Problem 5

5. Calculate the pH of a 0.240 M solution of sodium bicarbonate (NaHCO₃).

This will dissociate to produce the \(\ce{HCO3-}\) ion, which is in the carbonic acid system.

\[\begin{align}
&\ce{H2CO3 + H2O \leftrightarrow HCO3- + H3O+} &&\mathrm{K_{a1}}\nonumber\\
&\ce{HCO3- + H2O \leftrightarrow CO3^2- + H3O+} &&\mathrm{K_{a2}}\nonumber
\end{align}\nonumber\]

What we observe in this case is that the \(\ce{HCO3-}\) is an intermediate and we essentially have “all” of an intermediate. Some small amount of H₂CO₃ and \(\ce{CO3^2-}\) will form, since the two equilibrium constants have finite values, but not enough of either one will form to constitute a buffer. The bicarbonate ion has the ability to react as an acid (reaction 2) or a base (reaction 1), and it might be tempting to determine whether it’s a stronger acid or base (by comparing the relative magnitude of K_a2 to the K_b of K_a1) and use that reaction to calculate the pH. The situation is complicated by the fact that any H₂CO₃ and \(\ce{CO3^2-}\) that are formed can neutralize each other.

What we need to do in this case is write an expression for [H₃O⁺] in terms of species in the carbonate system. Remember, there is some H₃O⁺ around from dissociation of water, however, this value will be overcome by the carbonate system and can be ignored in this calculation.

We need to realize that we have a lot of \(\ce{HCO3-}\) in solution and consider what each of the reactions above does to the concentration of H₃O⁺ in solution. At the start, before the system equilibrates, there is no H₂CO₃ or \(\ce{CO3^2-}\).

If we look at the second reaction (K_a2), we see that for every \(\ce{CO3^2-}\) that is produced we need to produce one equivalent of H₃O⁺. If this was the only reaction that occurred, it would allow us to write the following expression:

\[\mathrm{[H_3O^+] = [CO_3^{2-}]}\nonumber\]

If we look at the first reaction (K_a1), we see that in order to produce a molecule of H₂CO₃, we actually need to remove an H₃O⁺ species. Therefore, every H₂CO₃ that we find in the final solution subtracts an H₃O⁺. This would allow us to write:

\[\mathrm{[H_3O^+] =\, –[H_2CO_3]}\nonumber\]

Both \(\ce{CO3^2-}\) and H₂CO₃ will be present in the final solution, so we can combine these two equations and come up with the following expression for the concentration of H₃O⁺ in the final solution:

\[\mathrm{[H_3O^+] = [CO_3^{2-}] - [H_2CO_3]}\nonumber\]

Now we need to do some algebra. First, rearrange the above expression into the following:

\[\mathrm{[H_3O^+] + [H_2CO_3] = [CO_3^{2-}]}\label{1}\]

Now rearrange the K_a1 and K_a2 expressions to solve them for [H₂CO₃] and [\(\ce{CO3^2-}\)] respectively.

\[\mathrm{K_{a1}= \dfrac{[HCO_3^-][H_3O^+]}{[H_2CO_3]} \hspace{60px} [H_2CO_3]= \dfrac{[HCO_3^-][H_3O^+]}{K_{a1}}}\label{2}\]

\[\mathrm{K_{a2}= \dfrac{[CO_3^{2-}][H_3O^+]}{[HCO_3^-]} \hspace{60px} [CO_3^{2-}]=\dfrac{K_{a2}[HCO_3^-]}{[H_3O^+]}}\label{3}\]

Substitute the expressions for [H₂CO₃] and [\(\ce{CO3^2-}\)] into equation (\(\ref{1}\)).

\[\mathrm{[H_3O^+] + \dfrac{[HCO_3^-][H_3O^+]}{K_{a1}}= \dfrac{K_{a2}[HCO_3^-]}{[H_3O^+]}}\label{4}\]

Multiply each side through by K_a1[H₃O⁺] to remove all terms from the denominator.

\[\mathrm{K_{a1}[H_3O^+]^2 + [HCO_3^-][H_3O^+]^2 = K_{a1}K_{a2}[HCO_3^-]}\label{5}\]

Pull out an [H₃O⁺]² term from the left-hand side of the equation.

\[\mathrm{[H_3O^+]^2(K_{a1} + [HCO_3^-]) = K_{a1}K_{a2}[HCO_3^-]}\label{6}\]

Divide both sides by (K_a1 + [\(\ce{HCO3-}\)]) to give the following:

\[\mathrm{[H_3O^+]^2= \dfrac{K_{a1}K_{a2}[HCO_3^-]}{K_{a1}+[HCO_3^-]}}\label{7}\]

Now comes a critical assessment of the terms in the denominator, as we want to compare the magnitude of K_a1 to [\(\ce{HCO3-}\)]. Usually, we would anticipate that K_a1 is a fairly small number since this is a weak acid. For example, in the current problem, K_a1 is 4.47×10^-7. Usually the concentration of the intermediate (\(\ce{HCO3-}\) in this case) is fairly high (0.240 M in this case). In many situations the concentration of the intermediate is a lot larger than the K_a value.

\[\mathrm{[HCO_3^-] >> K_{a1}}\nonumber\]

In this case, we can ignore the K_a1 in the term (K_a1 + [\(\ce{HCO3-}\)]). That simplifies equation (\(\ref{7}\)) to the following:

\[\mathrm{[H_3O^+]^2= \dfrac{K_{a1}K_{a2}[HCO_3^-]}{[HCO_3^-]}}\label{8}\]

Notice how the [\(\ce{HCO3-}\)] terms now cancel out of the equation leaving:

\[\mathrm{[H_3O^+]^2 = K_{a1} K_{a2}}\label{9}\]

Using the properties of logs, this expression can be rewritten as follows:

\[\mathrm{pH=\dfrac{pK_{a1}+ pK_{a2}}{2}}\label{10}\]

There is a certain way in which this outcome seems to make sense. We stated at the onset that one of the problems was that the bicarbonate ion could act as an acid and a base. The pK_a1 value represents bicarbonate acting as a base, and pK_a2 represents bicarbonate acting as an acid. This equation essentially represents an average of these two values. That average will also reflect whether the intermediate is more likely to act as an acid or base, as the pH of the final solution will either be acidic or basic depending on the magnitudes of the two pK_a values. One other wonderful aspect of the pH of this solution is that it is independent of the concentration of bicarbonate. Of course, we need to remember that we made an approximation to come up with this simple form to get the pH. At very dilute concentrations of intermediate, that approximation breaks down and then the calculation becomes more complicated. We would need to use equation (\(\ref{7}\)) in that case.

In this case, we can now substitute in the two pK_a values for the carbonic acid system and determine the pH:

\[\mathrm{pH=\dfrac{pK_{a1}+ pK_{a2}}{2}=\dfrac{6.35+10.33}{2}=8.34}\nonumber\]

The value of 8.34 is slightly basic. Perhaps it is not surprising then that we could use a solution of sodium bicarbonate as an antacid if a night of pizzas, tacos, jalapeno poppers, and tequila sunrises (including the worm at the bottom of the bottle) left our stomach with excess acid.

It turns out that the generalized expression we derived in this case, in which the pH was equal to (pK_a1 + pK_a2)/2, can be applied to any intermediate in a polyprotic acid system. For example, consider the series of equations for phosphoric acid.

\[\begin{align}
&\ce{H3PO4 + H2O \leftrightarrow H2PO4- + H3O+} &&\mathrm{K_{a1}}\nonumber\\
&\ce{H2PO4- + H2O \leftrightarrow HPO4^2- + H3O+} &&\mathrm{K_{a2}}\nonumber\\
&\ce{HPO4^2- + H2O \leftrightarrow PO4^3- + H3O+} &&\mathrm{K_{a3}}\nonumber
\end{align}\nonumber\]

Suppose we had a solution that to a first approximation was “all” sodium dihydrogen phosphate (NaH₂PO₄). This would dissolve to produce \(\ce{H2PO4-}\), the first intermediate in this series of reactions. We can ignore the third reaction because it will be insignificant. We can perform a derivation analogous to what we did for bicarbonate and would come up with the following expression for the pH:

\[\mathrm{pH=\dfrac{pK_{a1}+ pK_{a2}}{2}}\nonumber\]

Suppose instead we had a solution that to a first approximation was “all” disodium hydrogen phosphate (Na₂HPO₄). This would dissolve to produce \(\ce{HPO4^2-}\), the second intermediate in this series of reactions. In this case, we can ignore the first reaction and, doing a derivation analogous to what we did for bicarbonate, we would come up with the following expression for the pH:

\[\mathrm{pH=\dfrac{pK_{a2}+ pK_{a3}}{2}}\nonumber\]

Something to note in this case is that if we examined the comparable equation to equation (\(\ref{7}\)), it would look as follows:

\[\mathrm{[H_3O^+]^2= \dfrac{K_{a2}K_{a3}[HPO_4^{2-}]}{K_{a2}+ [HPO_4^{2-}]}}\nonumber\]

Note that in this case, we are comparing the magnitude of [\(\ce{HPO4^2-}\)] to K_a2. Since K_a2 is always smaller than K_a1 (and usually much smaller), the likelihood that we can ignore the magnitude of K_a2 relative to the concentration of [\(\ce{HPO4^2-}\)] is improved, allowing us to use this very straight forward way of calculating the pH.

Contributors and Attributions

• Thomas Wenzel, Bates College ([e-mail])
• Sourced from the Analytical Sciences Digital Library