The solubility of CdSO_{4} in water is 8.0 ti | vedclass

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(A) In pure water,the solubility product constant $K_{sp}$ is given by $K_{sp} = S^{2} = (8.0 	imes 10^{-4})^{2} = 64 	imes 10^{-8}$.In $0.01 \ M \

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$64$

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(A) In pure water,the solubility product constant $K_{sp}$ is given by $K_{sp} = S^{2} = (8.0 	imes 10^{-4})^{2} = 64 	imes 10^{-8}$.In $0.01 \ M \ H_{2}SO_{4}$,the dissociation is $H_{2}SO_{4} ightarrow 2H^{+} + SO_{4}^{2-}$.The concentration of $SO_{4}^{2-}$ from $H_{2}SO_{4}$ is $0.01 \ M$.Let the solubility of $CdSO_{4}$ in this solution be $x \ mol \ L^{-1}$.$CdSO_{4} ightleftharpoons Cd^{2+} + SO_{4}^{2-}$.The equilibrium concentrations are $[Cd^{2+}] = x$ and $[SO_{4}^{2-}] = (x + 0.01) \ M$.Since $x \ll 0.01$,we can approximate $(x + 0.01) \approx 0.01$.$K_{sp} = [Cd^{2+}][SO_{4}^{2-}] = x(0.01) = 64 	imes 10^{-8}$.$x = \frac{64 	imes 10^{-8}}{0.01} = 64 	imes 10^{-6} \ mol \ L^{-1}$.Thus,the value is $64$.

The solubility of $CdSO_{4}$ in water is $8.0 \times 10^{-4} \ mol \ L^{-1}$. Its solubility in $0.01 \ M \ H_{2}SO_{4}$ solution is $..... \ \times 10^{-6} \ mol \ L^{-1}$.
(Round off to the Nearest integer) (Assume that solubility is much less than $0.01 \ M$)

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The solubility of $CdSO_{4}$ in water is $8.0 \times 10^{-4} \ mol \ L^{-1}$. Its solubility in $0.01 \ M \ H_{2}SO_{4}$ solution is $..... \ \times 10^{-6} \ mol \ L^{-1}$.(Round off to the Nearest integer) (Assume that solubility is much less than $0.01 \ M$)