What is the concentration of Ba^{2+} when BaF | vedclass

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(C) The solubility equilibrium for $BaF_2$ is given by: $BaF_{2(s)} \rightleftharpoons Ba^{2+}_{(aq)} + 2F^{-}_{(aq)}$ The solubility product expressi

Vedclass · Accepted Answer

$1.1 	imes 10^{-5} \, M$

Vedclass · Answer

(C) The solubility equilibrium for $BaF_2$ is given by: $BaF_{2(s)} ightleftharpoons Ba^{2+}_{(aq)} + 2F^{-}_{(aq)}$ The solubility product expression is: $K_{sp} = [Ba^{2+}] [F^{-}]^2$ Given $K_{sp} = 1.0 	imes 10^{-6}$ and $[F^{-}] = 0.30 \, M$. Substituting the values into the expression: $1.0 	imes 10^{-6} = [Ba^{2+}] 	imes (0.30)^2$ $[Ba^{2+}] = \frac{1.0 	imes 10^{-6}}{0.09}$ $[Ba^{2+}] = 1.11 	imes 10^{-5} \, M \approx 1.1 	imes 10^{-5} \, M$

What is the concentration of $Ba^{2+}$ when $BaF_2$ $(K_{sp} = 1.0 \times 10^{-6})$ begins to precipitate from a solution that is $0.30 \, M \ F^{-}$?

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