A saturated solution of Ag_2SO_4 shows a 0.00 | vedclass

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Question

(A) The elevation in boiling point is given by $\Delta T_b = i K_b m$. For $Ag_2SO_4$,the dissociation is $Ag_2SO_4 \rightleftharpoons 2Ag^+ + SO_4^{2

Vedclass · Accepted Answer

$3.2 	imes 10^{-11}$

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(A) The elevation in boiling point is given by $\Delta T_b = i K_b m$. For $Ag_2SO_4$,the dissociation is $Ag_2SO_4 ightleftharpoons 2Ag^+ + SO_4^{2-}$,so the van't Hoff factor $i = 3$. Given $\Delta T_b = 0.003 \ K$,$K_b = 5 \ K \ kg \ mol^{-1}$,and $m = S$ (solubility). Substituting the values: $0.003 = 3 	imes 5 	imes S$. $S = \frac{0.003}{15} = 2 	imes 10^{-4} \ mol \ L^{-1}$. The solubility product $K_{sp}$ for $Ag_2SO_4$ is $K_{sp} = [Ag^+]^2 [SO_4^{2-}] = (2S)^2 (S) = 4S^3$. $K_{sp} = 4 	imes (2 	imes 10^{-4})^3 = 4 	imes 8 	imes 10^{-12} = 3.2 	imes 10^{-11}$.

$A$ saturated solution of $Ag_2SO_4$ shows a $0.003 \ K$ rise in boiling point. Calculate the $K_{sp}$ of $Ag_2SO_4$. Given: $K_b = 5 \ K \ kg \ mol^{-1}$ and $1 \ m = 1 \ M$.

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