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?
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:
