The solubility of PbCl when it is 80% ionized is — Ionic Equilibrium Chemistry Question

💡 Solution & Explanation

Let the solubility at 100% ionization be $S_{100}$. $K_{sp} = S_{100}(2S_{100})^2 = 4S_{100}^3$. If it is only 80% ionized, the concentration of the dissolved ions produced relative to the total dissolved solid $S_{80}$ decreases. $[\text{Pb}^{2+}] = 0.8 S_{80}$ and $[\text{Cl}^-] = 1.6 S_{80}$. $K_{sp} = (0.8 S_{80})(1.6 S_{80})^2 = 2.048 S_{80}^3$. Since $K_{sp}$ is a constant, $4S_{100}^3 = 2.048 S_{80}^3$, which means $S_{80}^3 > S_{100}^3$, and therefore $S_{80} > S_{100}$. The salt must dissolve more (greater total solubility) to generate the same requisite ionic product.