 9.67.

Solubility product:

In chemistry, we use solubility products in order to quantify the amount of precipitate a solid will be leftover after dissolving it. Thus, we can just write the equilibrium expressions and solve for x, molar solubility, for each hydroxide as follows:

$Ksp=[M^+][OH ^-]\\\\2.15x10^{-12}=(0.001+x)(x)\\\\x=2.15x10 ^{-9}M\\\\Ksp=[M^2^+][OH ^-]^2\\\\2.15x10^{-12}=(0.001+x)(x)^2\\\\x=4.535x10 ^{-5}M$

Then, since both x's contribute to the concentration of hydroxide ions, but just the second one is predominant, we add them together to get the total:

$[OH^-]=4.535x10^{-5}M+2.15x10^{-9}M=4.535x10^{-5}M$

Finally, we calculate the pOH and pH with:

$pOH=-log(4.535x10^{-5})=4.343\\\\pH=14-pOH=9.67$

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