Chapter 16, Problem 150AP

Calculate the concentrations of all the species in a $\text{0}\text{.100 }M{\text{ Na}}_{\text{2}}{\text{CO}}_{\text{3}}$ solution.

Interpretation Introduction

Interpretation:

The concentration of all the species present in the sodium carbonate solution is to be calculated.

Concept introduction:

Salt is a strong electrolyte that dissociates completely when added to water.

When a salt contains an anion thatcomes from a weak acid, then the anion recombineswith water to produce a weak acid and hydroxide ions which ultimately forms a basic solution. Ifthe cation comes from a strong base,it does not recombine and is present in the solution as a free ion, without having any effect on the pH of the solution.

The reaction of the salt $\left(\text{BA}\right)$ is as follows:

$\text{BA }\to {\text{ B}}^{+}+{\text{ A}}^{-}$

${\text{A}}^{-}\left(aq\right)+{\text{H}}_2\text{O}\left(l\right)\rightleftharpoons \text{HA}\left(aq\right)+{\text{OH}}^{-}\left(aq\right)$

Here, ${\text{A}}^{-}$ comes from the weak acid $\text{HA}$ and ${\text{B}}^{+}$ comes from strong base $\text{BOH}$. The pH of this solution is determined by the $\left[{\text{OH}}^{-}\right]$

The relationship between $K_b$, $K_a$, and ${\text{K}}_{\text{w}}$ gives the quantitative basis of the reciprocal relationship between the strength of an acid and its conjugate base or vice-versa.

$K_a\times K_b={\text{K}}_{\text{w}}$ …… (1)

Where,${\text{K}}_{\text{a}}{\text{and K}}_{\text{b}}$ are the equilibrium constants of the acid and base respectively.

$K_b$ is the measure of dissociation of a base and is known as base-ionization constant that is specific at a specific temperatures.

$K_b=\frac{\left[{\text{OH}}^{-}\right]\left[\text{HA}\right]}{\left[{\text{A}}^{-}\right]}$…… (2)as [H₂O] = 1

Answer to Problem 150AP

Solution:

$\begin{array}{l}\left[{\text{Na}}^{+}\right]=0.200\text{ M},\left[{\text{CO}}_3{}^{2-}\right]=0.095\text{ M},\left[{\text{HCO}}_3{}^{-}\right]=4.6\times {10}^{-3}\text{M},\\ \left[{\text{H}}_2{\text{CO}}_3\right]=2.4\times {10}^{-8}\text{ M},\left[{\text{OH}}^{-}\right]=4.6\times {10}^{-3}\text{M},\left[{\text{H}}^{+}\right]=2.2\times {10}^{-12}\text{M}\end{array}$

Explanation of Solution

Given information:

The concentration of the solution that contains sodium carbonate $\left({\text{Na}}_2{\text{CO}}_3\right)$ is $0.100\text{ M}$.

All the species present in the salt solution are ${\text{Na}}^{+},{\text{ CO}}_3{}^{2-},{\text{ HCO}}_3{}^{-},{\text{ H}}_2{\text{CO}}_3,{\text{ OH}}^{-},{\text{ and H}}^{+}$.

The $0.100\text{ M}$ salt ${\text{Na}}_2{\text{CO}}_3$ is dissociated into $0.200\text{ M}$${\text{Na}}^{+}$ and $0.100\text{ M}$${\text{CO}}_3{}^{2-}$ when dissolved in water. ${\text{Na}}^{+}$ comes from a strong base $\text{NaOH}$, thus ${\text{Na}}^{+}$ does not recombine with water to form base and remains as a free ion in the solution.

But the anion ${\text{CO}}_3{}^{2-}$ comes from the weak carbonic acid $\left({\text{H}}_2{\text{CO}}_3\right)$, which is a diprotic acid, thus the anion recombines with water in two steps to produce the acid and hydroxide ions.

First step takes place as follows:

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

${\text{CO}}_3{}^{2-}\left(aq\right)+{\text{H}}_2\text{O}\left(l\right)\rightleftharpoons {\text{HCO}}_3{}^{-}\left(aq\right)+{\text{OH}}^{-}\left(aq\right)$

From table $16.8$, $K_{a2}$ for ${\text{HCO}}_3{}^{-}$ is $4.8\times {10}^{-11}$.

$K_{b2}$ of ${\text{CO}}_3{}^{2-}$, which is the conjugate base of ${\text{HCO}}_3{}^{-}$,is calculated by equation (1) by substituting the values of $K_{a2}$ and ${\text{K}}_{\text{w}}$ as follows:

$\begin{array}{c}\left(4.8\times {10}^{-11}\right)\times K_{b2}=1.0\times {10}^{-14}\\ K_{b2}=\frac{1.0\times {10}^{-14}}{4.8\times {10}^{-11}}\\ K_{b2}=2.1\times {10}^{-4}\end{array}$

Now, prepare an initial change equilibrium table and represent each of the species in terms of $x$ as follows:

$\begin{array}{ccccc}& {\text{CO}}_3{}^{2-}\left(aq\right)& {\text{H}}_2\text{O}\left(l\right)& {\text{OH}}^{-}\left(aq\right)& {\text{HCO}}_3{}^{-}\left(aq\right)\\ \text{Initial concentration}\left(\text{M}\right)& 0.100& -& 0& 0\\ \text{Change in concentration}\left(\text{M}\right)& -x& -& +x& +x\\ \text{Equilibrium concentration}\left(\text{M}\right)& 0.100-x& -& x& x\end{array}$

Now, substitute these concentrations in equation (2) as follows:

$K_{b2}=\frac{\left(x\right)\left(x\right)}{\left(0.100-x\right)}$

Approximate the value $\left(0.100-x\right)$ as $0.100$ because the value of $x$ is too small.

Substitute the value of $K_{b2}$ in the above equation,

$\begin{array}{c}2.1\times {10}^{-4}=\frac{\left(x\right)\left(x\right)}{\left(0.100\right)}\\ x^2=2.1\times {10}^{-5}\\ x=\sqrt{2.1\times {10}^{-5}}\\ x=4.6\times {10}^{-3}\end{array}$

Thus,

$\begin{array}{c}\left[{\text{HCO}}_3{}^{-}\right]=4.6\times {10}^{-3}\text{ M}\\ \left[{\text{OH}}^{-}\right]=4.6\times {10}^{-3}\text{ M}\end{array}$

Now,thesecond step proceeds as:

${\text{HCO}}_3{{}^{-}}_{\left(\text{aq}\right)}+{\text{H}}_2{\text{O}}_{\left(\text{l}\right)}\rightleftharpoons {\text{H}}_2{\text{CO}}_3{}_{\left(\text{aq}\right)}+{\text{OH}}^{-}{}_{\left(\text{aq}\right)}$

From table $16.8$, $K_{a1}$ for ${\text{H}}_2{\text{CO}}_3$ is $4.2\times {10}^{-7}$.

$K_{b1}$ of ${\text{HCO}}_3{}^{-}$, which is the conjugate base of ${\text{H}}_2{\text{CO}}_3$, is calculated byequation (1) and substitute the values of $K_{a1}$ and ${\text{K}}_{\text{w}}$ as follows:

$\begin{array}{c}\left(4.2\times {10}^{-7}\right)\times K_{b1}=1.0\times {10}^{-14}\\ K_{b1}=\frac{1.0\times {10}^{-14}}{4.2\times {10}^{-7}}\\ K_{b1}=2.4\times {10}^{-8}\end{array}$

Now, prepare an initial change equilibrium table and represent each of the species in terms of $y$ as follows:

$\begin{array}{ccccc}& {\text{HCO}}_3{{}^{-}}_{\left(\text{aq}\right)}& {\text{H}}_2{\text{O}}_{\left(\text{l}\right)}& {\text{OH}}^{-}{}_{\left(\text{aq}\right)}& {\text{H}}_2{\text{CO}}_3{}_{\left(\text{aq}\right)}\\ \text{Initial concentration}\left(M\right)& 4.6\times {10}^{-3}& -& 4.6\times {10}^{-3}& 0\\ \text{Change in concentration}\left(M\right)& -y& -& 4.6\times {10}^{-3}+y& +y\\ \text{Equilibrium concentration}\left(M\right)& 4.6\times {10}^{-3}-y& -& 4.6\times {10}^{-3}+y& y\end{array}$

Now, substitute these concentrations in equation (2) as follows:

$K_{b1}=\frac{\left(y\right)\left(4.6\times {10}^{-3}+y\right)}{\left(4.6\times {10}^{-3}-y\right)}$

Approximate the value $\left(4.6\times {10}^{-3}\pm x\right)$ as $4.6\times {10}^{-3}$ because the value of $x$ is too small.

Substitute the value of $K_{b1}$ in above equation,

$\begin{array}{c}2.4\times {10}^{-8}=\frac{\left(y\right)\left(\cancel{4.6\times {10}^{-3}}\right)}{\left(\cancel{4.6\times {10}^{-3}}\right)}\\ y=2.4\times {10}^{-8}\end{array}$

Thus,

$\begin{array}{c}\left[{\text{H}}_2{\text{CO}}_3\right]=2.4\times {10}^{-8}M\\ \left[{\text{OH}}^{-}\right]=2.4\times {10}^{-8}M\end{array}$

The concentration of sodium and carbonate ions is as follows:

$\begin{array}{c}\left[{\text{Na}}^{+}\right]=0.200M\\ \left[{\text{CO}}_3{}^{2-}\right]=0.100-4.6\times {10}^{-3}\\ =0.095M\end{array}$

The concentration of ${\text{HCO}}_3{}^{-}$ and ${\text{H}}_2{\text{CO}}_3$ is as follows:

$\begin{array}{c}\left[{\text{HCO}}_3{}^{-}\right]=\left(4.6\times {10}^{-3}\right)-\left(2.4\times {10}^{-8}\right)\\ \approx 4.6\times {10}^{-3}M\\ \left[{\text{H}}_2{\text{CO}}_3\right]=2.4\times {10}^{-8}M\end{array}$

The concentration of hydroxide and hydrogen ions is as follows:

$\begin{array}{c}\left[{\text{OH}}^{-}\right]=\left(4.6\times {10}^{-3}\right)+\left(2.4\times {10}^{-8}\right)\\ \approx 4.6\times {10}^{-3}M\\ \left[{\text{H}}^{+}\right]=\frac{1.0\times {10}^{-14}}{4.6\times {10}^{-3}}M\\ =2.2\times {10}^{-12}M\end{array}$

Conclusion

The concentration of all the given species present in the salt solution as follows:

$\begin{array}{l}\left[{\text{Na}}^{+}\right]=0.200\text{ M},\left[{\text{CO}}_3{}^{2-}\right]=0.095\text{ M},\left[{\text{HCO}}_3{}^{-}\right]=4.6\times {10}^{-3}\text{M},\\ \left[{\text{H}}_2{\text{CO}}_3\right]=2.4\times {10}^{-8}\text{ M},\left[{\text{OH}}^{-}\right]=4.6\times {10}^{-3}\text{M},\left[{\text{H}}^{+}\right]=2.2\times {10}^{-12}\text{M}\end{array}$