Chapter 16, Problem 16.131QP

Interpretation Introduction

Interpretation:

The relation that ${\text{[OH}}^{\text{-}}\text{]=}\frac{K_w}{K_a}$ for the given acid and salt has to be shown.

Concept Information:

Acid ionization constant ${\text{K}}_{\text{a}}$:

The equilibrium expression for the reaction ${\text{HA}}_{\text{(aq)}}{\text{+H}}_{\text{2}}\text{O }\to {\text{H}}_{\text{3}}{\text{O}}^{\text{+}}{}_{\text{(aq)}}{\text{+A}}^{\text{-}}{}_{\text{(aq)}}$ is given below.

$K_a=\frac{[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}{\text{][A}}^{\text{-}}\text{]}}{[\text{HA]}}$

Where,

$K_a$ is acid ionization constant,

$[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\text{]}$  is concentration of hydronium ion,

$[{\text{A}}^{\text{-}}\text{]}$  is concentration of acid anion,

$[\text{HA]}$ is concentration of the acid

Autoionization of water:

The equation of equilibrium for autoionization of water is,

${\text{K}}_{\text{w}}\text{=}{\text{[H}}^{\text{+}}{\text{][OH}}^{\text{-}}\text{]}$

The equilibrium expression for water at $\text{25}{}^{\text{o}}\text{C}$ is,

${\text{[H}}^{\text{+}}{\text{][OH}}^{\text{-}}\text{]= 1}\times {\text{10}}^{\text{-14}}$

Taking negative logarithm on both sides, we get

$\begin{array}{l}-\log ({\text{[H}}^{\text{+}}{\text{][OH}}^{\text{-}}\text{])= -log(1}\times {\text{10}}^{\text{-14}})\\ (-\log {\text{[H}}^{\text{+}}{\text{])+(-log[OH}}^{\text{-}}\text{])= 14})\end{array}$