The solubility of PbCl_2 at 25 ^oC is 6.3 ti | vedclass

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(C) The dissociation of $PbCl_2$ is given by: $PbCl_2(s) ightleftharpoons Pb^{2+}(aq) + 2Cl^-(aq)$.Let the solubility of $PbCl_2$ be $S = 6.3 	imes

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$(6.3 	imes 10^{-3}) 	imes (12.6 	imes 10^{-3})^2$

Vedclass · Answer

(C) The dissociation of $PbCl_2$ is given by: $PbCl_2(s) ightleftharpoons Pb^{2+}(aq) + 2Cl^-(aq)$.Let the solubility of $PbCl_2$ be $S = 6.3 	imes 10^{-3} \ mol/L$.At equilibrium,the concentration of $Pb^{2+}$ is $S$ and the concentration of $Cl^-$ is $2S$.Therefore,$[Pb^{2+}] = 6.3 	imes 10^{-3} \ mol/L$ and $[Cl^-] = 2 	imes (6.3 	imes 10^{-3}) = 12.6 	imes 10^{-3} \ mol/L$.The solubility product $K_{sp}$ is defined as: $K_{sp} = [Pb^{2+}][Cl^-]^2$.Substituting the values: $K_{sp} = (6.3 	imes 10^{-3}) 	imes (12.6 	imes 10^{-3})^2$.

The solubility of $PbCl_2$ at $25 \ ^oC$ is $6.3 \times 10^{-3} \ mol/L$. Its solubility product at that temperature is:

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