# Problem: What is the molar solubility of Pb(OH)2 in a 0.138-M solution of CH3NH2?

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

###### FREE Expert Solution

CH3NH2 → weak base

Step 1: Calculate OH in the solution

Kb of CH3NH2 = 4.4×10‒4

${\mathbf{K}}_{\mathbf{b}}\mathbf{=}\frac{\left[{\mathrm{CH}}_{3}{{\mathrm{NH}}_{3}}^{+}\right]\left[{\mathrm{OH}}^{-}\right]}{\left[{\mathrm{CH}}_{3}{\mathrm{NH}}_{2}\right]}\phantom{\rule{0ex}{0ex}}{\mathbf{K}}_{\mathbf{b}}\mathbf{=}\frac{\left[\mathbf{x}\right]\left[\mathbf{x}\right]}{\left[\mathbf{0}\mathbf{.}\mathbf{138}\mathbf{-}\mathbf{x}\right]}\phantom{\rule{0ex}{0ex}}{\mathbf{K}}_{\mathbf{b}}\mathbf{=}\frac{{\mathbf{x}}^{\mathbf{2}}}{\mathbf{0}\mathbf{.}\mathbf{138}\mathbf{-}\mathbf{x}}$

null

$\mathbf{4}\mathbf{.}\mathbf{4}\mathbf{×}{\mathbf{10}}^{\mathbf{-}\mathbf{4}}\mathbf{=}\frac{{\mathbf{x}}^{\mathbf{2}}}{\mathbf{0}\mathbf{.}\mathbf{138}\mathbf{-}\mathbf{x}}\phantom{\rule{0ex}{0ex}}\mathbf{4}\mathbf{.}\mathbf{4}\mathbf{×}{\mathbf{10}}^{\mathbf{-}\mathbf{4}}\left(0.138-x\right)\mathbf{=}{\mathbf{x}}^{\mathbf{2}}\phantom{\rule{0ex}{0ex}}\mathbf{6}\mathbf{.}\mathbf{072}\mathbf{×}{\mathbf{10}}^{\mathbf{-}\mathbf{5}}\mathbf{-}\mathbf{4}\mathbf{.}\mathbf{4}\mathbf{×}{\mathbf{10}}^{\mathbf{-}\mathbf{4}}\mathbf{x}\mathbf{=}{\mathbf{x}}^{\mathbf{2}}\phantom{\rule{0ex}{0ex}}\mathbf{0}\mathbf{=}{\mathbf{x}}^{\mathbf{2}}\mathbf{+}\mathbf{4}\mathbf{.}\mathbf{4}\mathbf{×}{\mathbf{10}}^{\mathbf{-}\mathbf{4}}\mathbf{x}\mathbf{-}\mathbf{6}\mathbf{.}\mathbf{072}\mathbf{×}{\mathbf{10}}^{\mathbf{-}\mathbf{5}}$

$\overline{){\mathbf{x}}{\mathbf{=}}\frac{\mathbf{-}\mathbf{b}\mathbf{±}\sqrt{{\mathbf{b}}^{\mathbf{2}}\mathbf{-}\mathbf{4}\mathbf{a}\mathbf{c}}}{\mathbf{2}\mathbf{a}}}$

$\mathbf{x}\mathbf{=}\frac{\mathbf{-}\mathbf{\left(}\mathbf{4}\mathbf{.}\mathbf{4}\mathbf{×}{\mathbf{10}}^{\mathbf{-}\mathbf{4}}\mathbf{\right)}\mathbf{±}\sqrt{\mathbf{\left(}\mathbf{4}\mathbf{.}\mathbf{4}\mathbf{×}{\mathbf{10}}^{\mathbf{-}\mathbf{4}}{\mathbf{\right)}}^{\mathbf{2}}\mathbf{-}\mathbf{4}\mathbf{\left(}\mathbf{1}\mathbf{\right)}\mathbf{\left(}\mathbf{-}\mathbf{6}\mathbf{.}\mathbf{072}\mathbf{×}{\mathbf{10}}^{\mathbf{-}\mathbf{5}}\mathbf{\right)}}}{\mathbf{2}\mathbf{\left(}\mathbf{1}\mathbf{\right)}}\phantom{\rule{0ex}{0ex}}\mathbf{x}\mathbf{=}\frac{\mathbf{-}\mathbf{4}\mathbf{.}\mathbf{4}\mathbf{×}{\mathbf{10}}^{\mathbf{-}\mathbf{4}}\mathbf{±}\mathbf{0}\mathbf{.}\mathbf{01559}}{\mathbf{2}}$

$\mathbf{x}\mathbf{=}\frac{\mathbf{-}\mathbf{4}\mathbf{.}\mathbf{4}\mathbf{×}{\mathbf{10}}^{\mathbf{-}\mathbf{4}}\mathbf{+}\mathbf{0}\mathbf{.}\mathbf{01559}}{\mathbf{2}}$

x = 7.5754 × 10–3

$\mathbf{x}\mathbf{=}\frac{\mathbf{-}\mathbf{4}\mathbf{.}\mathbf{4}\mathbf{×}{\mathbf{10}}^{\mathbf{-}\mathbf{4}}\mathbf{-}\mathbf{0}\mathbf{.}\mathbf{01559}}{\mathbf{2}}$

x = –8.0154 × 10–3

x = [OH-]

[OH-] = 7.5754 × 10–3 M

Step 2:

Pb(OH)2(s)  Pb2+(aq) + 2 OH(aq)

Ksp = 1.43×10‒20

###### Problem Details

What is the molar solubility of Pb(OH)2 in a 0.138-M solution of CH3NH2?