Chapter 15, Problem 15.64QP

Compare the pH of a 0.040 M HCl solution with that of a 0.040 M H₂SO₄ solution. (Hint: H₂SO₄ is a strong acid; K_a for ${\text{HSO}}_4^{-}=1.3\times {10}^{-2}$.)

Interpretation Introduction

Interpretation:

The pH of a 0.040 M $\text{HCl}$ solution with that of a 0.040 M ${\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}$ solution has to be compared.

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

$\text{pH}$ 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 Compare: The pH of a 0.040 M $\text{HCl}$ solution with that of a 0.040 M ${\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}$ solution

Answer to Problem 15.64QP

• The $\text{pH}$ 0.040 M $\text{HCl}$ solution is 1.40
• The $\text{pH}$ 0.040 M ${\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}$ solution is 1.31

Explanation of Solution

Record the given data

${\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}$ is a strong acid

The concentration of $\text{HCl}$ solution is 0.040 M

The concentration of ${\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}$ solution is 0.040 M

The ${\text{K}}_{\text{a}}$ for ${\text{HSO}}^{\text{-}}{}_{\text{4}}$ is $\text{1}\text{.3}\times {\text{10}}^{\text{-2}}$

${\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}$ is a strong acid. The concentration of sulphuric acid solution and concentration of hydrochloric acid solution is given. From the known concentrations and acid ionization constant, the pH of $\text{HCl}$ and ${\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}$ has to be calculated. ${\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}$ is a diprotic acid.

Calculation of pH of $\text{HCl}$

$\text{HCl}$ is a strong acid, therefore the concentration of hydrogen ion will be equal to the concentration of  the acid.

The concentration of the hydrogen ion is 0.040 M

The pH can be calculated as follows,

$\begin{array}{l}\text{pH}\text{=}{\text{-log[H}}_{\text{3}}{\text{O}}^{\text{+}}\text{]}\\ \text{pH}\text{=}\text{-log(0}\text{.040)}\\ \text{=}1.40\end{array}$

First ionisation of

${\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}$ is a strong acid.  It is a diprotic acid.  The first ionization stage goes to completion. The ionization of ${\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}$ is as follows,

${\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}{}_{\text{(aq)}}\to {\text{H}}^{\text{+}}{}_{\text{(aq)}}{\text{+HSO}}_{\text{4}}{{}^{\text{-}}}_{\text{(aq)}}$

The concentration of all the species before and after ionization can be represented as follows,

	${\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}{}_{\text{(aq)}}\to {\text{H}}^{\text{+}}{}_{\text{(aq)}}{\text{+HSO}}_{\text{4}}{{}^{\text{-}}}_{\text{(aq)}}$
Initial $(M)$	$0.040$ $-0.040$ 0	$0.00$	$0.00$
Change $(M)$		$+0.040$	$+0.040$
Equilibrium $(M)$		$0.040$	$0.040$

Second ionization of ${\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}$

Consider the second stage of ionization ${\text{HSO}}_{\text{4}}{}^{-}$ is a weak acid.  Set up a table showing the concentrations for the second ionization stage.

Let x be the change in concentration.

Note the initial concentration of ${\text{H}}^{\text{+}}$ is 0.010 M from the first ionization.

	${\text{HSO}}_{\text{4}}{{}^{\text{-}}}_{\text{(aq)}}\to {\text{H}}^{\text{+}}{}_{\text{(aq)}}{\text{+SO}}_{\text{4}}{{}^{\text{2-}}}_{\text{(aq)}}$
Initial $(M)$	$0.040$ $-x$ $0.040-x$	$0.00$	$0.00$
Change $(M)$		$+x$	$+x$
Equilibrium $(M)$		$0.040+x$	$x$

Calculation of ${\text{H}}^{\text{+}}$ion concentration for ${\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}$

Write the ionization constant for $K_a$ . Then, solve for x.

The ${\text{K}}_{\text{a}}$ for ${\text{HSO}}^{\text{-}}{}_{\text{4}}$ is $\text{1}\text{.3}\times {\text{10}}^{\text{-2}}$

$\begin{array}{l}{K}_a=\frac{[{\text{H}}^{\text{+}}{\text{][SO}}_{\text{4}}{}^{\text{2-}}\text{]}}{[{\text{HSO}}_{\text{4}}{}^{\text{-}}\text{]}}\\ 1.3\times {10}^{-2}=\frac{(0.040+x)(x)}{(0.040-x)}\end{array}$

Since ${\text{K}}_{\text{a}}$ is quite large, we cannot make the assumptions that

$0.040-x\approx 0.040\text{and}\text{0}\text{.040+x}\approx \text{0}\text{.040}$

Therefore, we must solve a quadratic equation.

$\begin{array}{l}{x}^2+0.053x-(5.2\times {10}^{-4})=0\\ x=\frac{-0.053\pm \sqrt{{(0.053)}^2-4(1)(-5.2\times {10}^{-4})}}{2(1)}\\ x=\frac{-0.053\pm 0.070}{2}\\ x=8.5\times {10}^{-3}M\text{or}x=-0.062M\end{array}$

Neglect the negative answer.

Having solved for x, we can calculate the ${\text{H}}^{\text{+}}$ concentration at equilibrium.

$\begin{array}{l}[{\text{H}}^{\text{+}}]=0.040M+x\\ =[0.040+(8.5\times {10}^{-3})]M\\ =0.049M\end{array}$

Therefore, the ${\text{[H}}^{\text{+}}\text{]}$ ion concentration is 0.049 M

Calculation of pH of ${\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}$

The $\text{pH}$ of ${\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}$ can be calculated as follows,

$\begin{array}{l}\text{pH}\text{=}{\text{-log[H}}_{\text{3}}{\text{O}}^{\text{+}}\text{]}\\ \text{pH}\text{=}\text{-log(0}\text{.049)}\\ \text{=}1.31\end{array}$

Conclusion

The pH of a 0.040 M $\text{HCl}$ solution with that of a 0.040 M ${\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}$ solution was compared.