The solubility product of BaCl_2 is 4 times 1 | vedclass

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Question

(A) The dissociation of $BaCl_2$ is given by: $BaCl_2(s) \rightleftharpoons Ba^{2+}(aq) + 2Cl^-(aq)$.Let the solubility be $S \ mol/L$.Then,$[Ba^{2+}]

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$1 	imes 10^{-3}$

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(A) The dissociation of $BaCl_2$ is given by: $BaCl_2(s) ightleftharpoons Ba^{2+}(aq) + 2Cl^-(aq)$.Let the solubility be $S \ mol/L$.Then,$[Ba^{2+}] = S$ and $[Cl^-] = 2S$.The solubility product expression is: $K_{sp} = [Ba^{2+}][Cl^-]^2$.Substituting the values: $K_{sp} = (S)(2S)^2 = 4S^3$.Given $K_{sp} = 4 	imes 10^{-9}$.So,$4S^3 = 4 	imes 10^{-9}$.$S^3 = 10^{-9}$.$S = (10^{-9})^{1/3} = 10^{-3} \ mol/L$.

The solubility product of $BaCl_2$ is $4 \times 10^{-9}$. Its solubility in $mol/L$ is .......

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