What is the molarity of F^{-} ions in a satur | vedclass

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(C) The dissociation of $BaF_2$ is given by: $BaF_{2(s)} \rightleftharpoons Ba^{2+}_{(aq)} + 2F^{-}_{(aq)}$Let the solubility of $BaF_2$ be $S \ mol/L

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$1.26 	imes 10^{-2}$

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(C) The dissociation of $BaF_2$ is given by: $BaF_{2(s)} ightleftharpoons Ba^{2+}_{(aq)} + 2F^{-}_{(aq)}$Let the solubility of $BaF_2$ be $S \ mol/L$.Then,$[Ba^{2+}] = S$ and $[F^{-}] = 2S$.The solubility product expression is: $K_{sp} = [Ba^{2+}][F^{-}]^2 = (S)(2S)^2 = 4S^3$.Given $K_{sp} = 1.0 	imes 10^{-6}$,we have $4S^3 = 1.0 	imes 10^{-6}$.$S^3 = 0.25 	imes 10^{-6} = 250 	imes 10^{-9}$.$S = (250)^{1/3} 	imes 10^{-3} \approx 6.3 	imes 10^{-3} \ mol/L$.The concentration of $F^{-}$ ions is $[F^{-}] = 2S = 2 	imes 6.3 	imes 10^{-3} = 1.26 	imes 10^{-2} \ M$.

What is the molarity of $F^{-}$ ions in a saturated solution of $BaF_2$? $(K_{sp} = 1.0 \times 10^{-6})$

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