Question

${\mathbf{K}}_{\mathrm{sp}}$for $\mathbf{Pb}{\mathbf{\left(}\mathbf{OH}\mathbf{\right)}}_2$is $\mathbf{4}\mathbf{.}\mathbf{2}\mathbf{\times }{\mathbf{10}}^{-15}$,and${\mathbf{K}}_f$for$\mathbf{Pb}{{\mathbf{\left(}\mathbf{OH}\mathbf{\right)}}_3}^{\mathbf{-}}$ is $\mathbf{4}\mathbf{\times }{\mathbf{10}}^{14}$. Suppose a solution whose initial concentration of${\mathbf{Pb}}^{2+}\mathbf{\left(}\mathbf{aq}\mathbf{\right)}$ is 1.00 M is brought to pH 13.0 by addition of solid NaOH. Will solid$\mathbf{Pb}{\mathbf{\left(}\mathbf{OH}\mathbf{\right)}}_2$ precipitate, or will the lead be dissolved asdata-custom-editor="chemistry" $\mathbf{Pb}{{\mathbf{\left(}\mathbf{OH}\mathbf{\right)}}_3}^{-}$ ? What will be ${\mathbf{Pb}}^{2+}$and $\mathbf{Pb}{{\mathbf{\left(}\mathbf{OH}\mathbf{\right)}}_3}^{-}$ at equilibrium? Repeat the calculation for an initial${\mathbf{Pb}}^{\mathbf{2}\mathbf{+}}$concentration of 0.050 M.

Short answer

The assumption is incorrect. The precipitate of $\mathrm{Pb}{\left(\mathrm{OH}\right)}_2$is not formed at the given pH value that is 13.0.

Step-by-step solution

The given data

The ${\mathrm{K}}_{\mathrm{sp}}\mathrm{of}\mathrm{Pb}{\left(\mathrm{OH}\right)}_2\mathrm{is}4.2\times {10}^{-15}$

The pH of the solution is 13.0

Initial $\left[{\mathrm{Pb}}^{2+}\right]=1.00\mathrm{M}$

The ${\mathrm{K}}_{\mathrm{f}}\mathrm{of}\mathrm{Pb}{{\left(\mathrm{OH}\right)}_3}^{-}\mathrm{is}4\times {10}^{14}$

Assume that precipitate of Pb(OH)2 is formed

The pH of the solution is 13.0

So, pOH is 1.0.

$\left[{\mathrm{OH}}^{-}\right]=0.1\mathrm{mol}.{\mathrm{L}}^{-1}$

$\begin{array}{c}{\mathrm{K}}_{\mathrm{sp}}=\left[{\mathrm{Pb}}^{2+}\right]{\left[{\mathrm{OH}}^{-}\right]}^2\\ 4.2\times {10}^{-15}=\left[{\mathrm{Pb}}^{2+}\right]{\left(0.1\right)}^2\\ \left[{\mathrm{Pb}}^{2+}\right]=4.2\times {10}^{-13}\end{array}$

width="366" height="117" role="math">Kf=[Pb(OH)3−][Pb2+][OH−]3=Kf=5×1014[Pb(OH)3−]=(4×1014)×(4.2×10−13)(0.1)3=1.7×10−2 mol.L−1

Check the assumption:

$\left[{\mathrm{Pb}}^{2+}\right]+\left[\mathrm{Pb}{{\left(\mathrm{OH}\right)}_3}^{-}\right]<\mathrm{initial}\mathrm{concentration}\mathrm{of}{\mathrm{Pb}}^{2+}$

$4.2\times {10}^{-13}+0.017\cong 0.017\mathrm{mol}.{\mathrm{L}}^{-1}>\mathrm{initial}\left[{\mathrm{Pb}}^{2+}\right]\mathrm{that}\mathrm{is}1.00\mathrm{M}.$

Explanation

Hence, the assumption is incorrect. The precipitate of $\mathbf{Pb}{\mathbf{\left(}\mathbf{OH}\mathbf{\right)}}_{\mathbf{2}}$ is not formed at the given pH value that is 13.0.