Solubility of BaF_2 in a solution of Ba(NO_3) | vedclass

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(C) The dissolution of $BaF_2$ is given by: $BaF_2(s) \rightleftharpoons Ba^{2+}(aq) + 2F^{-}(aq)$.In the presence of $Ba(NO_3)_2$ (a strong electroly

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$\frac{1}{2}[F^{-}]$

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(C) The dissolution of $BaF_2$ is given by: $BaF_2(s) \rightleftharpoons Ba^{2+}(aq) + 2F^{-}(aq)$.In the presence of $Ba(NO_3)_2$ (a strong electrolyte),the concentration of $Ba^{2+}$ ions is increased due to the common ion effect.Let $s$ be the solubility of $BaF_2$ and $x$ be the concentration of $Ba(NO_3)_2$.Total $[Ba^{2+}] = s + x$ and $[F^{-}] = 2s$.From the solubility product expression: $K_{sp} = [Ba^{2+}][F^{-}]^2$.Substituting the values: $K_{sp} = (s + x)(2s)^2$.Since $2s = [F^{-}]$,we have $s = \frac{1}{2}[F^{-}]$.Thus,the solubility of $BaF_2$ is represented by $\frac{1}{2}[F^{-}]$.

Solubility of $BaF_2$ in a solution of $Ba(NO_3)_2$ will be represented by the concentration term?

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