# What is the reaction equation for a mixture of trichloroethanoic acid and dimethylpropanoic acid?

Write an equation for the reaction you might expect to take place if trichloroethanoic acid, $$\ce{Cl3CCO2H}$$, $$\mathrm{p}K_\mathrm{a} = 0.65$$, were added to dimethylpropanoic acid, $$\ce{(CH3)3CCO2H}$$, $$\mathrm{p}K_\mathrm{a} = 5.05$$. Explain your answer in terms of the Brønsted–Lowry theory.

I know that $$\ce{Cl3CCO2H}$$ is the stronger acid as it has a lower $$\mathrm{p}K_\mathrm{a}$$, however, how can dimethylpropanoic acid be a base?

• Organic Chemistry generally considers that anything can be an acid or a base. – Lighthart Mar 3 '15 at 23:22
• Even concentrated sulphuric acid can act as a base, e.g. $$\ce{2 H2SO4 <=> H3SO4+ + HSO4-}$$ – Poutnik Feb 6 at 8:19
• It can be a base; but it's such a poor one that you'd better get an ultra-strong acid to protonate it. Concentrated sulfuric acid might work. Trichloroacetic acid probably won't. – orthocresol Feb 6 at 13:48

TL;DR

Contrary to what the answers/comments have suggested, I would say that no reaction happens here. The acid–base reaction between the two given species is thermodynamically unfeasible, with an equilibrium constant $$K \sim 10^{-9}$$.

If you want an acid-base reaction to occur between A and B, it's no use comparing the $$\mathrm pK_\mathrm a$$ values of A and B. That tells you how acidic A and B are, but you're not interested in both of them acting as acids. You're interested in one of them being an acid, and the other being a base. So the relevant quantities are the $$\mathrm pK_\mathrm a$$ of A and the $$\mathrm pK_\mathrm b$$ of B!

In this context, let's keep trichloroacetic acid $$\ce{Cl3CCOOH}$$ as the acid; we already know that it has $$\mathrm pK_\mathrm a = 0.65$$. The question tells us that for the "base" $$\ce{Me3CCO2H}$$, its $$\mathrm pK_\mathrm a$$ is $$5.05$$. But we're not interested in that value; we're more interested in its $$\mathrm pK_\mathrm b$$.

To find that out, we need to look for data beyond the question text itself. The veritable Evans tables states that the $$\mathrm pK_\mathrm a$$ of $$\ce{PhC(OH)+OH}$$ is $$-7.8$$. That species is the conjugate acid of $$\ce{PhCOOH}$$, so the $$\mathrm pK_\mathrm b$$ of $$\ce{PhCOOH}$$ is $$14 - (-7.8) = 21.8$$. Recall that a larger $$\mathrm pK_\mathrm b$$ means that a species is a weaker base. This gigantic number should already tip you off to the fact that $$\ce{PhCOOH}$$ is a terrible base. The species in the question, $$\ce{Me3CCOOH}$$, will likely have a value that's broadly similar.

How do we use these values to find whether an acid-base reaction is plausible or not? Returning to the reaction

$$\ce{AH + B <=> A- + BH+,}$$

we have the equilibrium constant

\begin{align} K &= \frac{[\ce{A-}][\ce{BH+}]}{[\ce{AH}][\ce{B}]} \\ &= \frac{[\ce{A-}][\ce{H3O+}]}{[\ce{AH}]} \cdot \frac{[\ce{BH+}][\ce{OH-}]}{[\ce{B}]} \cdot \frac{1}{[\ce{H3O+}][\ce{OH-}]} \\ &= \frac{K_\mathrm a(\ce{HA}) \cdot K_\mathrm b(\ce{B})}{K_\mathrm w} \end{align}

For our molecules, we know that the relevant quantities are $$\mathrm pK_\mathrm a(\ce{HA}) = 0.65$$ and $$\mathrm pK_\mathrm b(\ce{B}) \approx 21.8$$. Putting in $$K_\mathrm a = 10^{-0.65}$$, $$K_\mathrm b = 10^{-21.8}$$, and $$K_\mathrm w = 10^{-14}$$, we get an equilibrium constant of

$$K = \frac{10^{-0.65} \cdot 10^{-21.8}}{10^{-14}} = 3.548 \times 10^{-9}$$

which is so small that it makes very little sense to claim that this reaction "happens".

[…] I know that $\ce{Cl3CCO2H}$ is the stronger acid

With other words: This is the Brønsted donor. Show some confidence in your knowledge and draw: • I wonder if subsequent dehydration and decarbonylation might occur. – Mithoron Mar 4 '15 at 11:35