K_{sp} for BaSO_4 at 25 , ^oC is 1.1 times 10 | vedclass

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(B) The solubility product constant $(K_{sp})$ for $BaSO_4$ is given by: $K_{sp} = [Ba^{2+}][SO_4^{2-}] = 1.1 	imes 10^{-10}$.Let the solubility of $

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$0.01$

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(B) The solubility product constant $(K_{sp})$ for $BaSO_4$ is given by: $K_{sp} = [Ba^{2+}][SO_4^{2-}] = 1.1 	imes 10^{-10}$.Let the solubility of $BaSO_4$ in the presence of $Na_2SO_4$ be $S = 1.1 	imes 10^{-8} \, M$.In the solution,$BaSO_4$ dissociates as $BaSO_4 ightleftharpoons Ba^{2+} + SO_4^{2-}$,so $[Ba^{2+}] = S = 1.1 	imes 10^{-8} \, M$.$Na_2SO_4$ is a strong electrolyte and dissociates completely as $Na_2SO_4 ightarrow 2Na^+ + SO_4^{2-}$. Let the concentration of $Na_2SO_4$ be $C$.Then,$[SO_4^{2-}] = S + C \approx C$ (since $S$ is very small compared to $C$).Substituting these into the $K_{sp}$ expression: $1.1 	imes 10^{-10} = (1.1 	imes 10^{-8}) 	imes C$.Solving for $C$: $C = \frac{1.1 	imes 10^{-10}}{1.1 	imes 10^{-8}} = 10^{-2} = 0.01 \, M$.

$K_{sp}$ for $BaSO_4$ at $25 \, ^oC$ is $1.1 \times 10^{-10}$. To make the new solubility equal to $1.1 \times 10^{-8} \, M$,it is necessary to use a solution of $Na_2SO_4$ of the following concentration $....... \ M$

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