Question

The solubility product of $BaS{O}_4$ and $BaCr{O}_4$ at $25$ are $1\times {10}^{-10}$ and $1\times {10}^{-10}$ respectively. Calculate the solubilities of $BaS{O}_4$ and $Ba{CrO}_4$.

Answer 1

${BaSO}_4⟶{Ba}^{2+}+{SO}_4^{2-}$

$K_{sp}=\left[{Ba}^{2+}\right]\left[{SO}_4^{2-}\right]=s^2$

$s=\sqrt{K_{sp}}=\sqrt{1\times {10}^{-10}}$

$={10}^{-5}$ mol ${elm}^{-3}$

${BaCrO}_4⟶{Ba}^{2+}+{CrO}_4^{2-}$

$K_{sp}=\left[{Ba}^{2+}\right]\left[{CrO}_4^{2-}\right]=s^2$

$s=\sqrt{K_{sp}}=\sqrt{1\times {10}^{-10}}={10}^{-5}$ mol ${dm}^{-3}$