At 298 K,the solubility product of PbCl_2 is | vedclass

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(A) The dissociation of $PbCl_2$ is given by: $PbCl_2(s) \rightleftharpoons Pb^{2+}(aq) + 2Cl^-(aq)$.Let the solubility be $s \ mol/L$.Then,$[Pb^{2+}]

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$6.3 	imes 10^{-3}$

Vedclass · Answer

(A) The dissociation of $PbCl_2$ is given by: $PbCl_2(s) ightleftharpoons Pb^{2+}(aq) + 2Cl^-(aq)$.Let the solubility be $s \ mol/L$.Then,$[Pb^{2+}] = s$ and $[Cl^-] = 2s$.The solubility product expression is $K_{sp} = [Pb^{2+}][Cl^-]^2$.Substituting the values: $K_{sp} = (s)(2s)^2 = 4s^3$.Given $K_{sp} = 1.0 	imes 10^{-6}$,we have $4s^3 = 1.0 	imes 10^{-6}$.$s^3 = \frac{1.0 	imes 10^{-6}}{4} = 0.25 	imes 10^{-6} = 250 	imes 10^{-9}$.$s = \sqrt[3]{250 	imes 10^{-9}} \approx 6.3 	imes 10^{-3} \ mol/L$.

At $298 \ K$,the solubility product of $PbCl_2$ is $1.0 \times 10^{-6}$. What will be the solubility of $PbCl_2$ in moles/litre?

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