Chapter 16, Problem 16.1P

Interpretation Introduction

Interpretation: The $\text{pH}$of a buffer that is $0.200\text{M}$in ${\text{HC}}_{\text{2}}{\text{H}}_{\text{3}}{\text{O}}_{\text{2}}$and $0.100\text{M}$in ${\text{NaC}}_{\text{2}}{\text{H}}_{\text{3}}{\text{O}}_{\text{2}}$is to be calculated.

Concept introduction: An aqueous solution of mixture of a weak acid and its conjugate base or mixture of weak base with its conjugate acid is known as the buffer solution. It resists in the change of the $\text{pH}$of a solution by neutralizing added amount of acid or base.

The acid in a buffer solution neutralizes the added base and the base present in buffer solution neutralizes the added acid.

To determine: The $\text{pH}$of a buffer that is $0.200\text{M}$in ${\text{HC}}_{\text{2}}{\text{H}}_{\text{3}}{\text{O}}_{\text{2}}$and $0.100\text{M}$in ${\text{NaC}}_{\text{2}}{\text{H}}_{\text{3}}{\text{O}}_{\text{2}}$ .

Answer to Problem 16.1P

Solution:

The $\text{pH}$of a buffer that is $0.200\text{M}$in ${\text{HC}}_{\text{2}}{\text{H}}_{\text{3}}{\text{O}}_{\text{2}}$and $0.100\text{M}$in ${\text{NaC}}_{\text{2}}{\text{H}}_{\text{3}}{\text{O}}_{\text{2}}$is $4.44$ .

Explanation of Solution

Given information: The concentration of ${\text{HC}}_{\text{2}}{\text{H}}_{\text{3}}{\text{O}}_{\text{2}}$is $0.200\text{M}$ .

The concentration of ${\text{NaC}}_{\text{2}}{\text{H}}_{\text{3}}{\text{O}}_{\text{2}}$is $0.100\text{M}$ .

The balanced equation for the ionization of acid ${\text{HC}}_{\text{2}}{\text{H}}_{\text{3}}{\text{O}}_{\text{2}}$in aqueous solution is,

${\text{HC}}_{\text{2}}{\text{H}}_{\text{3}}{\text{O}}_{\text{2}}\left(aq\right)+{\text{H}}_{\text{2}}\text{O}\left(l\right)\rightleftharpoons {\text{H}}_{\text{3}}{\text{O}}^{+}\left(aq\right)+{\text{C}}_{\text{2}}{\text{H}}_{\text{3}}{\text{O}}_{\text{2}}{}^{-}\left(aq\right)$

The concentration of ${\text{HC}}_{\text{2}}{\text{H}}_{\text{3}}{\text{O}}_{\text{2}}$is $0.200\text{M}$ .

The concentration of ${\text{NaC}}_{\text{2}}{\text{H}}_{\text{3}}{\text{O}}_{\text{2}}$or ${\text{C}}_{\text{2}}{\text{H}}_{\text{3}}{\text{O}}_{\text{2}}{}^{-}$is $0.100\text{M}$ .

The concentration of ${\text{H}}_{\text{3}}{\text{O}}^{+}$is supposed to be $x$ .

Thus, the concentration of ${\text{HC}}_{\text{2}}{\text{H}}_{\text{3}}{\text{O}}_{\text{2}}$after removal of ${\text{H}}_{\text{3}}{\text{O}}^{+}$ions is $0.200\text{M}-x$ .

The changed concentration of ${\text{C}}_{\text{2}}{\text{H}}_{\text{3}}{\text{O}}_{\text{2}}{}^{-}$after the addition of ${\text{H}}_{\text{3}}{\text{O}}^{+}$ is $0.100\text{M}+x$ .

The acid ionization constant is expressed as,

$K_{\text{a}}=\frac{\left[\text{C}\right]\left[\text{D}\right]}{\left[\text{A}\right]\left[\text{B}\right]}$

Where,

• $\left[\text{A}\right]$and $\left[\text{B}\right]$ are concentrations of reactants.
• $\left[\text{C}\right]$and $\left[\text{D}\right]$ are concentrations of products.
• $K_{\text{a}}$ is the acid ionization constant.
The standard value of $K_{\text{a}}$for acetic acid is $1.8\times {10}^{-5}$ .

Thus, for the given equation the acid ionization constant is expressed as,

$K_{\text{a}}=\frac{\left[{\text{H}}_{\text{3}}{\text{O}}^{+}\right]\left[{\text{C}}_{\text{2}}{\text{H}}_{\text{3}}{\text{O}}_{\text{2}}{}^{-}\right]}{\left[{\text{HC}}_{\text{2}}{\text{H}}_{\text{3}}{\text{O}}_{\text{2}}\right]}$

Substitute the respective values of concentrations and $K_{\text{a}}$in the above equation as,

$1.8\times {10}^{-5}=\frac{\left[x\right]\left[0.100\text{M}+x\right]}{\left[0.200\text{M}-x\right]}$

In the above expression, the value of $x$is very small. Thus, the value of $x$can be neglected from the above equation then the equation becomes as,

$\begin{array}{c}1.8\times {10}^{-5}=\frac{\left[x\right]\left[0.100\text{M}\right]}{\left[0.200\text{M}\right]}\\ x=\left(\frac{1.8\times {10}^{-5}}{0.5}\right)\\ x=3.6\times {10}^{-5}\end{array}$

Therefore, the concentration of ${\text{H}}_{\text{3}}{\text{O}}^{+}$is, $x=3.6\times {10}^{-5}$ .

The $\text{pH}$of the buffer is calculated by the expression.

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

Substitute the value of ${\text{H}}_{\text{3}}{\text{O}}^{+}$in the above expression as,

$\begin{array}{c}\text{pH}=-\log \left[3.6\times {10}^{-5}\right]\\ =4.44\end{array}$

Hence, the $\text{pH}$of a buffer that is $0.200\text{M}$in ${\text{HC}}_{\text{2}}{\text{H}}_{\text{3}}{\text{O}}_{\text{2}}$and $0.100\text{M}$in ${\text{NaC}}_{\text{2}}{\text{H}}_{\text{3}}{\text{O}}_{\text{2}}$is $4.44$ .

Conclusion

The $\text{pH}$of a buffer that is $0.200\text{M}$in ${\text{HC}}_{\text{2}}{\text{H}}_{\text{3}}{\text{O}}_{\text{2}}$and $0.100\text{M}$in ${\text{NaC}}_{\text{2}}{\text{H}}_{\text{3}}{\text{O}}_{\text{2}}$is $4.44$ .