Chapter 16, Problem 16.55QP

Determine the percent ionization of the following solutions of formic acid at 25°C: (a) 0.016 M, (b) 5.7 × 10⁻⁴ M, (C) 1.75 M

Interpretation Introduction

Interpretation:

The percent ionization of the given solutions of formic 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,

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

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 0.016 M solution of formic acid

Answer to Problem 16.55QP

Answer

The percent ionization of the given solution (a) of phenol is 10%

Explanation of Solution

Calculation for finding out hydrogen ions:

From the equilibrium table for given formic acid 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 $x$

	${\text{HCOOH}}_{\text{(aq)}}{\text{+H}}_{\text{2}}{\text{O}}_{\text{(l)}}\to {\text{HCOO}}^{\text{-}}{}_{\text{(aq)}}\text{+}{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}{}_{\text{(aq)}}$
Initial $(M)$	$0.016$ $-x$ $0.016-x$	$0.00$	$0.00$
Change $(M)$		$+x$	$+x$
Equilibrium $(M)$		$x$	$x$

The $K_a$ of formic acid is $1.7\times {10}^{-4}$

$\begin{array}{l}{K}_a=\frac{[{\text{H}}^{\text{+}}{\text{][HCOO}}^{\text{-}}\text{]}}{[\text{HCOOH]}}\\ 1.7\times {10}^{-4}\approx \frac{x^2}{(0.016-x)}\end{array}$

The approximate value of x is 10% of 0.016, so it is probably not appropriate to have assumed that $0.016-x\approx ~0.016$.  Instead, attempt to solve for x without making any approximations.

Start by clearing the denominator and then expand the result to get:

$x^2+1.7\times {10}^{-4}x-2.72\times {10}^{-6}=0$

This equation is quadratic equation of the form $a{x}^2+bx+c=0$, the solutions of which are given by,

$x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

For this case, $a=1,b=1.7\times {10}^{-4},\text{and}c=-2.72\times {10}^{-6}$ . Substituting these values into the quadratic formula gives:

$\begin{array}{l}x=\frac{-1.7\times {10}^{-4}\pm \sqrt{{(1.7\times {10}^{-4})}^2-4(1)(-2.72\times {10}^{-6})}}{(2)(1)}\\ =\frac{-1.7\times {10}^{-4}\pm 3.30\times {10}^{-3}}{2}\\ x=1.6\times {10}^{-3}M\text{or}x\text{=}\text{-1}\text{.7}\times {\text{10}}^{\text{-3}}M\end{array}$

The negative solution is not physically reasonable since there is no such things as a negative concentration.

Therefore, the ${\text{[H}}^{\text{+}}\text{]}$ ion concentration is $1.6\times {\text{10}}^{\text{-3}}M$

Calculation of percent ionization

The percent ionization can be calculated as follows,

$\begin{array}{l}\text{percent}\text{ionization}=\frac{{\text{[H}}^{\text{+}}\text{]}}{\text{[HA]}}\times 100\%\\ =\frac{{\text{[H}}^{\text{+}}\text{]}}{[\text{HCOOH}]}\times 100\%\\ =\frac{1.6\times {10}^{-3}}{0.016}\times 100\%\\ =10\%\end{array}$

Therefore, the percent ionization of given 0.016 M formic acid solution is 10%

Interpretation Introduction

Interpretation:

The percent ionization of the given solutions of formic 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,

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

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 $5.7\times {10}^{-4}$ solution of formic acid

Answer to Problem 16.55QP

Answer

The percent ionization of the given solution (b) of phenol is 42%

Explanation of Solution

Calculation for finding out hydrogen ions:

From the equilibrium table for given formic acid 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 $x$

	${\text{HCOOH}}_{\text{(aq)}}{\text{+H}}_{\text{2}}{\text{O}}_{\text{(l)}}\to {\text{HCOO}}^{\text{-}}{}_{\text{(aq)}}\text{+}{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}{}_{\text{(aq)}}$
Initial $(M)$	$5.7\times {10}^{-4}$ $-x$ $5.7\times {10}^{-4}-x$	$0.00$	$0.00$
Change $(M)$		$+x$	$+x$
Equilibrium $(M)$		$x$	$x$

The $K_a$ of formic acid is $1.7\times {10}^{-4}$

$\begin{array}{l}{K}_a=\frac{[{\text{H}}^{\text{+}}{\text{][HCOO}}^{\text{-}}\text{]}}{[\text{HCOOH]}}\\ 1.7\times {10}^{-4}\approx \frac{x^2}{(5.7\times {10}^{-4}-x)}\end{array}$

The approximate value of x is 10% of 0.016, so it is probably not appropriate to have assumed that $0.016-x\approx ~0.016$.  Instead, attempt to solve for x without making any approximations.

Start by clearing the denominator and then expand the result to get:

$x^2+1.7\times {10}^{-4}x-9.69\times {10}^{-8}=0$

This equation is quadratic equation of the form $a{x}^2+bx+c=0$, the solutions of which are given by,

$x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

For this case, $a=1,b=1.7\times {10}^{-4},\text{and}c=-2.72\times {10}^{-6}$.  Substituting these values into the quadratic formula gives:

$\begin{array}{l}x=\frac{-1.7\times {10}^{-4}\pm \sqrt{{(1.7\times {10}^{-4})}^2-4(1)(-9.69\times {10}^{-8})}}{(2)(1)}\\ =\frac{-1.7\times {10}^{-4}\pm 6.45\times {10}^{-4}}{2}\\ x=2.4\times {10}^{-4}M\text{or}x\text{=}\text{-4}\text{.1}\times {\text{10}}^{\text{-4}}M\end{array}$

The negative solution is rejected.

Therefore, the ${\text{[H}}^{\text{+}}\text{]}$ ion concentration is $2.4\times {\text{10}}^{\text{-4}}M$

Calculation of percent ionization

The percent ionization can be calculated as follows,

$\begin{array}{l}\text{percent}\text{ionization}=\frac{{\text{[H}}^{\text{+}}\text{]}}{\text{[HA]}}\times 100\%\\ =\frac{{\text{[H}}^{\text{+}}\text{]}}{[\text{HCOOH}]}\times 100\%\\ =\frac{2.4\times {10}^{-4}}{5.7\times {10}^{-4}}\times 100\%\\ =42\%\end{array}$

Therefore, the percent ionization of given 0.016 M formic acid solution is 42%

Interpretation Introduction

Interpretation:

The percent ionization of the given solutions of formic 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,

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

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 1.75 M solution of formic acid

Answer to Problem 16.55QP

Answer

The percent ionization of the given solution (c) of phenol is 0.97%

Explanation of Solution

Calculation for finding out hydrogen ions:

From the equilibrium table for given formic acid 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 $x$

	${\text{C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{OH}}_{\text{(aq)}}{\text{+H}}_{\text{2}}{\text{O}}_{\text{(l)}}\to {\text{C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{O}}^{\text{-}}{}_{\text{(aq)}}\text{+}{\text{H}}^{\text{+}}{}_{\text{(aq)}}$
Initial $(M)$	$1.75$ $-x$ $1.75-x$	$0.00$	$0.00$
Change $(M)$		$+x$	$+x$
Equilibrium $(M)$		$x$	$x$

The $K_a$ of formic acid is $1.7\times {10}^{-4}$

$\begin{array}{l}{K}_a=\frac{[{\text{H}}^{\text{+}}{\text{][HCOO}}^{\text{-}}\text{]}}{[\text{HCOOH]}}\\ 1.7\times {10}^{-4}\approx \frac{x^2}{(1.75-x)}\end{array}$

Assuming that x is small compared to 1.75, we neglect it in the denominator:

$\begin{array}{l}1.7\times {10}^{-4}=\frac{x^2}{1.75}\\ x=\sqrt{(1.7\times {10}^{-4})(1.75)}\\ =0.017M\end{array}$

Therefore, the ${\text{[H}}^{\text{+}}\text{]}$ ion concentration is 0.017 M

Calculation of percent ionization

The percent ionization can be calculated as follows,

$\begin{array}{l}\text{percent}\text{ionization}=\frac{{\text{[H}}^{\text{+}}\text{]}}{\text{[HA]}}\times 100\%\\ =\frac{{\text{[H}}^{\text{+}}\text{]}}{[\text{HCOOH}]}\times 100\%\\ =\frac{0.017}{1.75}\times 100\%\\ =0.97\%\end{array}$

Therefore, the percent ionization of given 0.56 M phenol solution is 0.97%