A solution is saturated with SrCO_3 and SrF_2 | vedclass

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(D) For $SrCO_3$, the solubility product expression is $K_{sp} = [Sr^{2+}][CO_3^{2-}]$. Given $[CO_3^{2-}] = 1.2 	imes 10^{-3} \ M$ and $K_{sp}(SrCO_

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$3.7 	imes 10^{-2} \ M$

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(D) For $SrCO_3$, the solubility product expression is $K_{sp} = [Sr^{2+}][CO_3^{2-}]$. Given $[CO_3^{2-}] = 1.2 	imes 10^{-3} \ M$ and $K_{sp}(SrCO_3) = 7.0 	imes 10^{-10}$, we calculate the concentration of $Sr^{2+}$: $[Sr^{2+}] = \frac{7.0 	imes 10^{-10}}{1.2 	imes 10^{-3}} = 5.83 	imes 10^{-7} \ M$. For $SrF_2$, the solubility product expression is $K_{sp} = [Sr^{2+}][F^{-}]^2$. Substituting the values: $7.9 	imes 10^{-10} = (5.83 	imes 10^{-7}) 	imes [F^{-}]^2$. $[F^{-}]^2 = \frac{7.9 	imes 10^{-10}}{5.83 	imes 10^{-7}} = 1.355 	imes 10^{-3}$. $[F^{-}] = \sqrt{1.355 	imes 10^{-3}} \approx 3.68 	imes 10^{-2} \ M \approx 3.7 	imes 10^{-2} \ M$.

$A$ solution is saturated with $SrCO_3$ and $SrF_2$. The $[CO_3^{2-}]$ is found to be $1.2 \times 10^{-3} \ M$. The concentration of $F^{-}$ in the solution would be:
Given: $K_{sp}(SrCO_3) = 7.0 \times 10^{-10}$, $K_{sp}(SrF_2) = 7.9 \times 10^{-10}$

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$A$ solution is saturated with $SrCO_3$ and $SrF_2$. The $[CO_3^{2-}]$ is found to be $1.2 \times 10^{-3} \ M$. The concentration of $F^{-}$ in the solution would be: Given: $K_{sp}(SrCO_3) = 7.0 \times 10^{-10}$, $K_{sp}(SrF_2) = 7.9 \times 10^{-10}$