Chapter 16, Problem 16.86QP

Interpretation Introduction

Interpretation:

The pH of a 0.25 M aqueous solution ${\text{H}}_{\text{3}}{\text{PO}}_{\text{4}}$ 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,

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

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

$\text{pH}$ definition:

The concentration of hydrogen ion is measured using $\text{pH}$ scale.  The acidity of aqueous solution is expressed by $\text{pH}$ scale.

The $\text{pH}$ of a solution is defined as the negative base-10 logarithm of the hydrogen or hydronium ion concentration.

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

Diprotic and polyprotic acids:

Acids having two or more hydrogen atoms are termed as diprotic or polyprotic acids.  These acids lose one proton at a time by undergoing successive ionizations.

For diprotic acids, the successive ionization constants are designated as $K_{a1}and{K}_{a2}$

For triprotic acids, the successive ionization constants are designated as $K_{a1},K_{a2}and{K}_{a3}$

To Calculate: The pH of a 0.25 M aqueous solution ${\text{H}}_{\text{3}}{\text{PO}}_{\text{4}}$