Chapter 17, Problem 5QP

Determine the pH of (a) a $0.40M{\text{ CH}}_{\text{3}}\text{COOH}$ solution and (b) a solution that is $0.40M{\text{ CH}}_{\text{3}}\text{COOH}$ and $0.20M{\text{ CH}}_{\text{3}}\text{COONa}\text{.}$.

Interpretation Introduction

Interpretation:

The $\text{pH}$ of $0.4\text{M}$ concentration of ${\text{CH}}_3\text{COOH}$ solution and $0.2\text{M}$ concentration of ${\text{CH}}_3\text{COOH}$ in $0.2\text{M}$ concentration of ${\text{CH}}_3\text{COONa}$ are to be determined.

Concept introduction:

The $\text{pH}$

measures the concentration of hydronium ions in a solution. The solution with high concentration of hydronium ions has a low $\text{pH}$ value and the solution with low concentration of hydronium ions has a high $\text{pH}$ value.

The $\text{pH}$ of the solution is calculated using the expression as follows:

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

The ionization of the weak acid takes place as:

$\text{HA}\left(aq\right)\rightleftharpoons {\text{H}}^{\text{+}}\left(aq\right)+{\text{A}}^{-}\left(aq\right)$

$K_{\text{a}}$ is the measure of ionization of an acid and is known as acid–ionization constant, which is specific at a particular temperature:

$K_{\text{a}}=\frac{\left[{\text{H}}^{+}\right]\left[{\text{A}}^{-}\right]}{\left[\text{HA}\right]}$

Answer to Problem 5QP

Solution:

a)

$2.57$

b)

$4.44$

Explanation of Solution

a) $\text{0}\text{.40 M}$

${\text{CH}}_{\text{3}}\text{COOH}$ the solution.

${\text{CH}}_{\text{3}}\text{COOH}$ is a weak acid.

Summarize the concentration at equilibrium as follows.

Consider $x$ to be the degree of dissociation.

$\begin{array}{l}{\text{ CH}}_{\text{3}}\text{COOH}\left(aq\right)\rightleftharpoons {\text{CH}}_{\text{3}}{\text{COO}}^{-}\left(aq\right)+{\text{H}}^{+}\left(aq\right)\\ \begin{array}{cccc}\text{Initial}(\text{M})& 0.4& 0& 0\\ \text{Change}(\text{M})& 0.4-x& +x& +x\\ \text{Equilibrium}(\text{M})& 0.4-x& x& x\end{array}\end{array}$

The equilibrium expression for a reaction is written as follows:

$K_{\text{a}}=\frac{\left[{\text{H}}^{+}\right]\left[{\text{A}}^{-}\right]}{\left[\text{HA}\right]}$

$K_{\text{a}}=\frac{{\text{[CH}}_{\text{3}}{\text{COO}}^{-}{\text{][H}}^{+}\text{]}}{{\text{[CH}}_{\text{3}}\text{OOH]}}$

Here, $K_{\text{a}}$ is the equilibrium constant, ${\text{[CH}}_{\text{3}}{\text{COO}}^{-}\text{]}$

is the concentration of acetate ion, ${\text{[CH}}_{\text{3}}\text{COOH]}$

is the concentration of acetic acid, and ${\text{[H}}^{+}\text{]}$ is the concentration of hydrogen ion.

Substitute the value of ${\text{[CH}}_{\text{3}}{\text{COO}}^{-}\text{]}$, the value of ${\text{[CH}}_{\text{3}}\text{COOH]}$, the value of ${\text{[H}}^{+}\text{]}$, and the $K_{\text{a}}$

value of ${\text{CH}}_{\text{3}}\text{COOH}$ as $1.8\times {10}^{-5}$ in the above expression,

$1.8\times {10}^{-5}=\frac{x^2}{(0.4-x)}$

The value of $x$ is very small as compared to $0.4$. It can be neglected.

$\text{(0}\text{.4}-\text{x)}\approx \text{0}\text{.4}$

$\begin{array}{l}1.8\times {10}^{-5}=\frac{x^2}{0.4}\\ x=2.7\times {10}^{-3}\text{M}\end{array}$

Concentration of ${\text{[H}}^{+}\text{] }=2.7\times {10}^{-3}\text{ M}$

The $\text{pH}$ of the solution is calculated using the expression as follows:

$\text{pH}=-{\text{log([H}}^{+}\text{]) }$

Substitute the value of ${\text{[H}}^{+}\text{]}$ in the above expression,

$\begin{array}{l}\text{pH}=-\text{log(2}\text{.7}\times {\text{10}}^{-3}\text{)}\\ \text{=}\text{2}\text{.57}\end{array}$

Hence, the $\text{pH}$ of the solution is 2.57.

b) A solution that is $\text{0}{\text{.40 M CH}}_{\text{3}}\text{COOH}$ and the $\text{0}{\text{.2M CH}}_{\text{3}}\text{COONa}$.

The acetate ions are formed from sodium acetate on dilution and from acetic acid by ionisation.

The equation for the ionisation of sodium acetate ion is as follows:

${\text{CH}}_{\text{3}}\text{COONa}(aq)\to {\text{CH}}_{\text{3}}{\text{COO}}^{-}(aq)+{\text{Na}}^{+}(aq)$

Here, acetate ion is formed on dilution in the solution. So, the concentration of the acetate ion from sodium acetate is 0.2 M and the sodium ion further does not take part in the reaction.

${\text{CH}}_{\text{3}}\text{COOH}$ is a weak acid.

Summarize the concentration at equilibrium as follows:

Consider $x$ to be the degree of dissociation.

$\begin{array}{l}{\text{CH}}_{\text{3}}\text{COOH}(aq)\rightleftharpoons {\text{CH}}_{\text{3}}{\text{COO}}^{-}(aq)+{\text{H}}^{+}(aq)\\ \begin{array}{cccc}\text{Initial(M)}& 0.4& \text{}\text{}\text{}\text{}\text{}\text{}\text{}0& 0.2\\ \text{Change}(\text{M})& (0.4-x)& +x& (0.2+x)\\ \text{Equilibrium}(\text{M})& (0.4-x)& +x& (0.2+x)\end{array}\end{array}$

The equilibrium expression for a reaction is written as:

$K_{\text{a}}=\frac{\left[{\text{H}}^{+}\right]\left[{\text{A}}^{-}\right]}{\left[\text{HA}\right]}$

$K_{\text{a}}=\frac{{\text{[CH}}_{\text{3}}{\text{COO}}^{-}{\text{][H}}^{+}\text{]}}{{\text{[CH}}_{\text{3}}\text{OOH]}}$

Here, ${\text{K}}_{\text{a}}$ is the equilibrium constant, ${\text{[CH}}_{\text{3}}{\text{COO}}^{-}\text{]}$ is the concentration of acetate ion, ${\text{[CH}}_{\text{3}}\text{COOH]}$ is the concentration of acetic acid, and ${\text{[H}}^{+}\text{]}$ is the concentration of hydrogen ion.

Substitute the value of ${\text{[CH}}_{\text{3}}{\text{COO}}^{-}\text{]}$, the value of ${\text{[CH}}_{\text{3}}\text{COOH]}$, the value of ${\text{[H}}^{+}\text{]}$, and the $K_{\text{a}}$

value of ${\text{CH}}_{\text{3}}\text{COOH}$ as $1.8\times {10}^{-5}$ in the above expression.

$1.8\times {10}^{-5}=\frac{x\left(0.2+x\right)}{(0.4-x)}$

The value of $x$ is very small as compared to $0.2$ and $0.4$. It can be neglected.

$\begin{array}{l}\text{(0}\text{.4}-x\text{)}\approx \text{0}\text{.4}\\ \text{(0}\text{.2}-x\text{)}\approx \text{0}\text{.2}\end{array}$

$\begin{array}{l}1.8\times {10}^{-5}=\frac{x\left(0.2\right)}{0.4}\\ x=3.6\times {10}^{-5}\text{M}\end{array}$

Concentration of ${\text{[H}}^{+}\text{] }=\text{3}{\text{.6×10}}^{-\text{5}}\text{ M}$

The $\text{pH}$ of the solution is calculated using the expression as follows:

$\text{pH}=-{\text{log([H}}^{+}\text{]) }$

Substitute the value of ${\text{[H}}^{+}\text{]}$ in the above expression,

$\begin{array}{l}\text{pH}=-\text{log(3}\text{.5}\times {\text{10}}^{-5}\text{)}\\ \text{=}4.\text{44}\end{array}$

Hence, the $\text{pH}$ of the solution is $4.44$.