Chapter 8, Problem 8.31P

Interpretation Introduction

Interpretation:

The value of $\text{pH}$ and fraction of association of $\text{0}\text{.026M}\text{NaOCl}$ has to be found.

Concept Introduction:

pH:

$\text{pH}$ is a scale used to specify how acidic or basic a solution is.  It ranges from 0-14. $\text{pH}$ 7.0 is considered as neutral solution, $\text{pH}$ more than 7.00 is taken as basic solution whereas $\text{pH}$ less than 7.0 is considered as acidic solution (at 25^oC).  It is the measurement of activity of free ${\text{H}}^{\text{+}}$ and ${\text{OH}}^{\text{-}}$ in solution.

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

  $\text{pOH}\text{=}-\log \left[{\text{OH}}^{-}\right]$

From ionization constant of water ${\text{K}}_{\text{w}}\text{=}\left[{\text{H}}^{\text{+}}\right]\left[{\text{OH}}^{\text{-}}\right]$

Ionization constant of water at ${\text{25}}^{\text{ο}}\text{C}$ ${\text{K}}_{\text{w}}\text{=}1\times {10}^{-14}$.

Less the $\text{pH}$ more is the acidity of the solution.

Explanation of Solution

The equation as follows,

  ${\text{OCl}}^{-}\left(\text{aq}\right)\text{+}{\text{H}}_{\text{2}}\text{O}\left(\text{l}\right)\rightleftharpoons \text{HOCl}\left(\text{aq}\right)\text{+}{\text{OH}}^{-}\left(\text{aq}\right)$

As known, ${\text{K}}_{\text{b}}\text{=}\frac{\left[\text{HOCl}\right]\left[{\text{OH}}^{-}\right]}{\left[{\text{OCl}}^{-}\right]}$

Dissociation of acid as shown below,

  $\text{HOCl}\left(\text{aq}\right)\rightleftharpoons {\text{H}}^{+}\left(\text{aq}\right)\text{+}{\text{OCl}}^{-}\left(\text{aq}\right)$

The value of ${\text{K}}_a$ as,

    $3.5\text{×}{\text{10}}^{\text{-8}}\text{=}\frac{\left[{\text{OCl}}^{-}\right]\left[{\text{H}}^{+}\right]}{\left[\text{HOCl}\right]}$

The value of ${\text{K}}_b$ is calculated as,

  $\begin{array}{c}{\text{K}}_b\text{=}\frac{{\text{K}}_{\text{w}}}{{\text{K}}_{\text{a}}}\\ \\ \text{=}\frac{1\text{×}{\text{10}}^{\text{-14}}}{3.5\text{×}{\text{10}}^{\text{-8}}}=2.86\text{×}{\text{10}}^{\text{-7}}\end{array}$

For the reaction,

  $\begin{array}{l}{\text{OCl}}^{\text{-}}\left(\text{aq}\right)\text{+}{\text{H}}_{\text{2}}\text{O}\left(\text{l}\right)\rightleftharpoons \text{HOCl }\left(\text{aq}\right)\text{+}{\text{OH}}^{\text{-}}\left(\text{aq}\right)\\ \left(\text{0}\text{.00573-x}\right)\text{x}\text{x}\end{array}$

From the above equation,

  $\begin{array}{l}{\text{K}}_{\text{b}}\text{=}\frac{\left[\text{HOCl}\right]\left[{\text{OH}}^{\text{-}}\right]}{\left[{\text{OCl}}^{\text{-}}\right]}\\ \\ 2.86\text{×}{\text{10}}^{\text{-7}}=\frac{x^2}{0.00573-x}\\ \\ \text{As}\text{'x'}\text{is}\text{negligible,}\\ \\ x=\sqrt{\left(2.86\text{×}{\text{10}}^{\text{-7}}\right)\left(0.00573\right)}=\sqrt{0.01639\text{×}{\text{10}}^{\text{-7}}}=4.05\text{×}{\text{10}}^{\text{-5}}.\end{array}$

Fraction of association of base:

  $\begin{array}{c}\frac{\left[B{H}^{+}\right]}{\left[B{H}^{+}\right]+\left[B\right]}=\frac{4.05\text{×}{\text{10}}^{\text{-5}}}{\left(4.05\text{×}{\text{10}}^{\text{-5}}\right)+0.00573}=\frac{4.05\text{×}{\text{10}}^{\text{-5}}}{5.7705\text{×}{\text{10}}^{\text{-3}}}\\ \\ =0.0070\end{array}$

The fraction of association is $\mathrm{0.0070.}$

The value of $\text{pOH}$ is calculated as,

  $\begin{array}{c}\text{pOH}\text{=}-\text{log}\left[{\text{OH}}^{\text{-}}\right]\\ \\ \text{=}-\text{log}\left(4.05\text{×}{\text{10}}^{\text{-5}}\right)\\ \\ \text{=}\mathrm{4.39.}\\ \\ \text{pH}\text{=}\text{14}-\text{pOH}\text{=}\text{14}-\text{4}\text{.39}\text{=}9.6\end{array}$

Therefore, the value of $\text{pH}$ is $9.6.$