Chapter 12, Problem 6Q

Use the Henderson-Hasselbalch equation and Table 12.1 to calculate the pH of the following solutions:

1. a. 0.05 M formic acid and 0.1 M sodium formate.
2. b. 0.2 M ammonium chloride and 0.1 M aqueous ammonia.
3. c. 0.1 M acetic acid and 0.1 M sodium acetate.

Interpretation Introduction

Interpretation:

pH of the solution containing $0.05M$ formic acid and $0.1M$ sodium formate has to be calculated using Henderson-Hasselbalch equation.

Concept Introduction:

The $\text{pH}$ of a solution is a figure that expresses the acidity or the alkalinity of a given solution.

If the value of $\text{pH}$ is less than $7$, then the solution is acidic whereas if the value of $\text{pH}$ is greater than $7$, then the solution is basic.

Henderson-Hasselbalch equation:

Henderson-Hasselbalch equation explains the relationship between $pH$ of solution and $p{K}_a$ of acid. For a dissociation of acid $\left(HA\right)$ in aqueous solution,

  $HA+H_{2}O\underset{}{\overset{}{\rightleftharpoons }}{H}_3{O}^{+}+A^{-}$

  $pH=p{K}_a+\log \frac{\left[A^{-}\right]}{\left[HA\right]}$

Explanation of Solution

Given information are shown below,

  $\begin{array}{l}\left[\text{formic acid}\right]\text{ = 0}\text{.05 M}\\ \\ \left[\text{sodium formate}\right]\text{ = 0}\text{.1 M}\\ \\ {\text{pK}}_a\text{ = 3}\text{.8}\end{array}$

pH of the solution can be determined using Henderson-Hasselbalch equation as given,

  $\begin{array}{c}pH=p{K}_a+\log \frac{\left[A^{-}\right]}{\left[HA\right]}\\ \\ =p{K}_a+\log \frac{\left[\text{formate}\right]}{\left[\text{formic acid}\right]}\\ \\ =3.8+\log \frac{\left(0.1\text{ M}\right)}{\left(0.05\text{ M}\right)}\\ \\ =\text{ 3}\text{.8 }+\text{ 0}\text{.30}\\ \\ \text{= 4}\text{.1}\end{array}$

pH of the solution containing $0.05M$ formic acid and $0.1M$ sodium formate is $4.1$.

Interpretation Introduction

Interpretation:

pH of the solution containing $\text{0}\text{.2 M}$ ammonium chloride and $0.1M$ aqueous ammonia has to be calculated using Henderson-Hasselbalch equation.

Concept Introduction:

The $\text{pH}$ of a solution is a figure that expresses the acidity or the alkalinity of a given solution.

If the value of $\text{pH}$ is less than $7$, then the solution is acidic whereas if the value of $\text{pH}$ is greater than $7$, then the solution is basic.

Henderson-Hasselbalch equation:

Henderson-Hasselbalch equation explains the relationship between $pH$ of solution and $p{K}_a$ of acid. For a dissociation of acid $\left(HA\right)$ in aqueous solution,

  $HA+H_{2}O\underset{}{\overset{}{\rightleftharpoons }}{H}_3{O}^{+}+A^{-}$

  $pH=p{K}_a+\log \frac{\left[A^{-}\right]}{\left[HA\right]}$

Explanation of Solution

Given information are shown below,

  $\begin{array}{l}\left[\text{ammonium}\right]\text{ = 0}\text{.2 M}\\ \\ \left[\text{ammonia}\right]\text{ = 0}\text{.1 M}\\ \\ {\text{pK}}_a\text{ = 9}\text{.3}\end{array}$

pH of the solution can be determined using Henderson-Hasselbalch equation as given,

  $\begin{array}{c}pH=p{K}_a+\log \frac{\left[A^{-}\right]}{\left[HA\right]}\\ \\ =p{K}_a+\log \frac{\left[\text{ammonia}\right]}{\left[\text{ammonium}\right]}\\ \\ =9.3+\log \frac{\left(0.1\text{ M}\right)}{\left(0.2\text{ M}\right)}\\ \\ =\text{ 9}\text{.3 }-\text{ 0}\text{.30}\\ \\ \text{= 9}\text{.0}\end{array}$

pH of the solution containing $\text{0}\text{.2 M}$ ammonium chloride and $0.1M$ aqueous ammonia is $9.0$.

Interpretation Introduction

Interpretation:

pH of the solution containing $\text{0}\text{.1 M}$ acetic acid and $0.1M$ sodium acetate has to be calculated using Henderson-Hasselbalch equation.

Concept Introduction:

The $\text{pH}$ of a solution is a figure that expresses the acidity or the alkalinity of a given solution.

If the value of $\text{pH}$ is less than $7$, then the solution is acidic whereas if the value of $\text{pH}$ is greater than $7$, then the solution is basic.

Henderson-Hasselbalch equation:

Henderson-Hasselbalch equation explains the relationship between $pH$ of solution and $p{K}_a$ of acid. For a dissociation of acid $\left(HA\right)$ in aqueous solution,

  $HA+H_{2}O\underset{}{\overset{}{\rightleftharpoons }}{H}_3{O}^{+}+A^{-}$

  $pH=p{K}_a+\log \frac{\left[A^{-}\right]}{\left[HA\right]}$

Explanation of Solution

Given information are shown below,

  $\begin{array}{l}\left[aceticacid\right]\text{ = 0}\text{.1 M}\\ \\ \left[sodiumacetate\right]\text{ = 0}\text{.1 M}\\ \\ {\text{pK}}_a\text{ = 4}\text{.8}\end{array}$

pH of the solution can be determined using Henderson-Hasselbalch equation as given,

  $\begin{array}{c}pH=p{K}_a+\log \frac{\left[A^{-}\right]}{\left[HA\right]}\\ \\ =p{K}_a+\log \frac{\left[acetate\right]}{\left[aceticacid\right]}\\ \\ =9.3+\log \frac{\left(0.1\text{ M}\right)}{\left(0.1\text{ M}\right)}\\ \\ =\text{ 4}\text{.8 }-\text{ 0}\\ \\ \text{= 4}\text{.8}\end{array}$

pH of the solution containing $\text{0}\text{.1 M}$ acetic acid and $0.1M$ sodium acetate is $4.8$.