Chapter 15, Problem 15.107QP

Interpretation Introduction

Interpretation:

The equilibrium constant for the given reaction has to be calculated.

Concept Information:

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

Acids ionize in water. Strong acids ionize completely whereas weak acids ionize to some limited extent.

The degree to which a weak acid ionizes depends on the concentration of the acid and the equilibrium constant for the ionization.

The ionization of a weak acid $\text{HA}$ can be given as follows,

${\text{HA}}_{\text{(aq)}}\to {\text{ H}}^{\text{+}}{}_{\text{(aq)}}{\text{+A}}^{\text{-}}{}_{\text{(aq)}}$

The equilibrium expression for the above reaction is given below.

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

Where,

$K_a$ is acid ionization constant,

$[{\text{H}}^{\text{+}}\text{]}$  is concentration of hydrogen 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,

${\text{H}}_{\text{2}}\text{O}\to {\text{H}}^{\text{+}}\text{+}{\text{OH}}^{\text{-}}$

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

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

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

The relationship between the hydronium ion concentration and the hydroxide ion concentration is given by the equation,

$\text{pH}\text{+}\text{pOH}\text{=}\text{14,}\text{at}\text{25}{}^{\text{o}}\text{C}$

As $\text{pOH}$ and $\text{pH}$ are opposite scale, the total of both has to be equal to 14.

Therefore,

$\begin{array}{l}{\text{K}}_{\text{w}}\text{=}{\text{[H}}^{\text{+}}{\text{][OH}}^{\text{-}}\text{]}\\ \text{=1}\times {\text{10}}^{\text{-14}}\end{array}$

To Calculate: The equilibrium constant for the given reaction