Chapter 16, Problem 83QP

Calculate the concentrations of ${\text{H}}_{\text{3}}{\text{O}}^{\text{-}}{\text{, HCO}}^{–}{}_{\text{3}}{\text{, and CO}}_{\text{3}}^{\text{2}–}\text{ in a 0}\text{.025 }M{\text{ H}}_{\text{2}}{\text{CO}}_{\text{3}}\text{ solution}\text{.}$

Interpretation Introduction

Interpretation:

The concentrations of ${\text{H}}_3{\text{O}}^{+}{\text{, HCO}}_3{}^{-}{\text{, and CO}}_3{}^{2-}$ in the solution of carbonic acid are to be determined.

Concept introduction:

The first ionization of the diprotic acid takes place as:

${\text{H}}_2{\text{A}}_{\left(\text{aq}\right)}+{\text{H}}_2{\text{O}}_{\left(\text{l}\right)}\rightleftharpoons {\text{H}}_{\text{3}}{\text{O}}^{\text{+}}{}_{\left(\text{aq}\right)}+{\text{HA}}^{-}{}_{\left(\text{aq}\right)}$

${\text{K}}_{\text{a1}}$ is the measure of the dissociation of the first proton of an acid and is known as the first acid-ionization constant, which is specific at a particular temperature.

${\text{K}}_{\text{a1}}=\frac{\left[{\text{H}}_3{\text{O}}^{+}\right]\left[{\text{HA}}^{-}\right]}{\left[{\text{H}}_2\text{A}\right]}$ …… (1)

The second ionization of the diprotic acid takes place as:

${\text{HA}}^{-}{}_{\left(\text{aq}\right)}+{\text{H}}_2{\text{O}}_{\left(\text{l}\right)}\rightleftharpoons {\text{H}}_{\text{3}}{\text{O}}^{\text{+}}{}_{\left(\text{aq}\right)}+{\text{A}}^{-}{}_{\left(\text{aq}\right)}$

${\text{K}}_{\text{a2}}$ is the measure of dissociation of the second proton of an acid and is known as the second acid-ionization constant, which is specific at a particular temperature.

${\text{K}}_{\text{a2}}=\frac{\left[{\text{H}}_3{\text{O}}^{+}\right]\left[{\text{A}}^{-}\right]}{\left[{\text{HA}}^{-}\right]}$ …… (2)

Percent ionization is the percentage of acid that gets dissociated upon addition in water. It depends on the hydronium ion concentration.

$\%\text{ dissociation}=\frac{{\left[{\text{H}}_3{\text{O}}^{+}\right]}_{\text{eq}}}{{\left[\text{HA}\right]}_{\text{o}}}\times 100\%$ …… (3)

Here, ${\left[{\text{H}}_3{\text{O}}^{+}\right]}_{\text{eq}}$ is hydronium ion concentration at equilibrium and ${\left[\text{HA}\right]}_{\text{o}}$ is the original acid concentration.

Answer to Problem 83QP

Solution:

$\begin{array}{c}\left[{\text{H}}_3{\text{O}}^{+}\right]=1.0\times {10}^{-4}M\\ \left[{\text{HCO}}_{\text{3}}{}^{-}\right]=1.0\times {10}^{-4}M\\ \left[{\text{CO}}_3{}^{-}\right]=4.8\times {10}^{-11}M\end{array}$

Explanation of Solution

Given information:

The concentration of carbonic acid $\left({\text{H}}_{\text{2}}{\text{CO}}_{\text{3}}\right)$ at $25{}^{\circ }\text{C}$ is $0.025M$.

Refer to table $16.8$ for ${\text{K}}_{\text{a1}}{\text{ and K}}_{\text{a2}}$ values of carbonic acid as:

$\begin{array}{l}{\text{K}}_{\text{a1}}=\text{ 4}\text{.2}\times {\text{10}}^{-7}\\ {\text{K}}_{\text{a2}}=4.8\times {10}^{-11}\end{array}$

When carbonic acid is dissolved in water, the dissociation takes place in two steps, as it is a diprotic acid. First, one proton is partially dissociated since carbonic acid is a weak acid. Thus, the concentration of ${\text{HCO}}_3{}^{-}$ is determined by ${\text{K}}_{\text{a1}}$ of the acid. The concentration ${\text{H}}_{\text{3}}{\text{O}}^{\text{+}}$ is determined by the contribution made by both the proton dissociations.

The reaction of the first proton dissociation of carbonic acid is depicted as:

${\text{H}}_2{\text{CO}}_3{}_{\left(\text{aq}\right)}+{\text{H}}_2{\text{O}}_{\left(\text{l}\right)}\rightleftharpoons {\text{H}}_{\text{3}}{\text{O}}^{\text{+}}{}_{\left(\text{aq}\right)}+{\text{HCO}}_3{{}^{-}}_{\left(\text{aq}\right)}$

First, prepare an equilibrium table and represent each of the species in terms of $x$ as:

$\begin{array}{ccccc}& {\text{H}}_2{\text{CO}}_3{}_{\left(\text{aq}\right)}& {\text{H}}_2{\text{O}}_{\left(\text{l}\right)}& {\text{H}}_{\text{3}}{\text{O}}^{\text{+}}{}_{\left(\text{aq}\right)}& {\text{HCO}}_3{{}^{-}}_{\left(\text{aq}\right)}\\ \text{Initial concentration}\left(M\right)& 0.025& -& 0& 0\\ \text{Change in concentration}\left(M\right)& -x& -& +x& +x\\ \text{Equilibrium concentration}\left(M\right)& 0.025-x& -& x& x\end{array}$

Now, substitute these concentrations in equation (1) as:

${\text{K}}_{\text{a1}}=\frac{\left(x\right)\left(x\right)}{\left(0.025-x\right)}$

Since the value of ${\text{K}}_{\text{a1}}$ is very small, the amount of acid dissociated is less. Therefore, $\left(0.025-x\right)$ can be approximated as $0.025$. Now, substitute the value of ${\text{K}}_{\text{a1}}$ in the above equation as:

$\begin{array}{c}4.2\times {10}^{-7}=\frac{\left(x\right)\left(x\right)}{0.025}\\ x^2=\left(4.2\times {10}^{-7}\right)0.025\\ x=\sqrt{1.05\times {10}^{-8}}\\ x=1.0\times {10}^{-4}\end{array}$

Thus, ${\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]}_{1\left(\text{eq}\right)}=1.0\times {10}^{-4}M$, ${\left[{\text{H}}_{\text{2}}{\text{CO}}_3\right]}_{\left(\text{O}\right)}=0.025M$

Calculate the percent dissociation from equation (3) as:

$\begin{array}{c}\%\text{ dissociation}=\frac{1.0\times {10}^{-4}}{\text{0}\text{.025}}\times 100\%\\ =0.41\%\end{array}$

Since, the percent dissociation is much less than $5\%$, the approximation taken is valid.

Thus,

$\begin{array}{l}\left[{\text{HCO}}_3{}^{-}\right]=1.0\times {10}^{-4}M\\ {\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]}_{\text{1}}=1.0\times {10}^{-4}M\end{array}$

Now, the reaction of the second proton dissociation of carbonic acid is depicted as:

${\text{HCO}}_3{{}^{-}}_{\left(\text{aq}\right)}+{\text{H}}_2{\text{O}}_{\left(\text{l}\right)}\rightleftharpoons {\text{H}}_{\text{3}}{\text{O}}^{\text{+}}{}_{\left(\text{aq}\right)}+{\text{CO}}_3{{}^{-}}_{\left(\text{aq}\right)}$

Prepare an equilibrium table and represent each of the species in terms of $y$ as:

$\begin{array}{ccccc}& {\text{HCO}}_3{{}^{-}}_{\left(\text{aq}\right)}& {\text{H}}_2{\text{O}}_{\left(\text{l}\right)}& {\text{H}}_{\text{3}}{\text{O}}^{\text{+}}{}_{\left(\text{aq}\right)}& {\text{CO}}_3{{}^{-}}_{\left(\text{aq}\right)}\\ \text{Initial concentration}\left(M\right)& 1.0\times {10}^{-4}& -& 1.0\times {10}^{-4}& 0\\ \text{Change in concentration}\left(M\right)& -y& -& +y& +y\\ \text{Equilibrium concentration}\left(M\right)& \left(1.0\times {10}^{-4}\right)-y& -& \left(1.0\times {10}^{-4}\right)+y& y\end{array}$

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

${\text{K}}_{\text{a2}}=\frac{\left(\left(1.0\times {10}^{-4}\right)+y\right)\left(y\right)}{\left(\left(1.0\times {10}^{-4}\right)-y\right)}$

Since the value of ${\text{K}}_{\text{a2}}$ is very small, the amount of acid dissociated is less. Therefore, $\left(\left(1.0\times {10}^{-4}\right)-y\right)$ and $\left(\left(1.0\times {10}^{-4}\right)+y\right)$ can be approximated as $1.0\times {10}^{-4}$. Now, substitute the value of ${\text{K}}_{\text{a2}}$ in the above equation as:

$\begin{array}{c}4.8\times {10}^{-11}=\frac{\left(\cancel{1.0\times {10}^{-4}}\right)\left(y\right)}{1.0\times {10}^{-4}}\\ y=4.8\times {10}^{-11}\end{array}$

Thus, ${\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]}_{2\left(\text{eq}\right)}=4.8\times {10}^{-11}M$, ${\left[{\text{HCO}}_{\text{3}}{}^{-}\right]}_{\left(\text{O}\right)}=1.0\times {10}^{-4}M$.

Calculate the percent dissociation from equation (3) as:

$\begin{array}{c}\%\text{ dissociation}=\frac{4.8\times {10}^{-11}}{\text{1}\text{.0}\times {\text{10}}^{-4}}\times 100\%\\ =0.000048\%\end{array}$

Since, the percent dissociation is much less than $5\%$, the approximation taken is valid.

$\begin{array}{c}\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]={\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]}_{1\left(\text{eq}\right)}+{\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]}_{2\left(\text{eq}\right)}\\ =1.0\times {10}^{-4}+4.8\times {10}^{-11}\\ \approx 1.0\times {10}^{-4}M\end{array}$

Therefore,

$\begin{array}{c}\left[{\text{H}}_3{\text{O}}^{+}\right]=1.0\times {10}^{-4}M\\ \left[{\text{HCO}}_{\text{3}}{}^{-}\right]=1.0\times {10}^{-4}M\\ \left[{\text{CO}}_3{}^{-}\right]=4.8\times {10}^{-11}M\end{array}$

Conclusion

The concentrations of the species ${\text{H}}_3{\text{O}}^{+}{\text{, HCO}}_3{}^{-}{\text{, and CO}}_3{}^{2-}$ in the carbonic acid solution are $1.{\text{0×10}}^{\text{-4}}\text{M, 1}{\text{.0×10}}^{\text{-4}}M,\text{ and }4.8\times {10}^{-11}M$, respectively.