Chemistry - Ionic Equilibrium Question with Solution | TestHub

Solubility in the buffer $=\frac{2.88\times 10^{-3}}{90}=3.2\times 10^{-5}\mathrm{M}$.

pH = 11 gives $\left[{\mathrm{OH}}^{-}\right]=10^{-3}\mathrm{M}$.

$K_{sp}=3.2\times 10^{-5}\times (10^{-3}{)}^2=3.2\times 10^{-11}$.

In pure water $K_{sp}=4s^3$, so $s^3=8\times 10^{-12}$ and $s=2.0\times 10^{-4}\mathrm{M}$.

Answer: (B).