# Problem: How many grams of Pb(OH)2 will dissolve in 500 mL of a 0.050-M PbCl2 solution (Ksp = 1.2 × 10–15)?

🤓 Based on our data, we think this question is relevant for Professor Petrovich's class at NCSU.

###### FREE Expert Solution

The ICE table for the dissociation of Pb(OH)2 is:

The Ksp expression for Pb(OH)2 is:

${\mathbf{K}}_{\mathbf{sp}}\mathbf{=}\frac{\mathbf{products}}{\overline{)\mathbf{reactants}}}\mathbf{=}\left[{\mathrm{Pb}}^{2+}\right]{\left[{\mathrm{OH}}^{-}\right]}^{\mathbf{2}}$

Plugging in the value for Ksp (1.2 × 10–15) and equilibrium concentrations:

$\mathbf{1}\mathbf{.}\mathbf{2}\mathbf{×}{\mathbf{10}}^{\mathbf{-}\mathbf{15}}\mathbf{=}\mathbf{\left[}\mathbf{0}\mathbf{.}\mathbf{05}\mathbf{+}\mathbf{x}\mathbf{\right]}{\mathbf{\left[}\mathbf{2}\mathbf{x}\mathbf{\right]}}^{\mathbf{2}}$

Since the value of Ksp is very small, we can remove +x from the equation to simplify it:

$\mathbf{1}\mathbf{.}\mathbf{2}\mathbf{×}{\mathbf{10}}^{\mathbf{-}\mathbf{15}}\mathbf{=}\mathbf{\left[}\mathbf{0}\mathbf{.}\mathbf{05}\overline{)\mathbf{+}\mathbf{x}}\mathbf{\right]}{\mathbf{\left[}\mathbf{2}\mathbf{x}\mathbf{\right]}}^{\mathbf{2}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathbf{1}\mathbf{.}\mathbf{2}\mathbf{×}{\mathbf{10}}^{\mathbf{-}\mathbf{15}}\mathbf{=}\mathbf{\left[}\mathbf{0}\mathbf{.}\mathbf{05}\mathbf{\right]}{\mathbf{\left[}\mathbf{2}\mathbf{x}\mathbf{\right]}}^{\mathbf{2}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathbf{1}\mathbf{.}\mathbf{2}\mathbf{×}{\mathbf{10}}^{\mathbf{-}\mathbf{15}}\mathbf{=}\mathbf{\left[}\mathbf{0}\mathbf{.}\mathbf{05}\mathbf{\right]}\mathbf{4}{\mathbf{x}}^{\mathbf{2}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}$

Solving for x:

$\frac{\mathbf{1}\mathbf{.}\mathbf{2}\mathbf{×}{\mathbf{10}}^{\mathbf{-}\mathbf{15}}}{\mathbf{0}\mathbf{.}\mathbf{2}}\mathbf{=}\frac{\overline{)\mathbf{0}\mathbf{.}\mathbf{2}}{\mathbf{x}}^{\mathbf{2}}}{\overline{)\mathbf{0}\mathbf{.}\mathbf{2}}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\sqrt{{\mathbf{x}}^{\mathbf{2}}}\mathbf{=}\sqrt{\frac{\mathbf{1}\mathbf{.}\mathbf{2}\mathbf{×}{\mathbf{10}}^{\mathbf{-}\mathbf{15}}}{\mathbf{0}\mathbf{.}\mathbf{2}}}$

###### Problem Details

How many grams of Pb(OH)2 will dissolve in 500 mL of a 0.050-M PbCl2 solution (Ksp = 1.2 × 10–15)?