Chapter 6, Problem 6D.13E

(a)

Interpretation Introduction

Interpretation:

The $\text{pH}$ of $\text{1}{\text{.0×10}}^{\text{-5}}\text{M}$ aqueous hydrochloric acid solution has to be calculated.

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.

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

On rearranging, the concentration of hydrogen ion ${\text{[H}}^{+}\text{]}$ can be calculated using pH as follows,

  ${\text{[H}}^{+}\text{]}\text{=}{\text{10}}^{\text{-pH}}$

Answer to Problem 6D.13E

The $\text{pH}$ of $\text{1}{\text{.0×10}}^{\text{-5}}\text{M}$ aqueous hydrochloric acid solution is 5.0.

Explanation of Solution

The ${\text{H}}_{\text{3}}{\text{O}}^{\text{+}}$ ion concentration is equal to the concentration of hydrochloric acid, because it is a strong acid therefore, it dissociates completely and the following equation is given below.

  ${\text{HCl+H}}_{\text{2}}\text{O}\to {\text{H}}_{\text{3}}{\text{O}}^{\text{+}}{\text{+Cl}}^{\text{-}}$

If consider, $\text{1}{\text{.0×10}}^{\text{-5}}\text{M HCl}$=$\text{1}{\text{.0×10}}^{\text{-5}}{\text{M [H}}_3{\text{O}}^{+}\text{]}$, using the information we can calculate $\text{pH}$ of the solution.

  $\begin{array}{l}\text{pH}\text{=}{\text{-log[H}}_{\text{3}}{\text{O}}^{\text{+}}\text{]}\\ \\ \text{=}{\text{-log(1×10}}^{\text{-5}}\text{)}\\ \\ \text{=}\text{5}\text{.0}\end{array}$

Therefore, the $\text{pH}$ of $\text{1}{\text{.0×10}}^{\text{-5}}\text{M}$ aqueous hydrochloric acid solution is 5.0

(b)

Interpretation Introduction

Interpretation:

The $\text{pH}$ of $\text{0}\text{.20}{\text{M CH}}_{\text{3}}{\text{NH}}_{\text{3}}\text{Cl}$ aqueous solution has to be calculated.

Concept introduction:

Refer to part (a).

Answer to Problem 6D.13E

The $\text{pH}$ of $\text{0}\text{.20}{\text{M CH}}_{\text{3}}{\text{NH}}_{\text{3}}\text{Cl}$ aqueous solution is 5.63.

Explanation of Solution

The equilibrium reaction of methylamine is given below.

  ${\text{CH}}_{\text{3}}{\text{NH}}_{\text{3}}{{}^{\text{+}}}_{\text{(aq)}}{\text{+H}}_{\text{2}}{\text{O}}_{\text{(l)}}\rightleftharpoons {\text{CH}}_{\text{3}}{\text{NH}}_{\text{2}}{}_{\text{(aq)}}{\text{+H}}_{\text{3}}{\text{O}}^{\text{+}}{}_{\text{(aq)}}$

The equilibrium expression for the above reaction is given below.

  ${\text{K}}_{\text{a}}\text{=}\frac{{\text{[CH}}_{\text{3}}{\text{NH}}_{\text{2}}\text{]}{\text{[H}}_{\text{3}}{\text{O}}^{\text{+}}\text{]}}{{\text{[CH}}_{\text{3}}{\text{NH}}_{\text{3}}{}^{\text{+}}\text{]}}$

	${\text{CH}}_{\text{3}}{\text{NH}}_{\text{3}}{}^{+}$	${\text{CH}}_{\text{3}}{\text{NH}}_2$	${\text{H}}_{\text{3}}{\text{O}}^{\text{+}}$
Initial concentration	0.20	0	0
Change in concentration	-x	+x	+x
Equilibrium concentration	0.20-x	x	x

The equilibrium concentration values obtained in the above table is substituted in the above equation and is given below.

  ${\text{K}}_{\text{a}}\text{=}\frac{\text{x×x}}{\text{0}\text{.20-x}}$

Methylamine ${\text{K}}_{\text{b}}$ value is $3.6\times {10}^{-4}$ which is given in table 6C.2 and using this ${\text{K}}_{\text{a}}$ value is calculated and the respective formula is given below,

$\begin{array}{l}{\text{K}}_{\text{a}}{\text{×K}}_{\text{b}}{\text{= K}}_{\text{w}}\\ \\ {\text{ K}}_{\text{a}}\text{=}\text{}\frac{{\text{K}}_{\text{w}}}{{\text{K}}_{\text{b}}}\\ \\ \text{=}\frac{\text{1}{\text{.0×10}}^{\text{-14}}}{\text{3}{\text{.6×10}}^{\text{-4}}}\\ \\ \text{=}\text{2}{\text{.78×10}}^{\text{-11}}\end{array}$

Therefore, the ${\text{K}}_{\text{a}}$ value of methylamine is $\text{2}{\text{.78×10}}^{\text{-11}}$.

The obtained ${\text{K}}_{\text{a}}$ value is substitute in below the equilibrium reaction.

  $\begin{array}{l}{\text{K}}_{\text{a}}\text{= }\frac{{\text{x}}^{\text{2}}}{\text{0}\text{.20-x}}\\ \\ 2.78\times {10}^{-11}=\frac{{\text{x}}^{\text{2}}}{\text{0}\text{.20-x}}\end{array}$

The above equation, assume that the x present in 0.20-x is very small than 0.20 then it can be negligible and as follows,

  $\begin{array}{l}\text{2}{\text{.78×10}}^{\text{-11}}\text{=}\frac{{\text{x}}^{\text{2}}}{\text{0}\text{.20}}\\ \\ {\text{x}}^{\text{2}}\text{=}\text{0}\text{.20×(2}{\text{.78×10}}^{\text{-11}}\text{)}\\ \\ \text{x}\text{=}\sqrt{\text{0}\text{.20×(2}{\text{.78×10}}^{\text{-11}}\text{)}}\\ \\ \text{=}2.36\times {10}^{-6}\text{(}\because \text{x}\text{=}{\text{[H}}_{\text{3}}{\text{O}}^{\text{+}}\text{])}\end{array}$

Now, the $\text{pH}$ of the solution is calculated using given equation.

  $\begin{array}{l}\text{pH}\text{=}{\text{-log[H}}_{\text{3}}{\text{O}}^{\text{+}}\text{] }\\ \\ \text{=}\text{-log(2}{\text{.36×10}}^{\text{-6}}\text{)}\\ \\ \text{=}\text{5}\text{.63}\end{array}$

Therefore, the calculated $\text{pH}$ value of $\text{0}\text{.20}{\text{M CH}}_{\text{3}}{\text{NH}}_{\text{3}}\text{Cl}$ aqueous solution is 5.63.

(c)

Interpretation Introduction

Interpretation:

The $\text{pH}$ of $\text{0}\text{.20}{\text{M CH}}_{\text{3}}\text{COOH}$ aqueous solution has to be calculated.

Concept introduction:

Refer to part (a).

Answer to Problem 6D.13E

The $\text{pH}$ of $\text{0}\text{.20}{\text{M CH}}_{\text{3}}\text{COOH}$ aqueous solution is 2.72.

Explanation of Solution

Acetic acid is a weak acid when it is dissolved in water it ionized as positive and negative ions and it is given below.

  ${\text{CH}}_{\text{3}}{\text{COOH}}_{\text{(s)}}{\text{+H}}_{\text{2}}{\text{O}}_{\text{(l)}}\rightleftharpoons {\text{CH}}_{\text{3}}{\text{COO}}^{\text{-}}{}_{\text{(aq)}}{\text{+H}}_{\text{3}}{\text{O}}^{\text{+}}{}_{\text{(aq)}}$

The equilibrium expression for the above reaction is given below.

  ${\text{K}}_{\text{a}}\text{=}\frac{{\text{[CH}}_{\text{3}}{\text{COO}}^{\text{-}}\text{]}{\text{[H}}_{\text{3}}{\text{O}}^{\text{+}}\text{]}}{{\text{[CH}}_{\text{3}}\text{COOH]}}$

	${\text{CH}}_{\text{3}}\text{COOH}$	${\text{H}}_{\text{3}}{\text{O}}^{\text{+}}$	${\text{CH}}_{\text{3}}{\text{COO}}^{\text{-}}$
Initial concentration	0.20	0	0
Change in concentration	-x	+x	+x
Equilibrium concentration	0.20-x	x	x

The equilibrium concentration values are obtained in the above table and is substituted in above equation and is given below.

  ${\text{K}}_{\text{a}}\text{=}\frac{\text{x}\times \text{x}}{\text{0}\text{.20-x}}$

Acetic acid ${\text{K}}_{\text{a}}$ value is $1.8\times {10}^{-5}$ which is given in table 6C.1 and it is substituted in above equation and then the value of x is calculated.

  $\text{1}\text{.8}\times {\text{10}}^{-5}\text{=}\frac{{\text{x}}^2}{\text{0}\text{.20-x}}$

The above equation, assume that the x present in 0.20-x is very small than 0.20 then it can be negligible and as follows,

  $\begin{array}{l}\text{1}{\text{.8×10}}^{\text{-5}}\text{=}\frac{{\text{x}}^{\text{2}}}{\text{0}\text{.20}}\\ \\ {\text{x}}^{\text{2}}\text{=}\text{0}\text{.20×(1}{\text{.8×10}}^{\text{-5}}\text{)}\\ \\ \text{x}\text{=}\sqrt{0.20\times (1.8\times {10}^{-5})}\\ \\ \text{=}1.9\times {\text{10}}^{-3}\text{(}\because \text{x}\text{=}{\text{[H}}_{\text{3}}{\text{O}}^{\text{+}}\text{])}\end{array}$

Now, the $\text{pH}$ of the solution is calculated using given equation.

  $\begin{array}{l}\text{pH}\text{=}{\text{-log[H}}_{\text{3}}{\text{O}}^{\text{+}}\text{] }\\ \\ \text{=}\text{-log(1}\text{.9}\times {10}^{-3}\text{)}\\ \\ \text{=}2.72\end{array}$

Therefore, the calculated $\text{pH}$ value of 0.20 M acetic acid is 2.72.

(d)

Interpretation Introduction

Interpretation:

The $\text{pH}$ of $\text{0}\text{.20}{\text{M C}}_6{\text{H}}_5{\text{NH}}_2$ aqueous solution has to be calculated.

Concept introduction:

Refer to part (a).

Answer to Problem 6D.13E

The $\text{pH}$ of $\text{0}\text{.20}{\text{M C}}_6{\text{H}}_5{\text{NH}}_2$ aqueous solution is 8.97.

Explanation of Solution

Aniline is a weak base when it is dissolved in water it ionized as positive and negative ions and it is given below.

  ${\text{C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{NH}}_{\text{2}}{}_{\text{(aq)}}{\text{+H}}_{\text{2}}{\text{O}}_{\text{(l)}}\rightleftharpoons {\text{C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{NH}}_{\text{3}}{{}^{\text{+}}}_{\text{(aq)}}{\text{+OH}}^{\text{-}}{}_{\text{(aq)}}$

The equilibrium expression for the above reaction is given below.

  ${\text{K}}_{\text{b}}\text{=}\frac{{\text{[C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{NH}}_{\text{3}}{}^{\text{+}}\text{]}{\text{[OH}}^{\text{-}}\text{]}}{{\text{[C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{NH}}_{\text{2}}\text{]}}$

	${\text{C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{NH}}_{\text{2}}$	${\text{C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{NH}}_3{}^{+}$	${\text{OH}}^{\text{-}}$
Initial concentration	0.20	0	0
Change in concentration	-x	+x	+x
Equilibrium concentration	0.20-x	x	x

The equilibrium concentration values are obtained in the above table and is substituted in above equation and is given below.

  ${\text{K}}_{\text{b}}\text{=}\frac{\text{x×x}}{\text{0}\text{.20-x}}$

Aniline ${\text{K}}_{\text{b}}$ value is $4.3\times {10}^{-10}$ which is given in table 6C.2 and it is substituted in above equation and then the value of x is calculated.

  $\text{4}\text{.3}\times {\text{10}}^{-10}\text{=}\frac{{\text{x}}^2}{\text{0}\text{.20-x}}$

The above equation, assume that the x present in 0.20-x is very small than 0.20 then it can be negligible and as follows,

  $\begin{array}{l}\text{4}{\text{.3×10}}^{\text{-10}}\text{=}\frac{{\text{x}}^{\text{2}}}{\text{0}\text{.20}}\\ \\ {\text{x}}^{\text{2}}\text{=}\text{0}\text{.20×(4}{\text{.3×10}}^{\text{-10}}\text{)}\\ \\ \text{x}\text{=}\sqrt{0.20\times (4.3\times {10}^{-10})}\\ \\ \text{=}9.27\times {\text{10}}^{-6}\text{(}\because \text{x}\text{=}{\text{[OH}}^{-}\text{])}\end{array}$

Therefore, the concentration of ${\text{OH}}^{\text{-}}$ is $\text{9}{\text{.27×10}}^{\text{-6}}$.

The $\text{pOH}$ of the solution is calculated using given equation.

  $\begin{array}{l}\text{pOH}\text{=}{\text{-log[OH}}^{-}\text{] }\\ \\ \text{=}\text{-log(9}\text{.27}\times {10}^{-6}\text{)}\\ \\ \text{=}5.03\end{array}$

Therefore, the calculated $\text{pOH}$ value of 0.20 M aniline is 5.03.

Using the $\text{pOH}$ value we can calculate the $\text{pH}$ of the solution and the following formula is given below.

  $\begin{array}{l}\text{pH+pOH}\text{=}\text{14}\\ \\ \text{pH}\text{=}\text{14-pOH}\\ \\ \text{=}\text{14-5}\text{.03}\\ \\ \text{=}\text{8}\text{.97}\end{array}$

Therefore, the $\text{pH}$ of the aqueous aniline solution is 8.97.

The solutions are ranked in the order of increasing pH and it is given below.

$\text{0}\text{.20}\text{M}{\text{CH}}_{\text{3}}\text{COOH<}\text{1}{\text{.0×10}}^{\text{-5}}\text{M}\text{HCl}\text{<}\text{0}\text{.20}\text{M}{\text{CH}}_{\text{3}}{\text{NH}}_{\text{3}}\text{Cl}\text{<}\text{0}\text{.20}\text{M}{\text{C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{NH}}_{\text{2}}$