# A student is asked to prepare a buffer solution with pH = 8.6, using one of the following weak acids: HA (Ka = 2.7 × 10^-3), HB (Ka = 4.4 × 10^-6) or HC (Ka = 2.6 × 10^-9). Which acid should she choose?

Mar 25, 2016

$H C$ should be the acid of choice.

#### Explanation:

The key to this question is understanding what we mean by $p H$ and $p {K}_{a}$, which are logarithmic functions. Students tend to have problems with the logarithmic function. I will introduce the subject briefly.

When I write ${\log}_{a} b = c$, I am asking to what power I raise the base $a$, to get $b$. Here ${a}^{c} = b$. Now, usually we use the bases $10$ or $e$. So ${\log}_{10} 100 = 2$, ${\log}_{10} 1000 = 3$, ${\log}_{10} 1000000 = 6$. Likewise ${\log}_{10} 0.1 = {\log}_{10} {10}^{- 1}$ $=$ $- 1$. In the days before electronic calculators (approx. 30-40 years), students would be issued log tables so that complicated calculations could be performed.

From the above $p H$ $=$ $- {\log}_{10} \left[{H}_{3} {O}^{+}\right]$, and $p {K}_{a}$ $=$ $- {\log}_{10} {K}_{a}$. These are simple functions that have been widely used in chemistry.

Now it is fact, that when a weak acid is mixed with its conjugate base in appreciable concentrations, a buffer solution is formed that tends to resist gross changes in $p H$.

We can write, $p H = p {K}_{a} + {\log}_{10} \left\{\frac{\left[{A}^{-}\right]}{\left[H A\right]}\right\}$.

It is clear that when $\left[H A\right]$ $=$ $\left[{A}^{-}\right]$, then $p H = p {K}_{a}$ because ${\log}_{10} \left\{\frac{\left[{A}^{-}\right]}{\left[H A\right]}\right\}$ $=$ ${\log}_{10} 1 = 0$

So we want an acid whose $p {K}_{a}$ $\cong$ $p H$. To get $p {K}_{a}$ I simply perform the function $- {\log}_{10} {K}_{a}$, on each of the acid dissociation constants.

$p {K}_{a}$ $H A$ $=$ $2.57$;

$p {K}_{a}$ $H B$ $=$ $5.36$;

$p {K}_{a}$ $H C$ $=$ $8.58$;

We want to maintain $p H$ at $8.6$, so it is clear that acid $H C$ is the acid of choice.