The K_{sp} for BaSO_4 at 25 ^circ C is 1.1 ti | vedclass

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(B) The solubility product expression for $BaSO_4$ is $K_{sp} = [Ba^{2+}][SO_4^{2-}]$.Given $K_{sp} = 1.1 	imes 10^{-10}$.Let the new solubility be $

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$0.01$

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(B) The solubility product expression for $BaSO_4$ is $K_{sp} = [Ba^{2+}][SO_4^{2-}]$.Given $K_{sp} = 1.1 	imes 10^{-10}$.Let the new solubility be $s = 1.1 	imes 10^{-8} \ M$.In the presence of $Na_2SO_4$,the concentration of $Ba^{2+}$ ions is $s = 1.1 	imes 10^{-8} \ M$.The total concentration of $SO_4^{2-}$ ions is $s + [Na_2SO_4] \approx [Na_2SO_4]$ because $s$ is very small compared to the concentration of $Na_2SO_4$.Substituting into the $K_{sp}$ expression: $1.1 	imes 10^{-10} = (1.1 	imes 10^{-8}) 	imes [Na_2SO_4]$.$[Na_2SO_4] = \frac{1.1 	imes 10^{-10}}{1.1 	imes 10^{-8}} = 10^{-2} = 0.01 \ M$.

The $K_{sp}$ for $BaSO_4$ at $25\ ^\circ C$ 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 (in $M$):

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