Chapter 16, Problem 16.85QP

Interpretation Introduction

Interpretation:

The concentrations of ${\text{HCO}}_{\text{3}}{}^{\text{-}},{\text{CO}}_3{}^{2-}\text{and}{\text{H}}^{\text{+}}$ ions in a 0.025 M ${\text{H}}_{\text{2}}{\text{CO}}_{\text{3}}$ solution 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

To Calculate: The concentrations of ${\text{HCO}}_{\text{3}}{}^{\text{-}},{\text{CO}}_3{}^{2-}\text{and}{\text{H}}^{\text{+}}$ ions in a 0.25 M ${\text{H}}_{\text{2}}{\text{CO}}_{\text{3}}$ solution

Answer to Problem 16.85QP

Answer

• The concentration of ${\text{HCO}}_{\text{3}}{}^{\text{-}}$ ion is $1.0\times {10}^{-4}$ M
• The concentration of ${\text{CO}}_3{}^{\text{2-}}$ ion is $4.8\times {10}^{-11}$ M
• The concentration of ${\text{H}}^{\text{+}}$ ion is $1.0\times {10}^{-4}$ M

Explanation of Solution

Record the given datas

The concentration of ${\text{H}}_{\text{2}}{\text{CO}}_{\text{3}}$ solution is 0.025 M

${\text{H}}_{\text{2}}{\text{CO}}_{\text{3}}$ is a weak acid with a concentration of 0.025 M. Using the ionization constant for ${\text{H}}_{\text{2}}{\text{CO}}_{\text{3}}$ the concentration of ${\text{HCO}}_{\text{3}}{}^{\text{-}},{\text{CO}}_3{}^{2-}\text{and}{\text{H}}^{\text{+}}$  can be calculated

First ionisation of ${\text{H}}_{\text{2}}{\text{CO}}_{\text{3}}$

Calculation of ${\text{[HCO}}_3{}^{-}]\text{and}[{\text{H}}^{\text{+}}]$

${\text{H}}_{\text{2}}{\text{CO}}_{\text{3}}$ is a weak acid. The first ionization stage of ${\text{H}}_{\text{2}}{\text{CO}}_{\text{3}}$ is as follows,

${\text{H}}_{\text{2}}{\text{CO}}_{\text{3}}{}_{(aq)}\rightleftarrows {\text{H}}^{\text{+}}{}_{(aq)}+{\text{HCO}}_{\text{3}}{{}^{-}}_{(aq)}$

The concentration of all the species before and after ionization can be represented as follows,

	${\text{H}}_{\text{2}}{\text{CO}}_{\text{3}}{}_{\text{(aq)}}\to {\text{H}}^{\text{+}}{}_{\text{(aq)}}{\text{+HCO}}_{\text{3}}{{}^{\text{-}}}_{\text{(aq)}}$
Initial $(M)$	$0.025$ $-x$ $0.025-x$	$0.00$	$0.00$
Change $(M)$		$+x$	$+x$
Equilibrium $(M)$		$+x$	$+x$

The acid ionization constant ${\text{H}}_{\text{2}}{\text{CO}}_{\text{3}}$ of is $1.0\times {10}^{-4}M$

$\begin{array}{l}{K}_a=\frac{[{\text{H}}^{\text{+}}{\text{][HCO}}_3{}^{\text{-}}\text{]}}{[{\text{H}}_{\text{2}}{\text{CO}}_3{}^{\text{-}}\text{]}}\\ 4.2\times {10}^{-7}=\frac{x^2}{(0.025-x)}\\ \approx \frac{x^2}{0.025}\\ x=1.0\times {10}^{-4}M\end{array}$

Therefore, the concentration of ${\text{[HCO}}_3{}^{-}]=[{\text{H}}^{\text{+}}]$= $1.0\times {10}^{-4}M$

Second stage ionization of ${\text{H}}_{\text{2}}{\text{CO}}_{\text{3}}$

Consider the second stage of ionization ${\text{HCO}}_{\text{3}}{}^{-}$ is a weak acid. Set up a table showing the concentrations for the second ionization stage.

Let y be the change in concentration.

Note the initial concentration of ${\text{H}}^{\text{+}}$ is $1.0\times {10}^{-4}M$ from the first ionization.

	${\text{HCO}}_{\text{3}}{{}^{-}}_{(aq)}\to {\text{H}}^{\text{+}}{}_{(aq)}+{\text{CO}}_3{{}^{2-}}_{(aq)}$
Initial $(M)$	$1.0\times {10}^{-4}$ $-y$ $1.0\times {10}^{-4}-y$	$0.00$	$0.00$
Change $(M)$		$+y$	$+y$
Equilibrium $(M)$		$1.0\times {10}^{-4}+y$	$y$

Calculation of $[{\text{CO}}_{\text{3}}{}^{\text{2-}}\text{]}$ ion concentration

Write the ionization constant $K_a$ . Then, solve for y.

The ${\text{K}}_{\text{a}}$ for ${\text{HCO}}_{\text{3}}{}^{-}$ is $\text{4}\text{.8}\times {\text{10}}^{\text{-11}}$

$\begin{array}{l}{K}_a=\frac{[{\text{H}}^{\text{+}}{\text{][CO}}_{\text{3}}{}^{\text{2-}}\text{]}}{[{\text{HCO}}_{\text{3}}{}^{\text{-}}\text{]}}\\ 4.8\times {10}^{-11}=\frac{[(1.0\times {10}^{-4})+y](y)}{1.0\times {10}^{-4}-y}\\ \approx \frac{(1.0\times {10}^{-4})(y)}{1.0\times {10}^{-4}}\\ y=4.8\times {10}^{-11}M\end{array}$

Since ${\text{HCO}}_{\text{3}}{}^{-}$ is a very weak acid, there is little ionization at this stage.

Therefore,  $[{\text{CO}}_{\text{3}}{}^{\text{2-}}\text{]}$ ion concentration is $4.8\times {10}^{-11}M$

Conclusion

The concentrations of ${\text{HCO}}_{\text{3}}{}^{\text{-}},{\text{CO}}_3{}^{2-}\text{and}{\text{H}}^{\text{+}}$ ions in a 0.025 M ${\text{H}}_{\text{2}}{\text{CO}}_{\text{3}}$ solution was calculated.