Question

The K_sp of Al(OH)₃ is 2.7*10^-11. Calculate the solubility in gL^-1 and also find out the pH of this solution.

Answer 1

Dear Student,

$$\begin{gathered} \mathrm{Al}(\mathrm{OH}{)}_3(\mathrm{aq})\to {\mathrm{Al}}^{3+}(\mathrm{aq})+3{\mathrm{OH}}^{-}(\mathrm{aq}) \\ \mathrm{s}3\mathrm{s} \\ {\mathrm{K}}_{\mathrm{sp}}=[{\mathrm{Al}}^{3+}\left]\right[{\mathrm{OH}}^{-}{]}^3=\left(\mathrm{s}\right)(3\mathrm{s}{)}^3=27{\mathrm{s}}^4 \\ \mathrm{or},2.7\times {10}^{-11}=27{\mathrm{s}}^4 \\ \mathrm{or},27{\mathrm{s}}^4=27\times {10}^{-12} \\ {\mathrm{s}}^4={10}^{-12} \\ \mathrm{or},\mathrm{s}=({10}^{-12}{)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$4$}\right.}={10}^{-3}\mathrm{M} \\ \mathrm{or},\left[{\mathrm{OH}}^{-}\right]=3\mathrm{s}=3\times {10}^{-3}\mathrm{M} \\ \mathrm{or},\mathrm{pOH}=-\log \left[{\mathrm{OH}}^{-}\right]=-\log (3\times {10}^{-3})=(-\log 3)-(-3\log 10)=2.52 \\ \mathrm{Now},\mathrm{pH}=14-\mathrm{pOH} \\ \mathrm{or},\mathrm{pH}=14-2.52=11.47 \end{gathered}$$