25.0 mL of 0.050 M Ba(NO_3)_2 is mixed wit | vedclass

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(C) Step $1$: Calculate the final concentrations of ions after mixing.Total volume $= 25.0 \ mL + 25.0 \ mL = 50.0 \ mL$.$[Ba^{2+}] = \frac{25.0 	ime

Vedclass · Accepted Answer

$5$

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(C) Step $1$: Calculate the final concentrations of ions after mixing.Total volume $= 25.0 \ mL + 25.0 \ mL = 50.0 \ mL$.$[Ba^{2+}] = \frac{25.0 	imes 0.050}{50.0} = 0.025 \ M$.$[F^{-}] = \frac{25.0 	imes 0.020}{50.0} = 0.010 \ M$.Step $2$: Calculate the ionic product $Q = [Ba^{2+}][F^{-}]^2$.$Q = (0.025) 	imes (0.010)^2 = 0.025 	imes 10^{-4} = 2.5 	imes 10^{-6}$.Step $3$: Calculate the ratio $\frac{Q}{K_{sp}}$.Given $K_{sp} = 0.5 	imes 10^{-6} = 5 	imes 10^{-7}$.Ratio $= \frac{2.5 	imes 10^{-6}}{0.5 	imes 10^{-6}} = \frac{2.5}{0.5} = 5$.

$25.0 \ mL$ of $0.050 \ M \ Ba(NO_3)_2$ is mixed with $25.0 \ mL$ of $0.020 \ M \ NaF$. The $K_{sp}$ of $BaF_2$ is $0.5 \times 10^{-6}$ at $298 \ K$. The ratio of $[Ba^{2+}][F^{-}]^2$ to $K_{sp}$ is $.......$. (Nearest integer)

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