Chapter 16, Problem 16.54QP

Determine the percent ionization of the following solutions of phenol: (a) 0.56 M, (b) 0.25 M, (c) l.8 × 10⁻⁶ M.

Interpretation Introduction

Interpretation:

The percent ionization of the given solutions of phenol 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,

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

To Calculate: The percent ionization of the given solutions of phenol

The percent ionization of given 0.56 M solution of phenol

Answer to Problem 16.54QP

Answer

The percent ionization of the given solution (a) of phenol is 0.0015%

Explanation of Solution

Record the given datas

The concentration of the given solutions of phenol:

 0.56 M,

From the given concentrations of phenol solution and $K_a$ value of phenol, 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 for all the concentrations of given phenol solution.

Calculation for finding out hydrogen ions:

From the equilibrium table for given phenol solution, the concentration of hydrogen ion can be found as follows,

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

	${\text{C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{OH}}_{(aq)}+{\text{H}}_{\text{2}}\text{O}(l)\to {\text{C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{O}}^{\text{-}}{}_{(aq)}+{\text{H}}^{\text{+}}{}_{(aq)}$
Initial $(M)$	$0.56$ $-x$ $0.56-x$	$0.00$	$0.00$
Change $(M)$		$+x$	$+x$
Equilibrium $(M)$		$x$	$x$

The $K_a$ of phenol is $1.3\times {10}^{-10}$

$\begin{array}{l}{K}_a=\frac{[{\text{H}}^{\text{+}}{\text{][C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{O}}^{\text{-}}\text{]}}{[{\text{C}}_{\text{6}}{\text{H}}_{\text{5}}\text{OH]}}\\ 1.3\times {10}^{-10}=\frac{x^2}{(0.56-x)}\end{array}$

Assuming that x is small compared to 0.56, we neglect it in the denominator:

$\begin{array}{l}1.3\times {10}^{-10}=\frac{x^2}{0.56}\\ x=\sqrt{(1.3\times {10}^{-10})(0.56)}\\ =8.5\times {10}^{-6}M\end{array}$

Therefore, the ${\text{[H}}^{\text{+}}\text{]}$ ion concentration is $8.5\times {\text{10}}^{\text{-6}}M$

Calculation of percent ionization

The percent ionization can be calculated as follows,

$\begin{array}{l}\text{percent}\text{ionization}=\frac{{\text{[H}}^{\text{+}}\text{]}}{\text{[HA]}}\times 100\%\\ =\frac{{\text{[H}}^{\text{+}}\text{]}}{[{\text{C}}_{\text{6}}{\text{H}}_{\text{5}}\text{OH}]}\times 100\%\\ =\frac{8.5\times {10}^{-6}}{0.56}\times 100\%\\ =0.0015\%\end{array}$

Therefore, the percent ionization of given 0.56 M phenol solution is 0.0015%

Interpretation Introduction

Interpretation:

The percent ionization of the given solutions of phenol 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,

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

To Calculate: The percent ionization of the given solutions of phenol

The percent ionization of given 0.25 M solution of phenol

Answer to Problem 16.54QP

Answer

The percent ionization of the given solution (b) of phenol is 0.0023%

Explanation of Solution

Record the given datas

The concentration of the given solutions of phenol:

0.25 M,

From the given concentrations of phenol solution and $K_a$ value of phenol, 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 for all the concentrations of given phenol solution.

Calculation for finding out hydrogen ions:

From the equilibrium table for given phenol solution, the concentration of hydrogen ion can be found as follows,

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

	${\text{C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{OH}}_{(aq)}+{\text{H}}_{\text{2}}\text{O}(l)\to {\text{C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{O}}^{\text{-}}{}_{(aq)}+{\text{H}}^{\text{+}}{}_{(aq)}$
Initial $(M)$	$0.25$ $-x$ $0.25-x$	$0.00$	$0.00$
Change $(M)$		$+x$	$+x$
Equilibrium $(M)$		$x$	$x$

The $K_a$ of phenol is $1.3\times {10}^{-10}$

$\begin{array}{l}{K}_a=\frac{[{\text{H}}^{\text{+}}{\text{][C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{O}}^{\text{-}}\text{]}}{[{\text{C}}_{\text{6}}{\text{H}}_{\text{5}}\text{OH]}}\\ 1.3\times {10}^{-10}=\frac{x^2}{(0.25-x)}\end{array}$

Assuming that x is small compared to 0.25, we neglect it in the denominator:

$\begin{array}{l}1.3\times {10}^{-10}=\frac{x^2}{0.25}\\ x=\sqrt{(1.3\times {10}^{-10})(0.25)}\\ =5.7\times {10}^{-6}M\end{array}$

Therefore, the ${\text{[H}}^{\text{+}}\text{]}$ ion concentration is $5.7\times {\text{10}}^{\text{-6}}M$

Calculation of percent ionization

The percent ionization can be calculated as follows,

$\begin{array}{l}\text{percent}\text{ionization}=\frac{{\text{[H}}^{\text{+}}\text{]}}{\text{[HA]}}\times 100\%\\ =\frac{{\text{[H}}^{\text{+}}\text{]}}{[{\text{C}}_{\text{6}}{\text{H}}_{\text{5}}\text{OH}]}\times 100\%\\ =\frac{5.7\times {10}^{-6}}{0.25}\times 100\%\\ =0.0023\%\end{array}$

Therefore, the percent ionization of given 0.25 M phenol solution is 0.0023%

Interpretation Introduction

Interpretation:

The percent ionization of the given solutions of phenol 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,

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

To Calculate: The percent ionization of the given solutions of phenol

The percent ionization of given $1.8\times {10}^{-6}M$ solution of phenol

Answer to Problem 16.54QP

Answer

The percent ionization of the given solution (c) of phenol is 0.89%

Explanation of Solution

Record the given datas

The concentration of the given solutions of phenol:

$1.8\times {10}^{-6}M$

From the given concentrations of phenol solution and $K_a$ value of phenol, 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 for all the concentrations of given phenol solution.

Calculation for finding out hydrogen ions:

From the equilibrium table for given phenol solution, the concentration of hydrogen ion can be found as follows,

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

	${\text{C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{OH}}_{(aq)}+{\text{H}}_{\text{2}}\text{O}(l)\to {\text{C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{O}}^{\text{-}}{}_{(aq)}+{\text{H}}^{\text{+}}{}_{(aq)}$
Initial $(M)$	$1.8\times {10}^{-6}$ $-x$ $1.8\times {10}^{-6}-x$	$0.00$	$0.00$
Change $(M)$		$+x$	$+x$
Equilibrium $(M)$		$x$	$x$

The $K_a$ of phenol is $1.3\times {10}^{-10}$

$\begin{array}{l}{K}_a=\frac{[{\text{H}}^{\text{+}}{\text{][C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{O}}^{\text{-}}\text{]}}{[{\text{C}}_{\text{6}}{\text{H}}_{\text{5}}\text{OH]}}\\ 1.3\times {10}^{-10}=\frac{x^2}{(1.8\times {10}^{-6}-x)}\end{array}$

Assuming that x is small compared to $1.8\times {10}^{-6}$, we neglect it in the denominator:

$\begin{array}{l}1.3\times {10}^{-10}=\frac{x^2}{1.8\times {10}^{-6}}\\ x=\sqrt{(1.3\times {10}^{-10})(1.8\times {10}^{-6})}\\ =1.6\times {10}^{-8}M\end{array}$

Therefore, the ${\text{[H}}^{\text{+}}\text{]}$ ion concentration is $1.6\times {\text{10}}^{\text{-8}}M$

Calculation of percent ionization

The percent ionization can be calculated as follows,

$\begin{array}{l}\text{percent}\text{ionization}=\frac{{\text{[H}}^{\text{+}}\text{]}}{\text{[HA]}}\times 100\%\\ =\frac{{\text{[H}}^{\text{+}}\text{]}}{[{\text{C}}_{\text{6}}{\text{H}}_{\text{5}}\text{OH}]}\times 100\%\\ =\frac{1.6\times {10}^{-8}}{1.8\times {10}^{-6}}\times 100\%\\ =0.89\%\end{array}$

Therefore, the percent ionization of given 0.56 M phenol solution is 0.89%