Acetic acid has a Kaof,${\mathbf{\text{1.8×10}}}^{\mathbf{\text{-3}}}$and ammonia has a Kbof.localid="1663325976145" ${\mathbf{\text{1.8×10}}}^{\mathbf{\text{-3}}}$Findlocalid="1663326029962" $\mathbf{\left[}{\mathbf{\text{H}}}_{\mathbf{\text{3}}}{\mathbf{\text{O}}}^{\mathbf{\text{+}}}\mathbf{\right]}$,[OH], pH, and pOH for

(a)$\mathbf{\text{0.240}}$M acetic acid and

(b)$\mathbf{\text{0.240}}$M ammonia.

(a)${\text{0.240MCH}}_{\text{3}}\text{COOH}$First, write the reaction equation for the dissociationof.

${\text{CH}}_3\text{COOH}$

${\text{CH}}_3\text{COOH}+{\text{H}}_2\text{O}\rightleftharpoons {\text{H}}_3{\text{O}}^{+}+{\text{CH}}_3{\text{COO}}^{-}$

Next, construct the ICE table to obtain the equation for

$\begin{array}{l}{\text{K}}_{\text{a}}\text{=}\frac{\left|{\text{CH}}_{\text{3}}\text{COO}\Vert {\text{H}}_{\text{3}}{\text{O}}^{\mid }\right|}{\left|{\text{CH}}_{\text{3}}\text{COH}\right|}\\ {\text{K}}_{\text{a}}\text{=}\frac{{\text{x}}^{\text{2}}}{\text{0.240-x}}\end{array}$

We knowthat$\text{x\hspace{0.17em}=\hspace{0.17em}}\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]\text{\hspace{0.17em}=\hspace{0.17em}}\left[{\text{CH}}_{\text{3}}\text{COO}\right]{\text{.SinceCH}}_{\text{3}}\text{COOH}$ isa weak acid, its${\text{K}}_{\text{a}}$must be very small. So, assume that the x has no effect on 0.240-xin the denominator. Then substitute the${\text{K}}_{\text{a}}$to solve for x.

$\begin{array}{l}{\text{K}}_{\text{a}}\text{=}\frac{{\text{x}}^{\text{2}}}{\text{0.240}}\\ {\text{x}}^{\text{2}}\text{=}\left({\text{K}}_{\text{a}}\right)\text{(0.240)}\\ \text{x=}\sqrt{\left({\text{K}}_{\text{a}}\right)\text{(0.240)}}\end{array}$

Since$\text{x\hspace{0.17em}=\hspace{0.17em}}\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]\text{=}\left[{\text{ClO}}^{\text{-}}\right]\text{,then}\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]{\text{=\hspace{0.17em}2.08×10}}^{\text{-3}}\text{M}$

Next, calculate the pH of the solution.

$\begin{array}{l}\text{pH=\hspace{0.17em}-log}\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]\\ \text{pH=\hspace{0.17em}-log}\left({\text{2.08×10}}^{\text{-3}}\right)\\ \text{pH=\hspace{0.17em}2.68}\end{array}$

Then, solve for the pOH.

$\begin{array}{l}\text{pH+pOH=14}\\ \text{pOH=14-pH}\\ \text{=14-2.68}\\ \text{pOH=11.32}\end{array}$

Then, solve for the.$\left[{\text{OH}}^{-}\right]$

$\begin{array}{l}\text{pOH\hspace{0.17em}=\hspace{0.17em}-log}\left[{\text{OH}}^{\text{-}}\right]\\ \left[{\text{OH}}^{\text{-}}\right]{\text{=\hspace{0.17em}10}}^{\text{-pOH}}\\ \left[{\text{OH}}^{\text{-}}\right]{\text{=10}}^{\text{-11.32}}\end{array}$

_{0.240M$C{H}_{3}COOH$} since the ${\text{K}}_{\text{a}}$of ${\text{NH}}_{\text{3}}$and of are the same with the values of their and$\left[{\text{OH}}^{\text{-}}\right]$would interchange .The same thing would happen to their pH and pOH.

$\begin{array}{l}\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]{\text{=4.81×10}}^{\text{-12}}\text{M}\\ \left[{\text{OH}}^{\text{-}}\right]{\text{=2.08×10}}^{\text{-3}}\end{array}$

$\left[{\text{OH}}^{\text{-}}\right]{\text{=4.81×10}}^{\text{-12}}\text{M}$

Hence

$\begin{array}{l}\text{pH\hspace{0.17em}=\hspace{0.17em}2.68}\\ \text{pOH\hspace{0.17em}=\hspace{0.17em}11.32}\end{array}$