Question

At 22℃ an excess amount of a generic metal hydroxide M(OH)2 is mixed with pure water. The resulting equilibrium solution has a pH of 10.26. What is the Ksp of the salt at 22℃?

Answer 1

\(\mathrm{M}(\mathrm{OH}) 2(\mathrm{~s})<->\mathrm{M}^{2+}(\mathrm{aq})+2 \mathrm{OH}^{-}(\mathrm{aq})\)

\(\mathrm{Ksp}=\left[\mathrm{M}^{2+}\right]\left[\mathrm{OH}^{-}\right]^{2}\)

1) Use the \(\mathrm{pH}\) to get the \(\mathrm{pOH}\)

\(14-10.26=3.74\)

2) Use the \(\mathrm{pOH}\) to get the \(\left[\mathrm{OH}^{-}\right]\)

\(\left[\mathrm{OH}^{-}\right]=10^{-} \mathrm{pOH}=10^{\wedge}-3.74=1.82 \times 10^{\wedge}-4\)

3) \(\left[\mathrm{M}^{2+}\right]\) is half the value of \(\left[\mathrm{OH}^{-}\right]\), therefore:

\(\left[\mathrm{M}^{2+}\right]=1.82 \times 10^{\wedge}-4\) divided by \(2=9.10 \times 10^{n}-5\)

4) Put values into Ksp expression

\(\mathrm{Ksp}=\left(9.10 \times 10^{n}-5\right)\left(1.82 \times 10^{n}-4\right)^{2}\)

\(\mathrm{Ksp}=3.01 \times 10^{\wedge}-12\)