Chapter 15, Problem 15.70QA

Interpretation Introduction

To find:

The $pH$ of $5.00\times {10}^{-4}M H_2{SO}_4$.

Answer to Problem 15.70QA

Solution:

$pH=3.02$

Explanation of Solution

1) Concept:

Sulfuric acid is a strong diprotic acid in that one hydrogen atom per molecule ionizes completely $(K_{a1}>>1)$, but the second hydrogen atom may not completely ionize for every molecule $(K_{a2}>>0.012)$.

We start with the first ionization reaction that goes to completion.

$H_2{SO}_{4\left(aq\right)}+ H_2{O}_{\left(l\right)}\to {HSO}_{4\left(aq\right)}^{-}+H_3{O}_{\left(aq\right)}^{+}$

Therefore, as the second ionization step begins ${[HSO}_4^{-}]= {[H}_3{O}^{+}]=5.00\times {10}^{-4}M.$

Ionization of ${HSO}_4^{-}$ then produces additional $H_3{O}^{+}$ ions

${HSO}_{4\left(aq\right)}^{-}+ H_2{O}_{\left(l\right)}⥨{SO}_{4\left(aq\right)}^{2-}+H_3{O}_{\left(aq\right)}^{+}$

The increase in ${[H}_3{O}^{+}]$ due to second ionization will be between $0$ to $5.00\times {10}^{-4}M$, which means the total ${[H}_3{O}^{+}]$ value will be between $5.00\times {10}^{-4} and 0.001M$ and the $pH$ of the solution at equilibrium should be greater than 1.

We have to set up a RICE table for the second ionization reaction. Initially ${[HSO}_4^{-}]= {[H}_3{O}^{+}]=5.00\times {10}^{-4}M$. The change in ${[H}_3{O}^{+}]$ during the second ionization step is $+x$ filling the other cells of the table based on the stoichiometry of the second ionization step.

2) Formula:

i) $pH= -\log \left[{H_{3}O}^{+}\right]$

ii) $x=\frac{\left(-b\pm \sqrt{b^2-4ac }\right)}{2a}$

3) Given:

Concentration of ${ H}_2{SO}_4 is 5.00\times {10}^{-4}M$

4) Calculations:

Reaction	${HSO}_{4\left(aq\right)}^{-}+ H_2{O}_{\left(l\right)} ⥨ {SO}_{4\left(aq\right)}^{2-}+ H_3{O}_{\left(aq\right)}^{+}$
	${[HSO}_4^{-}\left] \right(M)$	${[SO}_4^{2-}\left] \right(M)$	${[H}_3{O}^{+}\left]\right(M)$
Initial	$5.00\times {10}^{-4}$	$0$	$5.00\times {10}^{-4}$
Change	$-x$	$+x$	$+x$
Equilibrium	$(5.00\times {10}^{-4}-x)$	$x$	$(5.00\times {10}^{-4}+x)$

Inserting the equilibrium concentrations in the equilibrium constant expression for $K_{a_2}$.

$K_{a_2}= \frac{{[H}_3{O}^{+}\left] {[SO}_4^{2-}\right] }{{[HSO}_4^{-}]}$

$1.2\times {10}^{-2}= \frac{\left(5.00\times {10}^{-4}+x\right)\left(x\right)}{(5.00\times {10}^{-4}-x)}$

${6.0\times 10}^{-6}-0.012x= x^2+\left({5.0\times 10}^{-4}\right)x$

$x^2+0.0125x-{6\times 10}^{-6}=0$

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

Here, $a=1$, $b=0.0125$ and $c= -{6\times 10}^{-6}$.

$x=\frac{\left(-0.0125\pm \sqrt{{\left(0.0125\right)}^2-(4\times 1\times (-{6\times 10}^{-6}\left)\right) }\right)}{2\times 1}$

$x=\frac{\left(-0.0125\pm \sqrt{0.00018 }\right)}{2}$

Solving this equation for x yields a positive value and a negative value.

$x=0.0004628 and x= -0.01296$

The negative value for x has no physical meaning because it gives us a negative ${[SO}_4^{2-}]$ value. Therefore,

${[H}_3{O}^{+}]=\left(5.00\times {10}^{-4}+x\right)=\left(5.00\times {10}^{-4}+0.0004628\right)=0.0009628$

The corresponding $pH$ is

$pH= -\log \left[{H_{3}O}^{+}\right]= -\log \left(0.0009628\right)=3.02$

Conclusion:

The total hydrogen ion concentration in an $H_2{SO}_4$ solution is the result of a complete first dissociation and partial second dissociation,