The solubility product of BaSO_4 at 25 , ^cir | vedclass

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(C) The solubility product expression for $BaSO_4$ is $K_{sp} = [Ba^{2+}][SO_4^{2-}]$.Given $K_{sp} = 1.0 	imes 10^{-9}$ and $[Ba^{2+}] = 0.01 \, M =

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$10^{-7} \, M$

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(C) The solubility product expression for $BaSO_4$ is $K_{sp} = [Ba^{2+}][SO_4^{2-}]$.Given $K_{sp} = 1.0 	imes 10^{-9}$ and $[Ba^{2+}] = 0.01 \, M = 10^{-2} \, M$.For precipitation to occur,the ionic product must exceed the solubility product: $[Ba^{2+}][SO_4^{2-}] > K_{sp}$.Substituting the values: $(10^{-2}) 	imes [SO_4^{2-}] > 1.0 	imes 10^{-9}$.$[SO_4^{2-}] > \frac{1.0 	imes 10^{-9}}{10^{-2}} = 1.0 	imes 10^{-7} \, M$.Since $H_2SO_4$ is a strong acid,it dissociates as $H_2SO_4 ightarrow 2H^+ + SO_4^{2-}$,so $[H_2SO_4] = [SO_4^{2-}]$.Therefore,the concentration of $H_2SO_4$ must be greater than $10^{-7} \, M$.

The solubility product of $BaSO_4$ at $25 \, ^\circ C$ is $1.0 \times 10^{-9}$. What will be the concentration of $H_2SO_4$ required to precipitate $BaSO_4$ from a $0.01 \, M$ solution of $Ba^{2+}$ ions?

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