Chapter 20, Problem 20.91P

Interpretation Introduction

Interpretation:

The value of $\text{pH}$ for $\text{0}\text{.010 M}$ sulfurous acid has to be calculated.

Concept Introduction:

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

  $\text{pH}\text{=}\text{-log}\left[{\text{H}}^{\text{+}}\right]$

Answer to Problem 20.91P

The value of $\text{pH}$ is $\text{4}\text{.7}$.

Explanation of Solution

Given: $\text{0}\text{.010 M}$ sulfurous acid.

The sulfurous acid is a diprotic acid. Its first dissociation is as follows,

    ${\text{H}}_{\text{2}}{\text{SO}}_{\text{3}}\left(aq\right){\text{ + H}}_{\text{2}}\text{O}\left(l\right)\rightleftharpoons {\text{HSO}}_{\text{3}}^{\text{-}}\left(aq\right)+{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\left(aq\right)$

The concentration of hydronium ions present is calculated by using ICE table as follows,

    $\begin{array}{l}{\text{ H}}_{\text{2}}{\text{SO}}_{\text{3}}\left(aq\right){\text{ + H}}_{\text{2}}\text{O}\left(l\right)\rightleftharpoons {\text{HSO}}_{\text{3}}^{\text{-}}\left(aq\right)+{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\left(aq\right)\\ \\ Initial\text{ 0}\text{.010 0 0}\\ Change\text{ -x +x +x}\\ Equilibrium\text{ 0}\text{.010-x }\text{ +x +x}\\ \\ {\text{K}}_{\text{a}}\text{ = }\frac{\left[{\text{HSO}}_{\text{3}}^{\text{-}}\right]\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]}{\left[{\text{H}}_{\text{2}}{\text{SO}}_{\text{3}}\right]}\end{array}$

        $\begin{array}{c}{\text{K}}_{\text{a}}\text{ = }\frac{\left[{\text{HSO}}_{\text{3}}^{\text{-}}\right]\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]}{\left[{\text{H}}_{\text{2}}{\text{SO}}_{\text{3}}\right]}\\ \\ 1.4\times {10}^{-2}\text{ M}=\frac{x^2}{0.010-x}\\ \\ =0.0067\text{M}\end{array}$

    $\begin{array}{c}\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]\text{ =}\left[{\text{HSO}}_{\text{3}}^{\text{-}}\right]\text{= }0.0067\text{ M}\\ \\ \left[{\text{H}}_{\text{2}}{\text{SO}}_{\text{3}}\right]=0.010-x\\ \\ =0.010-0.0067\\ \\ \text{= 3}\text{.3}\times {\text{10}}^{\text{-3}}\text{M}\end{array}$

The second dissociation of ${\text{HSO}}_{\text{3}}^{\text{-}}\left(aq\right)$ is as follows,

    ${\text{HSO}}_{\text{3}}^{\text{-}}\left(aq\right){\text{ + H}}_{\text{2}}\text{O}\left(l\right)\rightleftharpoons {\text{SO}}_{\text{3}}^{\text{2-}}\left(aq\right)+{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\left(aq\right)$

The hydronium ions present after second dissociation is calculated as shown below,

    $\begin{array}{l}{\text{ HSO}}_{\text{3}}^{\text{-}}\left(aq\right){\text{ + H}}_{\text{2}}\text{O}\left(l\right)\rightleftharpoons {\text{SO}}_{\text{3}}^{\text{2-}}\left(aq\right)+{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\left(aq\right)\\ \\ Initial\text{ 0}\text{.0067 0 0}\\ Change\text{ -x +x +x}\\ Equilibrium\text{ 0}\text{.0067-x }\text{ +x +x}\\ \\ {\text{K}}_{\text{a}}\text{ = }\frac{\left[{\text{SO}}_{\text{3}}^{\text{2-}}\right]\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]}{\left[{\text{HSO}}_{\text{3}}^{\text{-}}\right]}\end{array}$

    $\begin{array}{c}{\text{K}}_{\text{a}}\text{ = }\frac{\left[{\text{SO}}_{\text{3}}^{\text{2-}}\right]\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]}{\left[{\text{HSO}}_{\text{3}}^{\text{-}}\right]}\\ \\ \text{6}\text{.31}\times {\text{10}}^{\text{-8}}\text{ = }\frac{{\text{x}}^{\text{2}}}{\text{0}\text{.0067-x}}\\ \\ x\text{ = }\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]=2.056\times {\text{10}}^{\text{-5}}\end{array}$

Now, the $\text{pH}$ value is calculated as follows,

    $\begin{array}{c}\text{pH = -log}\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]\\ \\ =\text{ -log}\left[2.056\times {\text{10}}^{\text{-5}}\right]\\ \\ =\text{ 4}\text{.7}\end{array}$