Chapter 16, Problem 16.114QP

Sodium benzoate is a salt of benzoic acid, C₆H₅COOH. A 0.15 M solution of this salt has a pOH of 5.31 at room temperature.

1. a Calculate the value for the equilibrium constant for the reaction

${\text{C}}_6{\text{H}}_5{\text{COO}}^{-}+{\text{H}}_2\text{O}\rightleftharpoons {\text{C}}_6{\text{H}}_5\text{COOH}+{\text{OH}}^{-}$

1. b What is the K_a value for benzoic acid?
2. c Benzoic acid has a low solubility in water. What is its molar solubility if a saturated solution has a pH of 2.83 at room temperature?

Interpretation Introduction

Interpretation:

A 0.15 M solution of sodium benzoate has a pOH of 5.31

The value for the equilibrium constant for the  given reaction ${\text{C}}_{\text{6}}{\text{H}}_5{\text{COO}}^{-}{\text{ + H}}_{\text{2}}\text{O }\stackrel{}{\rightleftharpoons }{\text{ C}}_{\text{6}}{\text{H}}_5{\text{COOH + OH}}^{-}$ has to be calculated

Concept Introduction:

Acid ionization constant ${\text{K}}_{\text{a}}$:

The ionization of a weak acid $\text{HA}$ can be given as follows,

${\text{HA}}_{(aq)}\rightleftarrows {\text{H}}^{+}{}_{(aq)}+{\text{A}}^{\text{-}}{}_{(aq)}$

The equilibrium expression for the above reaction is given below.

$K_a=\frac{[{\text{H}}^{\text{+}}{\text{][A}}^{\text{-}}\text{]}}{[\text{HA]}}$

Where,

$K_a$ is acid ionization constant,

$[{\text{H}}^{\text{+}}\text{]}$  is concentration of hydrogen ion

$[{\text{A}}^{\text{-}}\text{]}$  is concentration of acid anion

$[\text{HA]}$ is concentration of the acid

Relationship between ${\text{K}}_a$and ${\text{K}}_b$:

${\text{K}}_a\times {\text{ K}}_b{\text{ = K}}_w$

pH definition:

The pH of a solution is defined as the negative base-10 logarithm of the hydronium ion ${\text{[H}}_{\text{3}}{\text{O}}^{+}\text{]}$ concentration.

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

On rearranging,

${\text{[H}}_{\text{3}}{\text{O}}^{+}\text{]}\text{=}{10}^{-\text{pH}}$

Answer to Problem 16.114QP

The value for the equilibrium constant for the given reaction ${\text{C}}_{\text{6}}{\text{H}}_5{\text{COO}}^{-}{\text{ + H}}_{\text{2}}\text{O }\stackrel{}{\rightleftharpoons }{\text{ C}}_{\text{6}}{\text{H}}_5{\text{COOH + OH}}^{-}$ is $K_b=1.6\times 10^{-10}$

Explanation of Solution

To Calculate: The value for the equilibrium constant for the given reaction ${\text{C}}_{\text{6}}{\text{H}}_5{\text{COO}}^{-}{\text{ + H}}_{\text{2}}\text{O }\stackrel{}{\rightleftharpoons }{\text{ C}}_{\text{6}}{\text{H}}_5{\text{COOH + OH}}^{-}$

Given data:

Sodium benzoate is a salt of benzoic acid ${\text{(C}}_{\text{6}}{\text{H}}_5\text{COOH)}$

A 0.15 M solution of this salt has a pOH of 5.31

The given reaction is ${\text{C}}_{\text{6}}{\text{H}}_5{\text{COO}}^{-}{\text{ + H}}_{\text{2}}\text{O }\stackrel{}{\rightleftharpoons }{\text{ C}}_{\text{6}}{\text{H}}_5{\text{COOH + OH}}^{-}$

Equilibrium constant for the given reaction:

The hydroxide ion concentration is found from the given pOH as follows,

$\begin{array}{l}{\text{[OH}}^{-}\text{]}\text{=}{10}^{-\text{pOH}}\\ ={10}^{-5.31}\\ =4.90\times {10}^{-6}\text{ M}\end{array}$

Construct an equilibrium table for the given reaction:

The initial concentration of ${\text{C}}_{\text{6}}{\text{H}}_5{\text{COO}}^{-}$ is considered from the concentration of sodium benzoate salt (0.15 M)

	${\text{C}}_{\text{6}}{\text{H}}_5{\text{COO}}^{-}{\text{ + H}}_{\text{2}}\text{O }\stackrel{}{\rightleftharpoons }{\text{ C}}_{\text{6}}{\text{H}}_5{\text{COOH + OH}}^{-}$
Initial $(M)$	0.15 $-x$ 0.15-x	0.00	0.00
Change $(M)$		$+x$	$+x$
Equilibrium $(M)$		x	x

Here, x gives the concentration of hydroxide ion that reacted, $x=4.90\times {10}^{-6}\text{ M}$

Substitute the equilibrium concentrations into the equilibrium-constant expression:

$\begin{array}{l}{\text{K}}_b=\frac{[{\text{OH}}^{-}{\text{][C}}_{\text{6}}{\text{H}}_5\text{COOH]}}{{\text{[C}}_{\text{6}}{\text{H}}_5{\text{COO}}^{-}\text{]}}\\ =\frac{{\text{(}x\text{)}}^2}{(0.15-x)}\\ \text{Assume }x\text{ is small compared to 0}\text{.15 and neglect it}\text{.}\\ =\frac{{\text{(4}\text{.90}\times {\text{10}}^{-6}\text{)}}^2}{(0.15)}\\ =1.60\times {10}^{-10}\end{array}$

Therefore, the equilibrium-constant for the given reaction is $1.6\times {10}^{-10}$

Conclusion

The value for the equilibrium constant for the given reaction ${\text{C}}_{\text{6}}{\text{H}}_5{\text{COO}}^{-}{\text{ + H}}_{\text{2}}\text{O }\stackrel{}{\rightleftharpoons }{\text{ C}}_{\text{6}}{\text{H}}_5{\text{COOH + OH}}^{-}$ was calculated as $K_b=1.6\times 10^{-10}$

Interpretation Introduction

Interpretation:

A 0.15 M solution of sodium benzoate has a pOH of 5.31

The ${\text{K}}_a$ value for benzoic acid has to be calculated

Concept Introduction:

Acid ionization constant ${\text{K}}_{\text{a}}$:

The ionization of a weak acid $\text{HA}$ can be given as follows,

${\text{HA}}_{(aq)}\rightleftarrows {\text{H}}^{+}{}_{(aq)}+{\text{A}}^{\text{-}}{}_{(aq)}$

The equilibrium expression for the above reaction is given below.

$K_a=\frac{[{\text{H}}^{\text{+}}{\text{][A}}^{\text{-}}\text{]}}{[\text{HA]}}$

Where,

$K_a$ is acid ionization constant,

$[{\text{H}}^{\text{+}}\text{]}$  is concentration of hydrogen ion

$[{\text{A}}^{\text{-}}\text{]}$  is concentration of acid anion

$[\text{HA]}$ is concentration of the acid

Relationship between ${\text{K}}_a$and ${\text{K}}_b$:

${\text{K}}_a\times {\text{ K}}_b{\text{ = K}}_w$

pH definition:

The pH of a solution is defined as the negative base-10 logarithm of the hydronium ion ${\text{[H}}_{\text{3}}{\text{O}}^{+}\text{]}$ concentration.

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

On rearranging,

${\text{[H}}_{\text{3}}{\text{O}}^{+}\text{]}\text{=}{10}^{-\text{pH}}$

Answer to Problem 16.114QP

The ${\text{K}}_a$ value for benzoic acid is $6.3\times 10^{-5}$

Explanation of Solution

To Calculate: The ${\text{K}}_a$ value for benzoic acid

The ${\text{K}}_a$ value for benzoic acid is calculated using ${\text{K}}_w$ as follows,

$\begin{array}{l}{\text{K}}_a\times {\text{K}}_b{\text{= K}}_w\\ {\text{ K}}_a=\frac{{\text{K}}_w}{{\text{K}}_b}\\ =\frac{1.0\times {10}^{-14}}{1.6\times {10}^{-10}}\\ =6.3\times {10}^{-5}\end{array}$

Conclusion

The ${\text{K}}_a$ value for benzoic acid was calculated as $6.3\times 10^{-5}$

Interpretation Introduction

Interpretation:

A 0.15 M solution of sodium benzoate has a pOH of 5.31

Benzoic acid has a low solubility in water. If a saturated solution has a pH of 2.83, the molar solubility of the solution has to be calculated.

Concept Introduction:

Acid ionization constant ${\text{K}}_{\text{a}}$:

The ionization of a weak acid $\text{HA}$ can be given as follows,

${\text{HA}}_{(aq)}\rightleftarrows {\text{H}}^{+}{}_{(aq)}+{\text{A}}^{\text{-}}{}_{(aq)}$

The equilibrium expression for the above reaction is given below.

$K_a=\frac{[{\text{H}}^{\text{+}}{\text{][A}}^{\text{-}}\text{]}}{[\text{HA]}}$

Where,

$K_a$ is acid ionization constant,

$[{\text{H}}^{\text{+}}\text{]}$  is concentration of hydrogen ion

$[{\text{A}}^{\text{-}}\text{]}$  is concentration of acid anion

$[\text{HA]}$ is concentration of the acid

Relationship between ${\text{K}}_a$and ${\text{K}}_b$:

${\text{K}}_a\times {\text{ K}}_b{\text{ = K}}_w$

pH definition:

The pH of a solution is defined as the negative base-10 logarithm of the hydronium ion ${\text{[H}}_{\text{3}}{\text{O}}^{+}\text{]}$ concentration.

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

On rearranging,

${\text{[H}}_{\text{3}}{\text{O}}^{+}\text{]}\text{=}{10}^{-\text{pH}}$

Answer to Problem 16.114QP

Benzoic acid has a low solubility in water. If a saturated solution has a pH of 2.83, the molar solubility of the solution is 0.036 M

Explanation of Solution

To Calculate: Benzoic acid has a low solubility in water. If a saturated solution has a pH of 2.83, the molar solubility of the solution

Use equilibrium-constant expression and $K_a$ value to calculate molar solubility

The reaction is ${\text{C}}_{\text{6}}{\text{H}}_5{\text{COO}}^{-}{\text{ + H}}_{\text{2}}\text{O }\stackrel{}{\rightleftharpoons }{\text{ C}}_{\text{6}}{\text{H}}_5{\text{COOH + OH}}^{-}$

From the pH, calculate the hydronium ion and benzoate ion concentrations.

$\begin{array}{l}{\text{[H}}_{\text{3}}{\text{O}}^{\text{+}}\text{]}{\text{=10}}^{-\text{pH}}\\ ={\text{10}}^{-2.83}\\ =1.48\times {10}^{-3}\text{ M}\end{array}$

The concentration of benzoate ion is same as the concentration of hydronium ion

$[{\text{C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{COO}}^{\text{-}}\text{]=1}\text{.48}\times {\text{10}}^{-3}\text{ M}$

Substitute the above values in equilibrium-constant expression.

$\begin{array}{l}{\text{ K}}_a=\frac{[{\text{H}}_{\text{3}}{\text{O}}^{+}{\text{][C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{COO}}^{\text{-}}\text{]}}{{\text{[C}}_{\text{6}}{\text{H}}_{\text{5}}\text{COOH]}}\\ \text{ 6}\text{.3}\times {10}^{-5}=\frac{{\text{(1}\text{.48}\times {\text{10}}^{-3}\text{)}}^2}{y}\\ y=\frac{{\text{(1}\text{.48}\times {\text{10}}^{-3}\text{)}}^2}{\text{6}\text{.3}\times {10}^{-5}}\\ =0.0350\text{ M}\end{array}$

The molar solubility will be the total dissolved benzoic acid, molecular and dissociated:

$\begin{array}{l}0.0350\text{ + 0}\text{.00148 }\text{= 0}\text{.03648}\\ \approx 0.036\text{ M}\end{array}$

Conclusion

Benzoic acid has a low solubility in water. If a saturated solution has a pH of 2.83, the molar solubility of the solution was calculated as 0.036 M