Chapter 14, Problem 74E

A solution contains a mixture of acids: 0.50 M HA (K_a = 1 × 10⁻³), 0.20 M HB (K_a = 1.0 × 10⁻¹⁰), and 0.10 M HC (K_a = 1.0 × 10⁻¹²). Calculate the [H⁻] in this solution.

Interpretation Introduction

Interpretation: A solution prepared by mixing the given amount of ${\text{CH}}_{\text{3}}\text{COOH}$ and $\text{HCl}$ is given. The $\text{pH}$ and the acetate ion concentration of this solution is to be calculated.

Concept introduction: The $\text{pH}$ of a solution is define as a figure that expresses the acidity of the alkalinity of a given solution.

The $\text{pH}$ of a solution is calculated by the formula, $\text{pH}=-\log \left[{\text{H}}^{+}\right]$

The number of moles of a solute is calculated by the formula,

$\text{Moles}\text{of}\text{solute}=\text{Volume}\left(\text{L}\right)\times \text{Molarity}$

Answer to Problem 74E

The $\left[{\text{CH}}_3{\text{COO}}^{-}\right]$ is $\underset{\_}{1.11\times 10^{-3}M}$. The $\text{pH}$ of the given sample is $\underset{\_}{2.79}$.

The moles of ${\text{CH}}_3\text{COOH}$ are $\underset{\_}{0.01mol}$ and that of $\text{HCl}$ are $\underset{\_}{5.0\times 10^{-5}mol}$.

Explanation of Solution

Given

Volume of ${\text{CH}}_3\text{COOH}$ is $50.0\text{mL}\left(0.050\text{L}\right)$.

Concentration of ${\text{CH}}_3\text{COOH}$ is $0.200\text{M}$.

Volume of $\text{HCl}$ is $50.0\text{mL}\left(0.050\text{L}\right)$.

Concentration of $\text{HCl}$ is $1.00\times {10}^{-3}\text{M}$.

The number of moles of a solute is calculated by the formula,

$\text{Moles}\text{of}\text{solute}=\text{Volume}\left(\text{L}\right)\times \text{Molarity}$

Substitute the value of volume and molarity of ${\text{CH}}_3\text{COOH}$ and $\text{HCl}$ in the above expression.

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

$\begin{array}{c}\text{Moles}\text{of}{\text{CH}}_3\text{COOH}=0.050\text{L}\times \text{0}\text{.200}\text{M}\\ =\underset{\_}{0.01mol}\end{array}$

For $\text{HCl}$,

$\begin{array}{c}\text{Moles}\text{of}\text{HCl}=0.050\text{L}\times \left(\text{1}\text{.00}\times {\text{10}}^{-3}\right)\text{M}\\ =\underset{\_}{5.0\times 10^{-5}mol}\end{array}$

The $\left[{\text{H}}^{\text{+}}\right]$ from $\text{HCl}$ is $\underset{\_}{5.0\times 10^{-4}M}$.

The total volume of the solution is calculated by the formula,

$\text{Total}\text{volume}=\text{Volume}\text{of}{\text{CH}}_3\text{COOH}+\text{Volume}\text{of}\text{HCl}$

Substitute the value of the volume of ${\text{CH}}_3\text{COOH}$ and the volume of $\text{HCl}$ in the above expression.

$\begin{array}{c}\text{Total}\text{volume}=\left(\text{0}\text{.050}+\text{0}\text{.050}\right)\text{L}\\ =0.1L\end{array}$

$\text{HCl}$ is a strong acid. Therefore, it completely dissociates in water. Hence, the hydrogen ion concentration is equal to the concentration of $\text{HCl}$.

The concentration is calculated by the formula,

$\text{Concentration}=\frac{\text{Number}\text{of}\text{moles}}{\text{Volume}\left(\text{L}\right)}$

Substitute the value of the number of moles of $\text{HCl}$ and the volume in the above expression.

$\begin{array}{c}\text{Concentration}=\frac{\text{5}\text{.0}\times {\text{10}}^{-5}\text{mol}}{\text{0}\text{.10}\text{L}}\\ =\underset{\_}{5.0\times 10^{-4}M}\end{array}$

Therefore, the $\left[{\text{H}}^{\text{+}}\right]$ from $\text{HCl}$ is $\underset{\_}{5.0\times 10^{-4}M}$.

The initial $\left[{\text{CH}}_3\text{COOH}\right]$ is $\underset{\_}{0.1M}$.

The concentration is calculated by the formula,

$\text{Concentration}=\frac{\text{Number}\text{of}\text{moles}}{\text{Volume}\left(\text{L}\right)}$

Substitute the value of the number of moles of ${\text{CH}}_3\text{COOH}$ and the volume in the above expression.

$\begin{array}{c}\text{Concentration}=\frac{0.01\text{mol}}{\text{0}\text{.10}\text{L}}\\ =\underset{\_}{0.1M}\end{array}$

Therefore, the initial $\left[{\text{CH}}_3\text{COOH}\right]$ is $\underset{\_}{0.1M}$.

The equilibrium $\left[{\text{CH}}_3{\text{COO}}^{-}\right]$ is $\underset{\_}{1.1\times 10^{-3}M}$.

The change in concentration of ${\text{CH}}_3\text{COOH}$ is assumed to be $x$.

The ICE table for the stated reaction is,

$\begin{array}{cccccc}& {\text{CH}}_3\text{COOH}\left(aq\right)& \rightleftharpoons & {\text{H}}^{+}\left(aq\right)& +& {\text{CH}}_3{\text{COO}}^{-}\left(aq\right)\\ \text{Inititial}\text{concentration}& 0.1& & 5.0\times {10}^{-4}& & 0\\ \text{Change}& -x& & +x& & +x\\ \text{Equilibrium}\text{concentration}& 0.1-x& & \left(5.0\times {10}^{-4}\right)+x& & x\end{array}$

The equilibrium concentration of $\left[{\text{CH}}_3\text{COOH}\right]$ is $\left(0.1-x\right)\text{M}$.

The equilibrium concentration of $\left[{\text{H}}^{+}\right]$ is $\left(\left(5.0\times {10}^{-4}\right)+x\right)\text{M}$.

The equilibrium concentration of $\left[{\text{CH}}_3{\text{COO}}^{-}\right]$ is $x\text{M}$.

The $K_{\text{a}}$ for ${\text{CH}}_3\text{COOH}$ is $1.8\times {10}^{-5}$.

At equilibrium, the equilibrium constant expression is expressed by the formula,

$K_{\text{a}}=\frac{\text{Concentration}\text{of}\text{products}}{\text{Concentration}\text{of}\text{reactants}}$

Where,

• $K_{\text{a}}$ is the acid dissociation constant.

The equilibrium constant expression for the given reaction is,

$K_{\text{a}}=\frac{\left[{\text{H}}^{+}\right]\left[{\text{CH}}_3{\text{COO}}^{-}\right]}{\left[{\text{CH}}_3\text{COOH}\right]}$

Substitute the value of $K_{\text{a}}$, $\left[{\text{CH}}_3\text{COOH}\right]$, $\left[{\text{H}}^{+}\right]$ and $\left[{\text{CH}}_3{\text{COO}}^{-}\right]$ in the above expression.

$\begin{array}{c}1.8\times {10}^{-5}=\frac{\left[x\right]\left[\left(5.00\times {10}^{-4}\right)+x\right]}{\left[0.1-x\right]}\\ x=\underset{\_}{1.11\times 10^{-3}M}\end{array}$

Therefore, the equilibrium $\left[{\text{CH}}_3{\text{COO}}^{-}\right]$ is $\underset{\_}{1.1\times 10^{-3}M}$.

The equilibrium $\left[{\text{H}}^{+}\right]$ is $\underset{\_}{1.61\times 10^{-3}M}$.

The calculated value of $x$ is $1.1\times {10}^{-3}\text{M}$.

According to the ICE table formed,

The concentration of $\left[{\text{H}}^{+}\right]$ at equilibrium is calculated by the formula,

$\left[{\text{H}}^{+}\right]=\left(5.0\times {10}^{-4}\right)+x$

Substitute the calculated value of $x$ in the above expression.

$\begin{array}{c}\left[{\text{H}}^{+}\right]=\left(\left(5.0\times {10}^{-4}\right)+\left(1.1\times {10}^{-3}\right)\right)\text{M}\\ =\underset{\_}{1.61\times 10^{-3}M}\end{array}$

The $\text{pH}$ of the solution is $\underset{\_}{2.79}$.

The calculated value of $\left[{\text{H}}^{\text{+}}\right]$ is $\text{1}{\text{.61×10}}^{-\text{3}}\text{M}$.

The $\text{pH}$ of a solution is calculated by the formula,

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

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

$\begin{array}{c}\text{pH}=-\log \left[\text{1}{\text{.61×10}}^{-\text{3}}\right]\\ =\underset{\_}{2.79}\end{array}$

The $\text{pH}$ of the solution is $\underset{\_}{2.79}$.

Conclusion

The $\left[{\text{CH}}_3{\text{COO}}^{-}\right]$ is $\underset{\_}{1.11\times 10^{-3}M}$. The $\text{pH}$ of the given sample is $\underset{\_}{2.79}$.