Chapter 15, Problem 15.47QP

Calculate the percent ionization of benzoic acid having the following concentrations: (a) 0.20 M, (b) 0.00020 M.

Interpretation Introduction

Interpretation:

The percent ionization of the given solution of benzoic acid has to be calculated

Concept Information:

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

Acids ionize in water.  Strong acids ionize completely whereas weak acids ionize to some limited extent.

The degree to which a weak acid ionizes depends on the concentration of the acid and the equilibrium constant for the ionization.

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

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

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

Percent ionization:

A quantitative measure of the degree of ionization is percent ionization.

For a weak, monoprotic acid $\text{HA}$ percent ionization can be calculated as follows,

$\text{percent}\text{ionization}=\frac{{\text{[H}}_{\text{3}}{\text{O}}^{\text{+}}\text{]}}{\text{[HA]}}\times 100\%$

Answer to Problem 15.47QP

The percent ionization of given $0.20M$ benzoic acid solution is $1.8\%$

Explanation of Solution

From the given concentrations of benzoic acid solution and $K_a$ value of benzoic acid, the hydrogen ion concentration can be found out through equilibrium table for each of the given solution.  Then percent ionization can be calculated from the obtained hydrogen ion concentration as follows,

Construct an equilibrium table and express the equilibrium concentration of each species in terms of $x$

$\begin{array}{cccccc}& C_6{H}_{5}COO{H}_{\left(aq\right)}& \rightleftarrows & H_{\left(aq\right)}^{+}& +& C_6{H}_{5}CO{O}_{\left(aq\right)}^{-}\\ Initial\left(M\right)& 0.20& & 0& & 0\\ Change\left(M\right)& -x& & +x& & +x\\ Equilibrium\left(M\right)& \left(0.20-x\right)& & x& & x\end{array}$

The $K_a$ of benzoic acid is $6.5\times {10}^{-5}$

$\begin{array}{c}{\text{K}}_{\text{a}}\text{=}\frac{{\text{[H}}^{\text{+}}{\text{][C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{COO}}^{\text{-}}\text{]}}{{\text{[C}}_{\text{6}}{\text{H}}_{\text{5}}\text{COOH]}}\\ \text{6}{\text{.5×10}}^{\text{-5}}\text{=}\frac{{\text{x}}^{\text{2}}}{\text{(0}\text{.20-x)}}\end{array}$

Assuming that x is small compared to $0.20$ , we neglect it in the denominator:

$\begin{array}{c}\text{6}{\text{.5×10}}^{\text{-5}}\text{=}\frac{{\text{x}}^{\text{2}}}{\text{(0}\text{.20-x)}}\\ \text{x}\text{=}\sqrt{\text{0}\text{.20}\text{×}\text{6}\text{.5}\text{×}{\text{10}}^{\text{-5}}}\\ \text{=}\text{3}\text{.6}\text{×}{\text{10}}^{\text{-3}}\text{M}\\ \left[{\text{H}}^{\text{+}}\right]\text{=}\left[{\text{C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{COO}}^{\text{-}}\right]\text{=}\text{3}\text{.6}\text{×}{\text{10}}^{\text{-3}}\text{M}\end{array}$

The percent ionization can be calculated as follows,

$\begin{array}{l}\text{percent}\text{ionization=}\frac{{\text{[H}}^{\text{+}}\text{]}}{\text{[HA]}}\text{×100}\text{\%}\\ \text{=}\frac{{\text{[H}}^{\text{+}}\text{]}}{{\text{[C}}_{\text{6}}{\text{H}}_{\text{5}}\text{COOH]}}\text{×100}\text{\%}\\ \text{=}\frac{\text{3}\text{.6}\text{×}{\text{10}}^{\text{-3}}\text{M}}{\text{0}\text{.20}\text{M}}\text{×100}\text{\%}\\ \text{=1}\text{.8}\text{\%}\end{array}$

Therefore, the percent ionization of given $0.20M$ benzoic acid solution is $1.8\%$

Interpretation Introduction

Interpretation:

The percent ionization of the given solution of benzoic acid has to be calculated

Concept Information:

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

Acids ionize in water.  Strong acids ionize completely whereas weak acids ionize to some limited extent.

The degree to which a weak acid ionizes depends on the concentration of the acid and the equilibrium constant for the ionization.

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

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

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

Percent ionization:

A quantitative measure of the degree of ionization is percent ionization.

For a weak, monoprotic acid $\text{HA}$ percent ionization can be calculated as follows,

$\text{percent}\text{ionization}=\frac{{\text{[H}}_{\text{3}}{\text{O}}^{\text{+}}\text{]}}{\text{[HA]}}\times 100\%$

Answer to Problem 15.47QP

The percent ionization of given $0.00020M$ benzoic acid solution is $43\%$

Explanation of Solution

From the given concentrations of benzoic acid solution and $K_a$ value of benzoic acid, the hydrogen ion concentration can be found out through equilibrium table for each of the given solution.  Then percent ionization can be calculated from the obtained hydrogen ion concentration as follows,

Construct an equilibrium table and express the equilibrium concentration of each species in terms of $x$

$\begin{array}{cccccc}& C_6{H}_{5}COO{H}_{\left(aq\right)}& \rightleftarrows & H_{\left(aq\right)}^{+}& +& C_6{H}_{5}CO{O}_{\left(aq\right)}^{-}\\ Initial\left(M\right)& 0.00020& & 0.00000& & 0.00000\\ Change\left(M\right)& -x& & +x& & +x\\ Equilibrium\left(M\right)& \left(0.00020-x\right)& & x& & x\end{array}$

The $K_a$ of benzoic acid is $6.5\times {10}^{-5}$

$\begin{array}{c}{\text{K}}_{\text{a}}\text{=}\frac{{\text{[H}}^{\text{+}}{\text{][C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{COO}}^{\text{-}}\text{]}}{{\text{[C}}_{\text{6}}{\text{H}}_{\text{5}}\text{COOH]}}\\ \text{6}{\text{.5×10}}^{\text{-5}}\text{=}\frac{{\text{x}}^{\text{2}}}{\text{(}0.00020\text{-x)}}\end{array}$

$\begin{array}{l}{\text{x}}^{\text{2}}\text{+}\left(\text{6}\text{.5}\text{×}{\text{10}}^{\text{-5}}\right)\text{x-}\left(\text{1}\text{.3}\text{×}{\text{10}}^{\text{-8}}\right)\text{=}\text{0}\\ \text{solving}\text{this}\text{equation,}\\ \text{x}\text{=}\text{8}\text{.6}\text{×}{\text{10}}^{\text{-5}}\text{M}\end{array}$

The percent ionization can be calculated as follows,

$\begin{array}{l}\text{percent}\text{ionization=}\frac{{\text{[H}}^{\text{+}}\text{]}}{\text{[HA]}}\text{×100}\text{\%}\\ \text{=}\frac{{\text{[H}}^{\text{+}}\text{]}}{{\text{[C}}_{\text{6}}{\text{H}}_{\text{5}}\text{COOH]}}\text{×100}\text{\%}\\ \text{=}\frac{\text{8}\text{.6}\text{×}{\text{10}}^{\text{-5}}\text{M}}{\text{0}\text{.00020}\text{M}}\text{×100}\text{\%}\\ \text{=}43\%\end{array}$

Therefore, the percent ionization of given $0.00020M$ benzoic acid solution is $43\%$