Question

The solubility of BaCrO₄ in water at 298K is 1.45 x 10^-5 . Calculate the solubility product of BaCrO₄.

Answer 1

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$$\begin{gathered} \mathrm{Solubility},\mathrm{s}\mathrm{of}{\mathrm{BaCrO}}_4\mathrm{is}1.45\times {10}^{-5} \\ \mathrm{The}\mathrm{equilibrium}\mathrm{is} \\ {\mathrm{BaCrO}}_4\leftrightarrow {\mathrm{Ba}}^{2+}+{{\mathrm{CrO}}_4}^{2-} \\ \mathrm{s}\mathrm{s} \\ \mathrm{So},{\mathrm{K}}_{\mathrm{sp}}=\left[{\mathrm{Ba}}^{2+}\right]\left[{{\mathrm{CrO}}_4}^{2-}\right]=\mathrm{s}\times \mathrm{s}={\mathrm{s}}^2 \\ \mathrm{or},{\mathrm{K}}_{\mathrm{sp}}=(1.45\times {10}^{-5}{)}^2=2.1025\times {10}^{-10} \end{gathered}$$