# Problem 5

- Page ID
- 302700

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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)**5. Calculate the pH of a 0.240 M solution of sodium bicarbonate (NaHCO _{3}).**

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 H_{2}CO_{3} 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 H_{2}CO_{3} and \(\ce{CO3^2-}\) that are formed can neutralize each other.

What we need to do in this case is write an expression for [H_{3}O^{+}] in terms of species in the carbonate system. Remember, there is some H_{3}O^{+} 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 H_{3}O^{+} in solution. At the start, before the system equilibrates, there is no H_{2}CO_{3} 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 H_{3}O^{+}. 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 H_{2}CO_{3}, we actually need to remove an H_{3}O^{+} species. Therefore, every H_{2}CO_{3} that we find in the final solution subtracts an H_{3}O^{+}. This would allow us to write:

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

Both \(\ce{CO3^2-}\) and H_{2}CO_{3} 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 H_{3}O^{+} 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 [H_{2}CO_{3}] 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 [H_{2}CO_{3}] and [\(\ce{CO3^2-}\)] into equation (\(\ref{1}\)).

Multiply each side through by K_{a1}[H_{3}O^{+}] 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 [H_{3}O^{+}]^{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:

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:

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 (NaH_{2}PO_{4}). 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 (Na_{2}HPO_{4}). 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 (twenzel@bates.edu)
- Sourced from the Analytical Sciences Digital Library