A saturated solution of Ag_2SO_4 is 2.5 times | vedclass

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(A) The dissociation of $Ag_2SO_4$ is given by: $Ag_2SO_4(s) ightleftharpoons 2Ag^+(aq) + SO_4^{2-}(aq)$.If the solubility is $x = 2.5 	imes 10^{-2

Vedclass · Accepted Answer

$62.5 	imes 10^{-6}$

Vedclass · Answer

(A) The dissociation of $Ag_2SO_4$ is given by: $Ag_2SO_4(s) ightleftharpoons 2Ag^+(aq) + SO_4^{2-}(aq)$.If the solubility is $x = 2.5 	imes 10^{-2} \ M$,then the concentration of ions are $[Ag^+] = 2x$ and $[SO_4^{2-}] = x$.The solubility product expression is $K_{sp} = [Ag^+]^2 [SO_4^{2-}] = (2x)^2 \cdot x = 4x^3$.Substituting the value of $x$: $K_{sp} = 4 \cdot (2.5 	imes 10^{-2})^3$.$K_{sp} = 4 \cdot (15.625 	imes 10^{-6}) = 62.5 	imes 10^{-6}$.

$A$ saturated solution of $Ag_2SO_4$ is $2.5 \times 10^{-2} \ M$. The value of its solubility product is:

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