The molar solubility of CaF_2 (K_{sp} = 5.3 t | vedclass

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Question

(C) $CaF_2$ dissociates as: $CaF_{2(s)} \rightleftharpoons Ca^{2+}_{(aq)} + 2F^{-}_{(aq)}$$NaF$ is a strong electrolyte: $NaF_{(aq)} \rightarrow Na^{+

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$5.3 	imes 10^{-9} \ mol \ L^{-1}$

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(C) $CaF_2$ dissociates as: $CaF_{2(s)} ightleftharpoons Ca^{2+}_{(aq)} + 2F^{-}_{(aq)}$$NaF$ is a strong electrolyte: $NaF_{(aq)} ightarrow Na^{+}_{(aq)} + F^{-}_{(aq)}$Given $[NaF] = 0.1 \ M$,so $[F^-] = 0.1 \ M$ from $NaF$.Let $s$ be the molar solubility of $CaF_2$. Then $[Ca^{2+}] = s$ and total $[F^-] = (2s + 0.1) \approx 0.1 \ M$ (since $s$ is very small).$K_{sp} = [Ca^{2+}][F^-]^2$$5.3 	imes 10^{-11} = s 	imes (0.1)^2$$s = \frac{5.3 	imes 10^{-11}}{0.01} = 5.3 	imes 10^{-9} \ mol \ L^{-1}$

The molar solubility of $CaF_2$ $(K_{sp} = 5.3 \times 10^{-11})$ in $0.1 \ M$ solution of $NaF$ will be:

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