PbCl_2 has a maximum concentration of 1.0 tim | vedclass

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(A) The dissociation of $PbCl_2$ is given by: $PbCl_{2(s)} \rightleftharpoons Pb^{2+}_{(aq)} + 2Cl^{-}_{(aq)}$In a saturated solution,the solubility $

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$4 	imes 10^{-7} \ M$

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(A) The dissociation of $PbCl_2$ is given by: $PbCl_{2(s)} ightleftharpoons Pb^{2+}_{(aq)} + 2Cl^{-}_{(aq)}$In a saturated solution,the solubility $s = 1.0 	imes 10^{-3} \ M$.The solubility product $K_{sp}$ is calculated as: $K_{sp} = [Pb^{2+}][Cl^{-}]^2 = (s)(2s)^2 = 4s^3$.$K_{sp} = 4 	imes (1.0 	imes 10^{-3})^3 = 4 	imes 10^{-9}$.In a $0.1 \ M \ NaCl$ solution,the concentration of $Cl^-$ ions is dominated by $NaCl$ $([Cl^-] \approx 0.1 \ M)$.Using the $K_{sp}$ expression: $K_{sp} = [Pb^{2+}][Cl^-]^2$.$4 	imes 10^{-9} = s' 	imes (0.1)^2$,where $s'$ is the new solubility.$s' = \frac{4 	imes 10^{-9}}{0.01} = 4 	imes 10^{-7} \ M$.

$PbCl_2$ has a maximum concentration of $1.0 \times 10^{-3} \ M$ in its saturated aqueous solution at $25 \ ^oC$. Its solubility in a $0.1 \ M \ NaCl$ solution will be:

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