Chapter 20, Problem 20.56P

Interpretation Introduction

Interpretation:

The $\text{pH}$ value for $\text{0}\text{.50M}$ ${\text{H}}_{\text{2}}{\text{SO}}_{\text{3}}\left(aq\right)$ has to be calculated.

Answer to Problem 20.56P

The $\text{pH}$ value is $\text{1}\text{.1}$.

Explanation of Solution

Given: $\text{0}\text{.50M}$${\text{H}}_{\text{2}}{\text{SO}}_{\text{3}}\left(aq\right)$.

The given acid is diprotic acid. The concentration of species, ${\text{HSO}}_{\text{3}}{}^{\text{-}}\left(aq\right)$ after first dissociation is as follows,

    $\begin{array}{l}{\text{ H}}_{\text{2}}{\text{SO}}_{\text{3}}\left(aq\right){\text{ + H}}_{\text{2}}\text{O}\left(l\right)\rightleftharpoons {\text{ HSO}}_{\text{3}}{}^{\text{-}}\left(aq\right){\text{ + H}}_{\text{3}}{\text{O}}^{\text{+}}\left(aq\right)\\ \\ Initial\text{ 0}\text{.50 0 0}\\ Change\text{ -x +x +x}\\ Equilibrium\text{ 0}\text{.50-x }\text{ +x +x}\\ \\ {\text{K}}_{\text{a}}\text{ = }\frac{\left[{\text{HSO}}_{\text{3}}{}^{\text{-}}\right]\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]}{\left[{\text{H}}_{\text{2}}{\text{SO}}_{\text{3}}\right]}\end{array}$

  $\begin{array}{c}\Rightarrow {\text{ pK}}_{\text{a1}}{\text{ of H}}_{\text{2}}{\text{SO}}_{\text{3}}\text{ = 1}\text{.85}\\ \\ \Rightarrow {\text{K}}_{\text{a1}}{\text{ of H}}_{\text{2}}{\text{SO}}_{\text{3}}{\text{ = 10}}^{\text{-1}\text{.85}}\\ \\ \Rightarrow {\text{K}}_{\text{a1}}{\text{ of H}}_{\text{2}}{\text{SO}}_{\text{3}}\text{ = }0.0141\end{array}$

Now using the ${\text{K}}_{\text{a1}}$ value the concentration of ${\text{HSO}}_{\text{3}}{}^{\text{-}}\left(aq\right)$ ions are determined.

    $\begin{array}{c}{\text{K}}_{\text{a}}\text{ = }\frac{\left[{\text{HSO}}_{\text{3}}{}^{\text{-}}\right]\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]}{\left[{\text{H}}_{\text{2}}{\text{SO}}_{\text{3}}\right]}\\ \\ 0.0141=\frac{{\left[x\right]}^2}{\left[\text{0}\text{.50-}x\right]}\\ \\ 0.0141=\frac{{\left[x\right]}^2}{\left[\text{0}\text{.50}\right]}\\ \\ x\text{ = 0}{\text{.084 = HSO}}_{\text{3}}{}^{\text{-}}\left(aq\right){\text{ = H}}_{\text{3}}{\text{O}}^{\text{+}}\left(aq\right)\end{array}$

Now, the concentration of hydronium ions present in solution at second dissociation is calculated as follows,

  $\begin{array}{l}{\text{ HSO}}_{\text{3}}{}^{-}\left(aq\right){\text{ + H}}_{\text{2}}\text{O}\left(l\right)\rightleftharpoons {\text{ SO}}_{\text{3}}{{}^2}^{-}\left(aq\right){\text{ + H}}_{\text{3}}{\text{O}}^{\text{+}}\left(aq\right)\\ \\ Initial\text{ 0}\text{.084 0 0}\\ Change\text{ -x +x +x}\\ Equilibrium\text{ 0}\text{.084-x }\text{ +x +x}\\ \\ {\text{K}}_{\text{a}}\text{ = }\frac{\left[{\text{SO}}_{\text{3}}{{}^2}^{-}\right]\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]}{\left[{\text{HSO}}_{\text{3}}{}^{-}\right]}\end{array}$

    $\begin{array}{c}\Rightarrow {\text{ pK}}_{\text{a2}}{\text{ of H}}_{\text{2}}{\text{SO}}_{\text{3}}\text{ = 7}\text{.2}\\ \\ \Rightarrow {\text{K}}_{\text{a2}}{\text{ of H}}_{\text{2}}{\text{SO}}_{\text{3}}{\text{ = 10}}^{\text{-7}\text{.2}}\\ \\ \Rightarrow {\text{K}}_{\text{a2}}{\text{ of H}}_{\text{2}}{\text{SO}}_{\text{3}}\text{ = }6.31\times {10}^{-8}\end{array}$

  $\begin{array}{c}{\text{K}}_{\text{a}}\text{ = }\frac{\left[{\text{SO}}_{\text{3}}{{}^2}^{-}\right]\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]}{\left[{\text{HSO}}_{\text{3}}{}^{-}\right]}\\ \\ \text{6}\text{.31}\times {\text{10}}^{\text{-8}}=\frac{{\left[x\right]}^2}{\left[\text{0}\text{.084-}x\right]}\\ \\ \text{6}\text{.31}\times {\text{10}}^{\text{-8}}=\frac{{\left[x\right]}^2}{\left[\text{0}\text{.084}\right]}\\ \\ x\text{ = 7}\text{.28}\times {\text{10}}^{\text{-5}}{\text{ = H}}_{\text{3}}{\text{O}}^{\text{+}}\left(aq\right)\end{array}$

The total concentration of hydronium ions present in solution is shown as follows,

    $\begin{array}{l}\Rightarrow \text{0}\text{.084 + 7}\text{.28}\times {\text{10}}^{\text{-5}}\\ \\ \Rightarrow \text{ 0}\text{.0841}\end{array}$

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

    $\begin{array}{c}\text{pH = -log }\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]\\ \\ =\text{ -log }\left[0.0841\right]\\ \\ =\text{ 1}\text{.1}\end{array}$