Chapter 16.6, Problem 12PPA

Determine the pH and percent ionization for hydrocyanic acid (HCN) solutions of concentration (a) 0.25 M, (b) 0.0075 M, and (c) 8.3 × 10⁻⁵ M.

Interpretation Introduction

Interpretation:

The pH and percent ionization for the given concentrations of Hydrocyanic acid ( $\text{HCN}$)  solution 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,

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

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,

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

pH definition:

The concentration of hydrogen ion is measured using $\text{pH}$ scale.  The acidity of aqueous solution is expressed by $\text{pH}$ scale.

The $\text{pH}$ of a solution is defined as the negative base-10 logarithm of the hydrogen or hydronium ion concentration.

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

To calculate: The pH and percent ionization of 0.25 M Hydrocyanic acid solution

Answer to Problem 12PPA

Answer

The pH of 0.25 M Hydrocyanic acid solution is 4.96

The percent ionization for 0.25 M Hydrocyanic acid is 0.0044%

Explanation of Solution

Calculation for finding out hydrogen ions:

From the equilibrium table for given acetic acid solution, the concentration of hydrogen ions can be found as follows,

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

	${\text{HCN}}_{(aq)}+{\text{H}}_{\text{2}}\text{O}(l)\stackrel{}{\to }{\text{H}}^{\text{+}}{}_{(aq)}+{\text{CN}}^{-}{}_{(aq)}$
Initial $(M)$	0.25 -x 0.25-x	$0.00$	$0.00$
Change $(M)$		$+x$	$+x$
Equilibrium $(M)$		$x$	$x$

The $K_a$ of Hydrocyanic acid is $4.9\times {10}^{-10}$

$\begin{array}{l}{K}_a=\frac{[{\text{H}}^{\text{+}}{\text{][CN}}^{\text{-}}\text{]}}{[\text{HCN]}}\\ 4.9\times {10}^{-10}=\frac{x^2}{(0.25-x)}\end{array}$

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

$\begin{array}{l}4.9\times {10}^{-10}=\frac{x^2}{(0.25)}\\ x=\sqrt{(4.9\times {10}^{-10})(0.25)}\\ =1.10\times {10}^{-5}\end{array}$

Therefore, the ${\text{[H}}^{\text{+}}\text{]}$ ion concentration is $1.10\times {10}^{-5}$

Calculation of pH

The pH can be calculated as follows,

$\begin{array}{l}\text{pH}\text{=}{\text{-log[H}}_{\text{3}}{\text{O}}^{\text{+}}\text{]}\\ \text{ =-log(}1.10\times {10}^{-5}\text{)}\\ \text{=4}\text{.96}\end{array}$

Calculation of percent ionization

The percent ionization can be calculated as follows,

$\begin{array}{l}\text{percent}\text{ionization}=\frac{{\text{[H}}^{\text{+}}\text{]}}{\text{[HA]}}\times 100\%\\ =\frac{{\text{[H}}^{\text{+}}\text{]}}{[\text{HCN}]}\times 100\%\\ =\frac{1.10\times {10}^{-5}}{0.25}\times 100\%\\ =0.0044\%\end{array}$

Interpretation Introduction

Interpretation:

The pH and percent ionization for the given concentrations of Hydrocyanic acid ( $\text{HCN}$)  solution 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,

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

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,

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

pH definition:

The concentration of hydrogen ion is measured using $\text{pH}$ scale.  The acidity of aqueous solution is expressed by $\text{pH}$ scale.

The $\text{pH}$ of a solution is defined as the negative base-10 logarithm of the hydrogen or hydronium ion concentration.

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

To calculate: The pH and percent ionization of 0.0075 M Hydrocyanic acid solution

Answer to Problem 12PPA

Answer

The pH of 0.0075 M Hydrocyanic acid solution is 5.72

The percent ionization for 0.0075 M Hydrocyanic acid is 0.026%

Explanation of Solution

Calculation for finding out hydrogen ions:

From the equilibrium table for given acetic acid solution, the concentration of hydrogen ions can be found as follows,

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

	${\text{HCN}}_{(aq)}+{\text{H}}_{\text{2}}\text{O}(l)\stackrel{}{\to }{\text{H}}^{\text{+}}{}_{(aq)}+{\text{CN}}^{-}{}_{(aq)}$
Initial $(M)$	0.0075 -x 0.0075-x	$0.00$	$0.00$
Change $(M)$		$+x$	$+x$
Equilibrium $(M)$		$x$	$x$

The $K_a$ of Hydrocyanic acid is $4.9\times {10}^{-10}$

$\begin{array}{l}{K}_a=\frac{[{\text{H}}^{\text{+}}{\text{][CN}}^{\text{-}}\text{]}}{[\text{HCN]}}\\ 4.9\times {10}^{-10}=\frac{x^2}{(0.0075-x)}\end{array}$

Assuming that x is small compared to 0.0075, we neglect it in the denominator:

$\begin{array}{l}4.9\times {10}^{-10}=\frac{x^2}{(0.25)}\\ x=\sqrt{(4.9\times {10}^{-10})(0.0075)}\\ =0.19\times {10}^{-5}M\end{array}$

Therefore, the ${\text{[H}}^{\text{+}}\text{]}$ ion concentration is $0.19\times {10}^{-5}M$

Calculation of pH

The pH can be calculated as follows,

$\begin{array}{l}\text{pH}\text{=}{\text{-log[H}}_{\text{3}}{\text{O}}^{\text{+}}\text{]}\\ \text{ =-log(}0.19\times {10}^{-5}M\text{)}\\ \text{=5}\text{.72}\end{array}$

Calculation of percent ionization

The percent ionization can be calculated as follows,

$\begin{array}{l}\text{percent}\text{ionization}=\frac{{\text{[H}}^{\text{+}}\text{]}}{\text{[HA]}}\times 100\%\\ =\frac{{\text{[H}}^{\text{+}}\text{]}}{[\text{HCN}]}\times 100\%\\ =\frac{0.19\times {10}^{-5}M}{0.0075}\times 100\%\\ =0.026\%\end{array}$

Interpretation Introduction

Interpretation:

The pH and percent ionization for the given concentrations of Hydrocyanic acid ( $\text{HCN}$)  solution 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,

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

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,

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

pH definition:

The concentration of hydrogen ion is measured using $\text{pH}$ scale.  The acidity of aqueous solution is expressed by $\text{pH}$ scale.

The $\text{pH}$ of a solution is defined as the negative base-10 logarithm of the hydrogen or hydronium ion concentration.

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

To calculate: The pH and percent ionization of 0.0075 M Hydrocyanic acid solution

Answer to Problem 12PPA

Answer

The pH of $8.3\times {10}^{-5}M$ Hydrocyanic acid solution is 6.70

The percent ionization for $8.3\times {10}^{-5}M$ Hydrocyanic acid is 0.24%

Explanation of Solution

To calculate: The pH and percent ionization of Hydrocyanic acid solution

Calculation for finding out hydrogen ions:

From the equilibrium table for given acetic acid solution, the concentration of hydrogen ions can be found as follows,

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

	${\text{HCN}}_{(aq)}+{\text{H}}_{\text{2}}\text{O}(l)\stackrel{}{\to }{\text{H}}^{\text{+}}{}_{(aq)}+{\text{CN}}^{-}{}_{(aq)}$
Initial $(M)$	$8.3\times {10}^{-5}$ -x $8.3\times {10}^{-5}-x$	$0.00$	$0.00$
Change $(M)$		$+x$	$+x$
Equilibrium $(M)$		$x$	$x$

The $K_a$ of Hydrocyanic acid is $4.9\times {10}^{-10}$

$\begin{array}{l}{K}_a=\frac{[{\text{H}}^{\text{+}}{\text{][CN}}^{\text{-}}\text{]}}{[\text{HCN]}}\\ 4.9\times {10}^{-10}=\frac{x^2}{(8.3\times {10}^{-5}-x)}\end{array}$

Assuming that x is small compared to $8.3\times {10}^{-5}$, we neglect it in the denominator:

$\begin{array}{l}4.9\times {10}^{-10}=\frac{x^2}{(8.3\times {10}^{-5})}\\ x=\sqrt{(4.9\times {10}^{-10})(8.3\times {10}^{-5})}\\ =4.06\times {10}^{-8}M\end{array}$

Therefore, the ${\text{[H}}^{\text{+}}\text{]}$ ion concentration is $4.06\times {10}^{-8}M$

Calculation of pH

The pH can be calculated as follows,

$\begin{array}{l}\text{pH}\text{=}{\text{-log[H}}_{\text{3}}{\text{O}}^{\text{+}}\text{]}\\ \text{ =-log(}4.06\times {10}^{-8}\text{)}\\ \text{=7}\text{.39}\end{array}$

Calculation of percent ionization

The percent ionization can be calculated as follows,

$\begin{array}{l}\text{percent}\text{ionization}=\frac{{\text{[H}}^{\text{+}}\text{]}}{\text{[HA]}}\times 100\%\\ =\frac{{\text{[H}}^{\text{+}}\text{]}}{[\text{HCN}]}\times 100\%\\ =\frac{4.06\times {10}^{-8}}{8.3\times {10}^{-5}}\times 100\%\\ =0.24\%\end{array}$