Question

A certain acetic acid solution has.$\mathit{p}\mathit{H}\mathbf{=}\mathbf{2}\mathbf{.68}$ Calculate the volume of 0.0975 M KOH required to “neutralize” 25.0 mL of this solution.

Short answer

The volume of 0.0975 M KOH required to “neutralize” 25.0 mL of this solution is${\mathit{V}}_{\mathbf{K}\mathbf{O}\mathbf{H}}\mathbf{\simeq }\mathbf{65}\mathit{m}\mathit{L}$ .

Step-by-step solution

The concentration of H+ ions. 

The ion concentration can be calculated as,

$\begin{array}{c}pH=-\log \left[H^{+}\right]\\ 2.68=-\log \left[H^{+}\right]\end{array}$

And,

$\begin{array}{l}\left[H^{+}\right]={10}^{-2.68}\\ \left[H^{+}\right]=2.1\times {10}^{-3}M\end{array}$

 Step 2: The value of a.

The dissociation reaction of acetic acid can be written as,

$C{H}_{3}COOH\left(aq\right)+H_{2}O\left(l\right)\rightleftharpoons H_3{O}^{+}\left(aq\right)+C{H}_{3}COO\left(aq\right)$

The expression of acid dissociation constant is,

$\begin{array}{l}{K}_a=\frac{x.x}{(a-x)}\\ K_a=\frac{x.x}{a}\end{array}$

The value of K_a for acetic acid is$1.8\times {10}^{-5}.$

On substituting the values,

$1.8\times {10}^{-5}=\frac{{(2.1\times {10}^{-3})}^2}{(a-2.1\times {10}^{-3})}$

$(a-2.1\times {10}^{-3})=2.5\times {10}^{-1}$

Hence,

$\begin{array}{l}a=0.25+0.0021\\ a=0.2521M\end{array}$

The volume of KOH. 

As,

$\begin{array}{c}{M}_{aceticacid}{V}_{aceticacid}=M_{KOH}{V}_{KOH}\\ 0.2521M\times 25mL=0.0975M\times V_{KOH}\end{array}$

Therefore,

$\begin{array}{l}{V}_{KOH}=\frac{0.2521M\times 25mL}{0.0975M}\\ V_{KOH}=64.64mL\\ V_{KOH}\simeq 65mL\end{array}$