Chapter 20, Problem 20.71P

Interpretation Introduction

Interpretation:

The $\text{pH}$ value of $\text{3}\text{.00M}$ acetic acid solution and the $\text{pH}$ value on further dilution of the acid have to be calculated.

Concept introduction:

$\text{pH}$: The concentration of hydrogen ion is measured using $\text{pH}$ scale. The $\text{pH}$ of a solution is a figure that expresses the acidity or the alkalinity of a given solution.

It is defined as the negative base-10 logarithm of the hydrogen or hydronium ion concentration.

  $\text{pH}\text{=}{\text{-log[H}}_{\text{3}}{\text{O}}^{\text{+}}\text{]}$

Answer to Problem 20.71P

The $\text{pH}$ value of $\text{3}\text{.00M}$ acetic acid is $\text{4}\text{.7}$.

The $\text{pH}$ value of diluted solution of $\text{3}\text{.00M}$ acetic acid is $\text{3}\text{.2}$.

Explanation of Solution

Given: $\text{3}{\text{.00M CH}}_{\text{3}}\text{COOH}\left(aq\right)$ solution in which $\text{2}\text{.00 mL}$ is diluted to make solution of $\text{250}\text{.0mL}$. ${\text{K}}_{\text{a}}$ value of acetic acid is $\text{1}{\text{.8×10}}^{\text{-5}}\text{ M}$.

First, the ${\text{pK}}_{\text{a}}$ value is calculated using given ${\text{K}}_{\text{a}}$ value.

    $\begin{array}{c}{\text{pK}}_{\text{a}}\text{= -log }\left[{\text{K}}_{\text{a}}\right]\\ \\ =\text{ -log }\left[\text{1}{\text{.8×10}}^{\text{-5}}\right]\\ \\ =\text{ 4}\text{.7}\end{array}$

The ICE table for reaction of ${\text{CH}}_{\text{3}}\text{COOH}\left(aq\right)$ with water is used to determine the $\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]$ concentration.

    $\begin{array}{l}{\text{CH}}_{\text{3}}\text{COOH}\left(aq\right)+{\text{ H}}_{\text{2}}\text{O}\left(l\right)\rightleftharpoons {\text{CH}}_{\text{3}}{\text{COO}}^{-}\left(aq\right){\text{ + H}}_{\text{3}}{\text{O}}^{\text{+}}\left(aq\right)\\ \\ Initial3\text{ 0 0}\\ Change\text{ -x +x +x}\\ Equilibrium3\text{-x x x}\\ \\ {\text{K}}_{\text{a}}\text{ = }\frac{\left[{\text{CH}}_{\text{3}}{\text{COO}}^{-}\right]\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]}{\left[{\text{CH}}_{\text{3}}\text{COOH}\right]}\end{array}$

    $\begin{array}{c}{\text{K}}_{\text{a}}\text{ = }\frac{\left[{\text{CH}}_{\text{3}}{\text{COO}}^{-}\right]\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]}{\left[{\text{CH}}_{\text{3}}\text{COOH}\right]}\\ \\ 1.8\times {10}^{-5}\text{ M}=\frac{x^2}{\text{3-}x}\\ \\ x=7.35\times {10}^{-3}\text{M}\\ \\ \left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]\text{ = }7.35\times {10}^{-3}\text{M}\end{array}$

    $\begin{array}{c}\text{pH = -log }\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]\\ \\ =\text{ -log }\left[7.35\times {10}^{-3}\right]\\ \\ =\text{ 2}\text{.1}\end{array}$

Next, the concentration for the diluted solution is determined by using ${\text{N}}_{\text{1}}{\text{V}}_{\text{1}}{\text{ = N}}_{\text{2}}{\text{V}}_{\text{2}}$.

    $\begin{array}{c}{\text{N}}_{\text{1}}{\text{V}}_{\text{1}}{\text{ = N}}_{\text{2}}{\text{V}}_{\text{2}}\\ \\ \text{ 3}\text{.00M}\times \text{2}\text{.00mL}={\text{ N}}_{\text{2}}\times 250\text{mL}\\ \\ {\text{N}}_{\text{2}}\text{= }\frac{\text{3}\text{.00}\times \text{2}\text{.00mL}}{250\text{mL}}\\ \\ =\text{ 0}\text{.024M}\end{array}$

The ICE table for reaction of diluted ${\text{CH}}_{\text{3}}\text{COOH}\left(aq\right)$ with water is used to determine the $\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]$ concentration.

    $\begin{array}{l}{\text{CH}}_{\text{3}}\text{COOH}\left(aq\right)+{\text{ H}}_{\text{2}}\text{O}\left(l\right)\rightleftharpoons {\text{CH}}_{\text{3}}{\text{COO}}^{-}\left(aq\right){\text{ + H}}_{\text{3}}{\text{O}}^{\text{+}}\left(aq\right)\\ \\ Initial0.024\text{ 0 0}\\ Change\text{ -x +x +x}\\ Equilibrium0.024\text{-x x x}\\ \\ {\text{K}}_{\text{a}}\text{ = }\frac{\left[{\text{CH}}_{\text{3}}{\text{COO}}^{-}\right]\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]}{\left[{\text{CH}}_{\text{3}}\text{COOH}\right]}\end{array}$

    $\begin{array}{c}{\text{K}}_{\text{a}}\text{ = }\frac{\left[{\text{CH}}_{\text{3}}{\text{COO}}^{-}\right]\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]}{\left[{\text{CH}}_{\text{3}}\text{COOH}\right]}\\ \\ 1.8\times {10}^{-5}\text{ M}=\frac{x^2}{\text{0}\text{.024-}x}\\ \\ x=6.57\times {10}^{-4}\text{M}\\ \\ \left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]\text{ = }6.57\times {10}^{-4}\text{M}\end{array}$

    $\begin{array}{c}\text{pH = -log }\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]\\ \\ =\text{ -log }\left[6.57\times {10}^{-4}\right]\\ \\ =\text{ 3}\text{.2}\end{array}$