What would be the concentration of H_2SO_4 ne | vedclass

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Question

(C) The dissolution equilibrium for $BaSO_4$ is: $BaSO_4(s) ⇌ Ba^{2+}(aq) + SO_4^{2-}(aq)$.The solubility product expression is: $K_{sp} = [Ba^{2+}][S

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$10^{-7} \ M$

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(C) The dissolution equilibrium for $BaSO_4$ is: $BaSO_4(s) ⇌ Ba^{2+}(aq) + SO_4^{2-}(aq)$.The solubility product expression is: $K_{sp} = [Ba^{2+}][SO_4^{2-}]$.Given $[Ba^{2+}] = 0.01 \ M = 10^{-2} \ M$ and $K_{sp} = 1.0 	imes 10^{-9}$.To precipitate $BaSO_4$,the ionic product must exceed $K_{sp}$,so $[SO_4^{2-}] > \frac{K_{sp}}{[Ba^{2+}]}$.$[SO_4^{2-}] > \frac{1.0 	imes 10^{-9}}{10^{-2}} = 10^{-7} \ M$.Since $H_2SO_4$ is a strong acid that dissociates completely into $2H^+$ and $SO_4^{2-}$,the concentration of $SO_4^{2-}$ ions is equal to the concentration of $H_2SO_4$.Therefore,the concentration of $H_2SO_4$ must be greater than $10^{-7} \ M$.

What would be the concentration of $H_2SO_4$ necessary to precipitate $BaSO_4$ from a solution of $0.01 \ M \ Ba^{2+}$ ions $(K_{sp} = 1.0 \times 10^{-9})$?

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