We have to determine the solubility of PbCl_{2} in a 0.15 M HCl solution.

PbCl_{2} is a sparingly soluble salt which partially dissociates in water to produce Pb^{2+} and Cl^{-} ions.

PbCl_{2(s)} ⇌ Pb^{2+}_{(aq)} + 2 Cl^{-}_{(aq)}

We can write the K_{sp} expression for this equilibrium.

$\overline{){{\mathrm{K}}}_{{\mathrm{sp}}}{}{=}{}\frac{\left[\mathrm{Products}\right]}{\left[\mathrm{Reactants}\right]}{}{=}{}\frac{\left[{\mathrm{Pb}}^{2+}\right]{\left[{\mathrm{Cl}}^{-}\right]}^{2}}{\overline{)\overline{)\left[{\mathrm{PbCl}}_{2}\right]}}}{}{=}{}\left[{\mathrm{Pb}}^{2+}\right]{\left[{\mathrm{Cl}}^{-}\right]}^{{2}}}$

What is the solubility (in M) of PbCl _{2} in a 0.15 M solution of HCl? The Ksp of PbCl _{2} is 1.6 × 10^{-5}?

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