The value of K_{sp} for B_2CO_3 is 3.2 times | vedclass

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(D) The dissociation of $B_2CO_3$ is given by: $B_2CO_3(s) \rightleftharpoons 2B^+(aq) + CO_3^{2-}(aq)$.Let the solubility of $B_2CO_3$ be $S \ M$.The

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$4 	imes 10^{-2} \ M$

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(D) The dissociation of $B_2CO_3$ is given by: $B_2CO_3(s) ightleftharpoons 2B^+(aq) + CO_3^{2-}(aq)$.Let the solubility of $B_2CO_3$ be $S \ M$.Then,$[B^+] = 2S$ and $[CO_3^{2-}] = S$.The solubility product expression is: $K_{sp} = [B^+]^2 [CO_3^{2-}] = (2S)^2(S) = 4S^3$.Given $K_{sp} = 3.2 	imes 10^{-5}$,we have $4S^3 = 3.2 	imes 10^{-5}$.$S^3 = 0.8 	imes 10^{-5} = 8 	imes 10^{-6}$.Taking the cube root,$S = 2 	imes 10^{-2} \ M$.The concentration of $B^+$ is $[B^+] = 2S = 2(2 	imes 10^{-2} \ M) = 4 	imes 10^{-2} \ M$.

The value of $K_{sp}$ for $B_2CO_3$ is $3.2 \times 10^{-5}$. The concentration of $B^{+}$ in its saturated solution is :-

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