Weak Acids and Bases

What are weak acids and bases?

Acids and bases which do not dissociate completely. Some examples include: Acetic acids (CH$_{3}$COOH), water (H$_{2}$O), and hydrofluoric acid (HF).

How can you calculate the amount a weak acid (or base) dissociates?

Take for example the weak acid acetic acid:

CH$_{3}$COOH$_{(aq)}$ $\rightleftharpoons$ H$^{+}_{(aq)}$ + CH$_{3}$COO$^{-}_{(aq)}$

If this were a strong acid, the reaction would practically proceed to the right and to completion. In the case of this weak acid, we need to go back to our definition $K_{Acid}$. With this reaction, this would be:

$$K_{Acid} = \frac{[H_{aq}^{+}][CH_{3}COO^{-}]}{[CH_{3}COOH_{(aq)}]}$$ (33.9)

Now, one has to be given more information to proceed. OK, lets say the solution was 3.5 M CH$_{3}$COOH$_{(aq)}$. What would be $[H^{+}_{(aq)}]$ given a $K_{Acetic \ acid}$ = 0.000018?

$K_{Acid} = \frac{[x][x]}{[3.5 - x]}$ $0.000018 = \frac{[x^{2}]}{[3.5 - x]}$

At this point, things seem a little bleak with the possibility of solving a quadratic equation. Luckily, we can drop the x in the denominator under the assumption that it is relatively small. Rewind: Weak acids dissociate poorly, i.e. the amount of conjugate base formed is very small. Well, at least small enough not to impact the ``size'' of the denominator, and the $x$ can be dropped:

0.000018 = $\frac{[x^{2}]}{[3.5 - x]}$ becomes 0.000018 = $\frac{[x^{2}]}{[3.5]}$
$(3.5)(0.000018) = x^{2}$
x = 0.00793

Note: See intro chapter on MCAT math to solve for square roots without a calculator

How would you calculate the pH of this reaction?

pH = $-log[H^{+}]$

Note: See intro chapter on MCAT math to calculate a log without a calculator

What are two assumptions made in the above weak acid calculation?

1. The weak acid dissociation is small enough that the change from the reactant can be ignored in the equilibrium calculation.
2. The weak acid dissociation is still large enough to ignore the passive dissociation of water.