Chapter 8, Problem 8.30P

Interpretation Introduction

Interpretation:

The value of $\text{pH}$ $\text{0}\text{.050M}\text{NaCN}$ 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.

Answer to Problem 8.30P

The value of $\text{pH}$ is $10.9.$

Explanation of Solution

As known, ${\text{K}}_{\text{a}}\text{of}\text{HCN}$ is $\text{6}\text{.2}\times {\text{10}}^{-10}.$

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

    $\begin{array}{c}{\text{K}}_{\text{b}}\text{=}\frac{{\text{K}}_{\text{w}}}{{\text{K}}_{\text{a}}}\\ \\ \text{=}\frac{\text{1}\text{×}{\text{10}}^{\text{-14}}}{\text{6}\text{.2}\times {\text{10}}^{-10}}=1.613\text{×}{\text{10}}^{\text{-5}}.\end{array}$

In water:

  $\begin{array}{l}{\text{CN}}^{\text{-}}\text{+}{\text{H}}_{\text{2}}\text{O}\rightleftharpoons \text{HCN}\text{+}{\text{OH}}^{\text{-}}\\ \left(\text{0}\text{.050}\right)00\\ \left(\text{0}\text{.050}-\text{x}\right)xx\end{array}$

From the above equation,

  $\begin{array}{l}{\text{K}}_{\text{b}}\text{=}\frac{\left[\text{HCN}\right]\left[{\text{OH}}^{\text{-}}\right]}{\left[{\text{CN}}^{\text{-}}\right]}\\ \\ 1.613\text{×}{\text{10}}^{\text{-5}}=\frac{x^2}{0.050}\\ \\ x=\left(1.613\text{×}{\text{10}}^{\text{-5}}\right)\left(0.050\right)=\sqrt{0.08065\text{×}{\text{10}}^{\text{-5}}}=8.98\text{×}{\text{10}}^{\text{-4}}.\end{array}$

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

  $\begin{array}{c}\text{pOH}\text{=}-\text{log}\left[{\text{OH}}^{\text{-}}\right]\\ \\ \text{=}-\text{log}\left(8.98\text{×}{\text{10}}^{\text{-4}}\right)\\ \\ \text{=}\mathrm{3.0.}\\ \\ \text{pH}\text{=}\text{14}-\text{pOH}\text{=}\text{14}-\text{3}\text{.0}\text{=}11\end{array}$

Therefore, the value of $\text{pH}$ is $11$.