# Problem 2

2. Calculate the pH of a 0.089 M solution of sodium carbonate.

The carbonate ion is part of the carbonic acid system. If we look up carbonic acid in the table, we find that it is a diprotic acid. The two equilibria are as follows. Also, looking at the two pKa values (6.35 and 10.33) we notice that there is an appreciable difference between the two, suggesting that it may well be necessary to consider one of the two reactions in this problem.

$\ce{H2CO3 + H2O \leftrightarrow HCO3- + H3O+} \hspace{60px} \mathrm{K_{a1}}\nonumber$

$\ce{HCO3- + H2O \leftrightarrow CO3^2- + H3O+} \hspace{60px} \mathrm{K_{a2}}\nonumber$

Before continuing, we need to know whether sodium carbonate refers to NaHCO3 or Na2CO3. We will adopt a particular system in this course for naming these types of species, but in this case note that sodium carbonate refers to Na2CO3. The $$\ce{HCO3-}$$ ion is known as the bicarbonate ion and the species NaHCO3 is known as sodium bicarbonate (also known as baking soda). The name bicarbonate for $$\ce{HCO3-}$$ is not a systematic name, but a common name for this ion.

The naming system we will adopt can be demonstrated for the phosphoric acid (H3PO4) series of reactions. Phosphoric acid has three dissociable hydrogen ions, leading to the following possible species.

\begin{align} &\ce{H3PO4} &&\textrm{– phosphoric acid}\nonumber\\ &\ce{NaH2PO4} &&\textrm{– sodium dihydrogen phosphate}\nonumber\\ &\ce{Na2HPO4} &&\textrm{– disodium hydrogen phosphate}\nonumber\\ &\ce{Na3PO4} &&\textrm{– sodium phosphate}\nonumber \end{align}\nonumber

Note that the species sodium phosphate refers to the one in which all the dissociable hydrogen ions have been replaced with sodium cations. The others contain a prefix that tells you how many hydrogen atoms or sodium ions are involved in the salt. You must be careful when using or purchasing species like the intermediate ones since the names given above may not be used. Sometimes these are referred to as sodium phosphate monobasic and sodium phosphate dibasic. Presumably the label actually gives the formula so that you can be certain which species you actually have.

If we go back to our solution of sodium carbonate, this means that we have the $$\ce{CO3^2-}$$ species in solution. Remember, if you add sodium carbonate to water, the ions will dissociate to produce sodium ions and carbonate ions. Since the sodium ion is the cation of a strong base (sodium hydroxide – NaOH), this species won’t form and the sodium ion is what we call a spectator ion (it effectively watches things but does not get involved in any important reactions). Since the carbonate ion is the anion of a weak acid, it's actually a base and we can write the following reactions to describe what will occur in this solution.

$\ce{CO3^2- + H2O \leftrightarrow HCO3- + OH-}\hspace{60px}\mathrm{K_b\: of\: K_{a2}}\nonumber$

$\ce{HCO3- + H2O \leftrightarrow H2CO3 + OH-}\hspace{60px}\mathrm{K_b\: of\: K_{a1}}\nonumber$

Because of the significant distinction between the two Kb values, we only need to consider the first reaction in the series above to calculate the pH. The amount of hydroxide produced by the second reaction will be insignificant.

We can then treat this as a monobasic base, a process we have seen before.

\begin{align} & &&\ce{CO3^2-}\hspace{25px} +\hspace{25px} \ce{H2O} \hspace{25px}\leftrightarrow &&\ce{HCO3-} \hspace{25px} + &&\ce{OH-}\nonumber\\ &\ce{Initial} &&0.089 &&0 &&0\nonumber\\ &\ce{Equilibrium} &&0.089 - \ce{x} &&\ce{x} &&\ce{x}\nonumber\\ &\ce{Approximation} &&0.089 &&\ce{x} &&\ce{x}\nonumber \end{align}\nonumber

Plugging this into the appropriate Kb expression gives:

$\mathrm{K_b = \dfrac{[HCO_3^-][OH^-]}{[CO_3^{2-}]} = \dfrac{(x)(x)}{0.089} = 2.14\times10^{-4}}\nonumber$

$\mathrm{x = [OH^-] = 4.36\times10^{-3}}\nonumber$

$\mathrm{pOH = 2.37}\nonumber$

$\mathrm{pH = 11.63}\nonumber$

The pH value of 11.63 seems reasonable since this is a solution of a base. Checking the approximation shows that this was just under our 5% rule.

$\dfrac{4.36\times10^{-3}}{0.089} \times100=4.9\%\nonumber$