Chapter 16, Problem 16.113QP

Methylammonium chloride is a salt of methylamine, CH₃NH₂. A 0.10 M solution of this salt has a pH of 5.82.

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

   ${\text{CH}}_3{\text{NH}}_3{}^{+}+{\text{H}}_2\text{O}\rightleftharpoons {\text{CH}}_3{\text{NH}}_2+{\text{H}}_3{\text{O}}^{+}$

2. b What is the K_b value for methylamine?
3. c What is the pH of a solution in which 0.450 mol of solid methylammonium chloride is added to 1.00 L of a 0.250 M solution of methylamine? Assume no volume change.

Interpretation Introduction

Interpretation:

A 0.10 M solution of methylammonium chloride has a pH of 5.82

1. (a) The value for the equilibrium constant for the  given reaction ${\text{CH}}_{\text{3}}{\text{NH}}_{\text{3}}{}^{\text{+}}{\text{ + H}}_{\text{2}}\text{O }\stackrel{}{\rightleftharpoons }{\text{ CH}}_{\text{3}}{\text{NH}}_{\text{2}}{\text{ + H}}_{\text{3}}{\text{O}}^{\text{+}}$ has to be calculated
2. (b) The ${\text{K}}_{\text{b}}$ value for methylamine has to be calculated
3. (c) The pH of a solution in which 0.450 mol of methylammonium chloride is added to 1.00 L of a 0.250 M solution of methylamine 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 16.113QP

The value for the equilibrium constant for the given reaction ${\text{CH}}_{\text{3}}{\text{NH}}_{\text{3}}{}^{\text{+}}{\text{ + H}}_{\text{2}}\text{O }\stackrel{}{\rightleftharpoons }{\text{ CH}}_{\text{3}}{\text{NH}}_{\text{2}}{\text{ + H}}_{\text{3}}{\text{O}}^{\text{+}}$ is $K_a=2.3\times 10^{-11}$

Explanation of Solution

To Calculate: The value for the equilibrium constant for the given reaction ${\text{CH}}_{\text{3}}{\text{NH}}_{\text{3}}{}^{\text{+}}{\text{ + H}}_{\text{2}}\text{O }\stackrel{}{\rightleftharpoons }{\text{ CH}}_{\text{3}}{\text{NH}}_{\text{2}}{\text{ + H}}_{\text{3}}{\text{O}}^{\text{+}}$

Given data:

Methylammonium chloride is a salt of methylamine ${\text{(CH}}_{\text{3}}{\text{NH}}_{\text{2}}\text{)}$

A 0.10 M solution of this salt has a pH of 5.82

The given reaction is ${\text{CH}}_{\text{3}}{\text{NH}}_{\text{3}}{}^{\text{+}}{\text{ + H}}_{\text{2}}\text{O }\stackrel{}{\rightleftharpoons }{\text{ CH}}_{\text{3}}{\text{NH}}_{\text{2}}{\text{ + H}}_{\text{3}}{\text{O}}^{\text{+}}$

Equilibrium constant for the given reaction:

The hydronium ion concentration is found from the given pH as follows,

$\begin{array}{l}{\text{[H}}_{\text{3}}{\text{O}}^{+}\text{]}\text{=}{10}^{-\text{pH}}\\ ={10}^{-5.82}\\ =1.51\times {10}^{-6}\text{ M}\end{array}$

Construct an equilibrium table for the given reaction:

The initial concentration of ${\text{CH}}_{\text{3}}{\text{NH}}_{\text{3}}{}^{\text{+}}$ is considered from the concentration of methylammonium chloride salt (0.10 M)

	${\text{CH}}_{\text{3}}{\text{NH}}_{\text{3}}{}^{\text{+}}{\text{ + H}}_{\text{2}}\text{O }\stackrel{}{\rightleftharpoons }{\text{ CH}}_{\text{3}}{\text{NH}}_{\text{2}}{\text{ + H}}_{\text{3}}{\text{O}}^{\text{+}}$
Initial $(M)$	0.10 $-x$ 0.10-x	0.00	0.00
Change $(M)$		$+x$	$+x$
Equilibrium $(M)$		x	x

Here, x gives the concentration of hydronium ion that reacted, $x=1.51\times {10}^{-6}\text{ M}$

Substitute the equilibrium concentrations into the equilibrium-constant expression:

$\begin{array}{l}{\text{K}}_a=\frac{[{\text{H}}_{\text{3}}{\text{O}}^{+}{\text{][CH}}_{\text{3}}{\text{NH}}_{\text{2}}\text{]}}{{\text{[CH}}_{\text{3}}{\text{NH}}_3{}^{\text{+}}\text{]}}\\ =\frac{{\text{(}x\text{)}}^2}{(0.10-x)}\\ \text{Assume }x\text{ is small compared to 0}\text{.10 and neglect it}\text{.}\\ =\frac{{\text{(1}\text{.51}\times {\text{10}}^{-6}\text{)}}^2}{(0.10)}\\ =2.3\times {10}^{-11}\end{array}$

Therefore, the equilibrium-constant for the given reaction is $2.3\times {10}^{-11}$

Conclusion

The value for the equilibrium constant for the given reaction ${\text{CH}}_{\text{3}}{\text{NH}}_{\text{3}}{}^{\text{+}}{\text{ + H}}_{\text{2}}\text{O }\stackrel{}{\rightleftharpoons }{\text{ CH}}_{\text{3}}{\text{NH}}_{\text{2}}{\text{ + H}}_{\text{3}}{\text{O}}^{\text{+}}$ was calculated as $K_a=2.3\times 10^{-11}$

Interpretation Introduction

Interpretation:

A 0.10 M solution of methylammonium chloride has a pH of 5.82

1. (a) The value for the equilibrium constant for the  given reaction ${\text{CH}}_{\text{3}}{\text{NH}}_{\text{3}}{}^{\text{+}}{\text{ + H}}_{\text{2}}\text{O }\stackrel{}{\rightleftharpoons }{\text{ CH}}_{\text{3}}{\text{NH}}_{\text{2}}{\text{ + H}}_{\text{3}}{\text{O}}^{\text{+}}$ has to be calculated
2. (b) The ${\text{K}}_{\text{b}}$ value for methylamine has to be calculated
3. (c) The pH of a solution in which 0.450 mol of methylammonium chloride is added to 1.00 L of a 0.250 M solution of methylamine 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 16.113QP

The ${\text{K}}_{\text{b}}$ value for methylamine is $4.4\times 10^{-4}$

Explanation of Solution

To Calculate: The ${\text{K}}_{\text{b}}$ value for methylamine

The ${\text{K}}_{\text{b}}$ value for methylamine is calculated using ${\text{K}}_w$ as follows,

$\begin{array}{l}{\text{K}}_a\times {\text{K}}_b{\text{= K}}_w\\ {\text{ K}}_b=\frac{{\text{K}}_w}{{\text{K}}_a}\\ =\frac{1.0\times {10}^{-14}}{2.28\times {10}^{-11}}\\ =4.4\times {10}^{-4}\end{array}$

Conclusion

The ${\text{K}}_{\text{b}}$ value for methylamine was calculated as $4.4\times 10^{-4}$

Interpretation Introduction

Interpretation:

A 0.10 M solution of methylammonium chloride has a pH of 5.82

1. (a) The value for the equilibrium constant for the  given reaction ${\text{CH}}_{\text{3}}{\text{NH}}_{\text{3}}{}^{\text{+}}{\text{ + H}}_{\text{2}}\text{O }\stackrel{}{\rightleftharpoons }{\text{ CH}}_{\text{3}}{\text{NH}}_{\text{2}}{\text{ + H}}_{\text{3}}{\text{O}}^{\text{+}}$ has to be calculated
2. (b) The ${\text{K}}_{\text{b}}$ value for methylamine has to be calculated
3. (c) The pH of a solution in which 0.450 mol of methylammonium chloride is added to 1.00 L of a 0.250 M solution of methylamine 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 16.113QP

The pH of a solution in which 0.450 mol of methylammonium chloride is added to 1.00 L of a 0.250 M solution of methylamine is 10.39

Explanation of Solution

To Calculate: The pH of a solution in which 0.450 mol of methylammonium chloride is added to 1.00 L of a 0.250 M solution of methylamine

Use equilibrium-constant expression and $K_a$ value to calculate hydronium ion concentration.

Methylammonium chloride contains ${\text{CH}}_{\text{3}}{\text{NH}}_{\text{3}}{}^{\text{+}}$ and ${\text{Cl}}^{-}$ ions.

0.450 mol of methylammonium chloride gives 0.450 mol of ${\text{CH}}_{\text{3}}{\text{NH}}_{\text{3}}{}^{\text{+}}$ ion and 0.450 mol of ${\text{Cl}}^{-}$ ion

The concentration of ${\text{CH}}_{\text{3}}{\text{NH}}_{\text{3}}{}^{\text{+}}$ is:

$\begin{array}{l}{\text{ [CH}}_{\text{3}}{\text{NH}}_{\text{3}}{}^{\text{+}}]=\frac{0.450\text{ mol}}{1.00\text{ L}}\\ =0.450\text{ M}\end{array}$

The concentration of methylamine is:

${\text{[CH}}_{\text{3}}{\text{NH}}_2]=0.250\text{ M}$

The ${\text{K}}_{\text{a}}$ value for the methylammonium ion is $2.3\times {10}^{-11}$

Substitute the above values in equilibrium-constant expression.

$\begin{array}{l}{\text{ K}}_a=\frac{[{\text{H}}_{\text{3}}{\text{O}}^{+}{\text{][CH}}_{\text{3}}{\text{NH}}_{\text{2}}\text{]}}{{\text{[CH}}_{\text{3}}{\text{NH}}_3{}^{\text{+}}\text{]}}\\ 2.3\times {10}^{-11}=\frac{[{\text{H}}_{\text{3}}{\text{O}}^{+}\text{](0}\text{.250)}}{(0.450)}\\ [{\text{H}}_{\text{3}}{\text{O}}^{+}\text{]}=\frac{(2.3\times {10}^{-11})(0.450)}{(0.250)}\\ =4.10\times {10}^{-11}\text{ M}\end{array}$

Thus, the pH is calculated from the obtained hydronium ion concentration as follows,

$\begin{array}{l}\text{pH}\text{=}{\text{-log[H}}_{\text{3}}{\text{O}}^{+}\text{]}\\ \text{=-log(}4.10\times {10}^{-11})\\ =10.39\end{array}$

Therefore the pH of the given solution is 10.39

Conclusion

The pH of a solution in which 0.450 mol of methylammonium chloride is added to 1.00 L of a 0.250 M solution of methylamine was calculated as 10.39