The solubility of BaCl_2 is 4 times 10^{-6} | vedclass

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(C) For $BaCl_2$,the dissociation is: $BaCl_2 ightleftharpoons Ba^{2+} + 2Cl^-$. Given solubility $s = 4 	imes 10^{-6} \ M$. $K_{sp} = [Ba^{2+}][Cl

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$256 	imes 10^{-18}$

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(C) For $BaCl_2$,the dissociation is: $BaCl_2 ightleftharpoons Ba^{2+} + 2Cl^-$. Given solubility $s = 4 	imes 10^{-6} \ M$. $K_{sp} = [Ba^{2+}][Cl^-]^2 = (s)(2s)^2 = 4s^3$. $K_{sp} = 4 	imes (4 	imes 10^{-6})^3 = 4 	imes 64 	imes 10^{-18} = 256 	imes 10^{-18}$. In the presence of $10^{-2} \ M \ Ba(OH)_2$,the common ion is $Ba^{2+}$. $[Ba^{2+}] = 10^{-2} \ M$ (from $Ba(OH)_2$) and $[Cl^-] = 2s'$. $K_{sp} = [Ba^{2+}][Cl^-]^2 = (10^{-2})(2s')^2 = 4 	imes 10^{-2} 	imes (s')^2$. $256 	imes 10^{-18} = 4 	imes 10^{-2} 	imes (s')^2$. $(s')^2 = 64 	imes 10^{-16}$. $s' = 8 	imes 10^{-8} \ M$.

The solubility of $BaCl_2$ is $4 \times 10^{-6} \ M$. Calculate its $K_{sp}$. Also,find its new solubility in the presence of $10^{-2} \ M \ Ba(OH)_2$.

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