Chapter 16.9, Problem 16.9.1SR

Interpretation Introduction

Interpretation:

The equilibrium concentration of ${\text{CO}}_{\text{3}}{}^{\text{2-}}$ in a $\text{0}\text{.050 M}$ solution of carbonic acid 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,

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

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

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 equilibrium concentration of ${\text{CO}}_{\text{3}}{}^{\text{2-}}$ in a $\text{0}\text{.050 M}$ solution of carbonic acid