Chapter 14, Problem 68E

A solution is prepared by dissolving 0.56 g benzoic acid (C₆H₅CO₂H, K_a = 6.4 × 10⁻⁵) in enough water to make 1.0 L of solution. Calculate [C₆H₅CO₂H], [C₆H₅CO₂⁻], [H⁺], [OH⁻], and the pH of this solution.

Interpretation Introduction

Interpretation: The concentration of $\left[{\text{C}}_{\text{6}}{\text{H}}_{\text{5}}\text{COOH}\right]$, $\left[{\text{C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{CO}}_2^{-}\right]$, $\left[{\text{H}}^{+}\right]$, $\left[{\text{OH}}^{-}\right]$ and the $\text{pH}$ of the solution, prepared by dissolving $0.56\text{g}$ of benzoic acid in enough water o make $1.0\text{L}$ of the 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, $\text{pH}=-\log \left[{\text{H}}^{+}\right]$

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

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

The equilibrium constant for water is denoted by $K_{\text{w}}$ and is expressed as,

$K_{\text{w}}=\left[{\text{H}}^{+}\right]\left[{\text{OH}}^{-}\right]$

The percent dissociation of an acid is calculated by the formula,

$\text{Percent}\text{dissociation}=\frac{\text{Equilibrium}\text{concentration}\text{of}\left[{\text{H}}^{+}\right]}{\text{Initial}\text{concentration}\text{of}\text{the}\text{acid}}\times 100$

Answer to Problem 68E

Answer

The $\left[{\text{H}}^{\text{+}}\right]$ and $\left[{\text{C}}_6{\text{H}}_{\text{5}}{\text{COO}}^{-}\right]$ is $\underset{\_}{5.4\times 10^{-4}M}$. The $\left[{\text{C}}_6{\text{H}}_{\text{5}}\text{COOH}\right]$ is $\underset{\_}{4.1\times 10^{-3}M}$. The $\left[{\text{OH}}^{-}\right]$ is $\underset{\_}{1.8\times 10^{-11}M}$. The $\text{pH}$ of the solution is $\underset{\_}{3.27}$.

Explanation of Solution

Explanation

To determine: The concentration of $\left[{\text{C}}_{\text{6}}{\text{H}}_{\text{5}}\text{COOH}\right]$, $\left[{\text{C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{CO}}_2^{-}\right]$, $\left[{\text{H}}^{+}\right]$, $\left[{\text{OH}}^{-}\right]$ and the $\text{pH}$ of the solution.

The dissociation of ${\text{C}}_{\text{6}}{\text{H}}_{\text{5}}\text{COOH}$ is, ${\text{C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{COOH}}_{\left(\text{aq}\right)}\rightleftharpoons {\text{ H}}^{+}{}_{\left(\text{aq}\right)}{\text{ + C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{COO}}_{{}^{\text{-}}\left(\text{aq}\right)}$

${\text{C}}_{\text{6}}{\text{H}}_{\text{5}}\text{COOH}$ is a weak acid. Hence, it does not completely dissociate in water.

${\text{C}}_{\text{6}}{\text{H}}_{\text{5}}\text{COOH}$ is a comparatively stronger acid than ${\text{H}}_{\text{2}}\text{O}$.

The dominant equilibrium reaction for the given case is,

${\text{C}}_{\text{6}}{\text{H}}_{\text{5}}\text{COOH}\left(aq\right)\rightleftharpoons {\text{H}}^{+}\left(aq\right)+{\text{C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{COO}}^{-}\left(aq\right)$

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,

$K_{\text{a}}=\frac{\left[{\text{H}}^{+}\right]\left[{\text{C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{COO}}^{-}\right]}{\left[{\text{C}}_{\text{6}}{\text{H}}_{\text{5}}\text{COOH}\right]}$ (1)

The number of moles of ${\text{C}}_{\text{6}}{\text{H}}_{\text{5}}\text{COOH}$ is $\underset{\_}{4.59\times 10^{-3}mol}$.

Given

The given mass of benzoic acid $\left({\text{C}}_{\text{6}}{\text{H}}_{\text{5}}\text{COOH}\right)$ is $0.56\text{g}$.

The molar mass of ${\text{C}}_{\text{6}}{\text{H}}_{\text{5}}\text{COOH}$ $\begin{array}{l}=7\text{C}+\text{6H}+\text{2O}\\ =\left(\left(7\times 12\right)+\left(6\times 1\right)+\left(2\times 16\right)\right)\text{g/mol}\\ =122\text{g/mol}\end{array}$

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

$\text{Number}\text{of}\text{moles}=\frac{\text{Given}\text{mass}}{\text{Molar}\text{mass}}$

Substitute the value of the given mass and the molar mass of ${\text{C}}_{\text{6}}{\text{H}}_{\text{5}}\text{COOH}$ in the above expression.

$\begin{array}{c}\text{Number}\text{of}\text{moles}=\frac{\text{0}\text{.56}\text{g}}{\text{122}\text{g/mol}}\\ =\underset{\_}{4.59\times 10^{-3}mol}\end{array}$

The initial $\left[{\text{C}}_{\text{6}}{\text{H}}_{\text{5}}\text{COOH}\right]$ is $\underset{\_}{0.0046M}$.

The calculated number of moles of ${\text{C}}_{\text{6}}{\text{H}}_{\text{5}}\text{COOH}$ is $4.59\times {10}^{-3}\text{mol}$.

The total volume is $1.0\text{L}$.

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 ${\text{C}}_{\text{6}}{\text{H}}_{\text{5}}\text{COOH}$ and the volume in the above expression.

$\begin{array}{c}\text{Concentration}=\frac{\text{4}\text{.59}\times {\text{10}}^{-3}\text{mol}}{\text{1}\text{L}}\\ =\underset{\_}{0.0046M}\end{array}$

The $\left[{\text{H}}^{\text{+}}\right]$ and $\left[{\text{C}}_6{\text{H}}_{\text{5}}{\text{COO}}^{-}\right]$ is $\underset{\_}{5.4\times 10^{-4}M}$

The change in concentration of ${\text{C}}_{\text{6}}{\text{H}}_{\text{5}}\text{COOH}$ is assumed to be $x$.

The ICE table for the stated reaction is,

$\begin{array}{cccccc}& {\text{C}}_{\text{6}}{\text{H}}_{\text{5}}\text{COOH}\left(aq\right)& \rightleftharpoons & {\text{H}}^{+}\left(aq\right)& +& {\text{C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{COO}}^{-}\left(aq\right)\\ \text{Inititial}\text{concentration}& 0.0046& & 0& & 0\\ \text{Change}& -x& & +x& & +x\\ \text{Equilibrium}\text{concentration}& 0.0046-x& & x& & x\end{array}$

The equilibrium concentration of $\left[{\text{C}}_{\text{6}}{\text{H}}_{\text{5}}\text{COOH}\right]$ is $\left(0.0046-x\right)\text{M}$.

The equilibrium concentration of $\left[{\text{H}}^{+}\right]$ is $x\text{M}$.

The equilibrium concentration of $\left[{\text{C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{COO}}^{-}\right]$ is $x\text{M}$.

The $K_{\text{a}}$ for ${\text{C}}_{\text{6}}{\text{H}}_{\text{5}}\text{COOH}$ is $6.4\times {10}^{-5}$.

Substitute the value of $K_{\text{a}}$, $\left[{\text{C}}_{\text{6}}{\text{H}}_{\text{5}}\text{COOH}\right]$, $\left[{\text{H}}^{+}\right]$ and $\left[{\text{C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{COO}}^{-}\right]$ in equation (1).

$\begin{array}{c}6.4\times {10}^{-5}=\frac{\left[x\right]\left[x\right]}{\left[0.0046-x\right]}\\ 6.4\times {10}^{-5}=\frac{{\left[x\right]}^2}{\left[0.0046-x\right]}\end{array}$

The value of $x$ will be very small as compared to $0.0046$. Hence, it is ignored from the term $\left[0.0046-x\right]$.

Simplify the above expression.

$\begin{array}{c}6.4\times {10}^{-5}=\frac{{\left[x\right]}^2}{\left[0.0046\right]}\\ {\left[x\right]}^2=\left(2.9\times {10}^{-7}\right)\\ \left[x\right]=\underset{\_}{5.4\times 10^{-4}M}\end{array}$

Therefore, the $\left[{\text{H}}^{\text{+}}\right]$ and $\left[{\text{C}}_6{\text{H}}_{\text{5}}{\text{COO}}^{-}\right]$, that is equal to $x$ is $\underset{\_}{5.4\times 10^{-4}M}$.

The $\left[{\text{C}}_6{\text{H}}_{\text{5}}\text{COOH}\right]$ is $\underset{\_}{4.1\times 10^{-3}M}$.

According to the ICE table formed,

The equilibrium concentration of ${\text{C}}_6{\text{H}}_{\text{5}}\text{COOH}$ is calculated by the formula,

$\left[{\text{C}}_6{\text{H}}_{\text{5}}\text{COOH}\right]=0.0046-x$

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

$\begin{array}{c}\left[{\text{C}}_6{\text{H}}_{\text{5}}\text{COOH}\right]=\left(0.0046-5.4\times {10}^{-4}\right)\text{M}\\ =\underset{\_}{4.1\times 10^{-3}M}\end{array}$

The $\left[{\text{OH}}^{-}\right]$ is $\underset{\_}{1.8\times 10^{-11}M}$.

The equilibrium constant for water is denoted by $K_{\text{w}}$ and is expressed as,

$K_{\text{w}}=\left[{\text{H}}^{+}\right]\left[{\text{OH}}^{-}\right]$

The value of $K_{\text{w}}$ is $1.0\times {10}^{-14}$.

The $\left[{\text{H}}^{\text{+}}\right]$ is $\text{5}{\text{.4×10}}^{\text{-4}}\text{M}$.

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

$\begin{array}{c}1.0\times {10}^{-14}=\left[\text{5}{\text{.4×10}}^{\text{-4}}\text{M}\right]\left[{\text{OH}}^{-}\right]\\ \left[{\text{OH}}^{-}\right]=\frac{1.0\times {10}^{-14}}{\left[\text{5}{\text{.4×10}}^{-\text{4}}\text{M}\right]}\\ =\underset{\_}{1.8\times 10^{-11}M}\end{array}$

The $\text{pH}$ of the solution is $\underset{\_}{3.27}$.

The calculated value of $\left[{\text{H}}^{\text{+}}\right]$ is $\text{5}{\text{.4×10}}^{-\text{4}}\text{M}$.

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.

$\begin{array}{c}\text{pH}=-\log \left[\text{5}{\text{.4×10}}^{-\text{4}}\right]\\ =\underset{\_}{3.27}\end{array}$

Conclusion

Conclusion

The $\left[{\text{H}}^{\text{+}}\right]$ and $\left[{\text{C}}_6{\text{H}}_{\text{5}}{\text{COO}}^{-}\right]$ is $\underset{\_}{5.4\times 10^{-4}M}$.

The $\left[{\text{C}}_6{\text{H}}_{\text{5}}\text{COOH}\right]$ is $\underset{\_}{4.1\times 10^{-3}M}$.

The $\left[{\text{OH}}^{-}\right]$ is $\underset{\_}{1.8\times 10^{-11}M}$.

The $\text{pH}$ of the solution is $\underset{\_}{3.27}.$