Chapter 16, Problem 16.150QP

Interpretation Introduction

Interpretation:

The concentration of all species in a 0.100 M ${\text{Na}}_{\text{2}}{\text{CO}}_{\text{3}}$ solution has to be calculated

Concept Information:

Base  ionization constant $K_b$

The equilibrium expression for the ionization of weak base $\text{B}$ will be,

${\text{B}}_{(aq)}+{\text{H}}_{\text{2}}{\text{O}}_{(l)}\stackrel{}{\to }{\text{HB}}^{\text{+}}{}_{(aq)}+{\text{OH}}^{\text{-}}{}_{(aq)}$

${\text{K}}_b=\frac{[{\text{HB}}^{\text{+}}][{\text{OH}}^{\text{-}}]}{[\text{B}]}$

Where $K_b$ is base ionization constant, $[{\text{OH}}^{-}\text{]}$ is concentration of hydroxide ion, $[{\text{HB}}^{\text{+}}\text{]}$ is concentration of conjugate acid, $[\text{B]}$ is concentration of the base

To Calculate: The concentration of all species in a 0.100 M ${\text{Na}}_{\text{2}}{\text{CO}}_{\text{3}}$ solution

Answer to Problem 16.150QP

Answer

The concentration of all species in a 0.100 M ${\text{Na}}_{\text{2}}{\text{CO}}_{\text{3}}$ solution is:

$\begin{array}{l}{\text{[Na}}^{\text{+}}\text{]}\text{=}\text{0}\text{.200 M}\\ {\text{[CO}}_{\text{3}}{}^{\text{2-}}\text{]}\text{=}0.095M\\ [{\text{HCO}}_{\text{3}}{}^{\text{-}}\text{]}\text{=}4.6\times {10}^{-3}M\\ [{\text{H}}_{\text{2}}{\text{CO}}_{\text{3}}\text{]}\text{=}\text{2}\text{.4}\times {\text{10}}^{\text{-8}}M\\ [{\text{OH}}^{\text{-}}\text{]}\text{=}4.6\times {10}^{-3}M\\ [\text{H+}]=2.2\times {10}^{-12}M\end{array}$

Explanation of Solution

Given data:

The concentration of ${\text{Na}}_{\text{2}}{\text{CO}}_{\text{3}}$ solution is 0.100 M

Ionization of ${\text{Na}}_{\text{2}}{\text{CO}}_{\text{3}}$

${\text{Na}}_{\text{2}}{\text{CO}}_{\text{3}}$ is a weak base with a concentration of 0.100 M.

$0.100M{\text{Na}}_{\text{2}}{\text{CO}}_{\text{3}}\to 0.200M{\text{Na}}^{\text{+}}+0.100M{\text{CO}}_{\text{3}}{}^{\text{2-}}$

Using the ionization constant for ${\text{Na}}_{\text{2}}{\text{CO}}_{\text{3}}$ the concentration of ${\text{Na}}^{\text{+}}{\text{, HCO}}_{\text{3}}{}^{\text{-}},{\text{CO}}_3{}^{2-},{\text{H}}_{\text{2}}{\text{CO}}_{\text{3}}{\text{, OH}}^{\text{-}}\text{ and }{\text{H}}^{\text{+}}$ can be calculated as follows.

First stage:

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

	${\text{CO}}_{\text{3}}{{}^{2-}}_{(aq)}+{\text{H}}_{\text{2}}{\text{O}}_{(l)}\stackrel{}{\to }{\text{HCO}}_{\text{3}}{{}^{-}}_{(aq)}+{\text{OH}}^{\text{-}}{}_{(aq)}$
Initial $(M)$	0.100 $-x$ 0.100-x	$0.00$	$0.00$
Change $(M)$		$+x$	$+x$
Equilibrium $(M)$		$x$	$x$

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

$\begin{array}{l}{K}_1=\frac{{\text{K}}_{\text{w}}}{{\text{K}}_{\text{2}}}\\ =\frac{1.0\times {10}^{-14}}{4.8\times {10}^{-11}}\\ =2.1\times {10}^{-4}\end{array}$

$\begin{array}{l}{K}_1=\frac{{\text{[HCO}}_3{}^{\text{-}}{\text{][OH}}^{\text{-}}\text{]}}{[{\text{CO}}_3{}^{\text{2-}}\text{]}}\\ 2.1\times {10}^{-4}=\frac{x^2}{(0.100-x)}\\ \approx \frac{x^2}{0.100}\\ x=4.6\times {10}^{-3}M\\ {\text{[HCO}}_{\text{3}}{}^{\text{-}}]=[{\text{OH}}^{\text{-}}]=4.6\times {10}^{-3}M\end{array}$

Therefore, the concentration of ${\text{[HCO}}_{\text{3}}{}^{\text{-}}]=[{\text{OH}}^{\text{-}}]=4.6\times {10}^{-3}M$

Second stage:

Let y be the change in concentration.

Note the initial concentration of ${\text{[HCO}}_{\text{3}}{}^{\text{-}}]$ is $4.6\times {10}^{-3}M$ from the first ionization.

	${\text{HCO}}_{\text{3}}{{}^{-}}_{(aq)}+{\text{H}}_{\text{2}}{\text{O}}_{(l)}\stackrel{}{\to }{\text{H}}_{\text{2}}{\text{CO}}_{\text{3}}{}_{(aq)}+{\text{ OH}}^{-}{}_{(aq)}$
Initial $(M)$	$4.6\times {10}^{-3}$ $-y$ $(4.6\times {10}^{-3})-y$	$0.00$	$0.00$
Change $(M)$		$+y$	$+y$
Equilibrium $(M)$		y	$(4.6\times {10}^{-3})+y$

$\begin{array}{l}{\text{K}}_{\text{2}}=\frac{[{\text{H}}_{\text{2}}{\text{CO}}_{\text{3}}{\text{][OH}}^{\text{-}}\text{]}}{[{\text{HCO}}_{\text{3}}{}^{\text{-}}\text{]}}\\ 2.4\times {10}^{-8}=\frac{y[(4.6\times {10}^{-3})+y]}{(4.6\times {10}^{-3})-y}\\ \approx \frac{(4.6\times {10}^{-3})(y)}{4.6\times {10}^{-3}}\\ y=2.4\times {10}^{-8}M\end{array}$

At equilibrium:

$\begin{array}{l}{\text{[Na}}^{\text{+}}\text{]}\text{=}\text{0}\text{.200 M}\\ {\text{[CO}}_{\text{3}}{}^{\text{2-}}\text{]}\text{=}\text{(0}\text{.100-4}\text{.6}\times {\text{10}}^{\text{-3}})M\\ =0.095M\\ [{\text{HCO}}_{\text{3}}{}^{\text{-}}\text{]}\text{=}\text{(4}\text{.6}\times {\text{10}}^{\text{-3}})M-(2.4\times {10}^{-8})M\\ \approx 4.6\times {10}^{-3}M\\ [{\text{H}}_{\text{2}}{\text{CO}}_{\text{3}}\text{]}\text{=}\text{2}\text{.4}\times {\text{10}}^{\text{-8}}M\\ [{\text{OH}}^{\text{-}}\text{]}\text{=}\text{(4}\text{.6}\times {\text{10}}^{\text{-3}})M+(2.4\times {10}^{-8})M\\ \approx 4.6\times {10}^{-3}M\\ [\text{H+}]=\frac{1.0\times {10}^{-14}}{4.6\times {10}^{-3}}\\ =2.2\times {10}^{-12}M\end{array}$

Conclusion

The concentration of all species in a 0.100 M ${\text{Na}}_{\text{2}}{\text{CO}}_{\text{3}}$ solution was calculated