# How does pKa affect acidity?

Mar 26, 2016

$p {K}_{a}$ is a MEASURE of acidity. The more negative the $p {K}_{a}$, the STRONGER the acid.

#### Explanation:

The generalized acid dissociation reaction in water is;

$H A + {H}_{2} O r i g h t \le f t h a r p \infty n s {H}_{3} {O}^{+} + {A}^{-}$.

And ${K}_{a}$ $=$ $\frac{\left[{H}_{3} {O}^{+}\right] \left[{A}^{-}\right]}{\left[H A\right]}$.

This is an arithmetic equation, that we can multiply, divide, subtract, AS LONG AS we do it to both sides of the equation. We can also take logarithms to the base 10 of both sides:

${\log}_{10} {K}_{a}$ $=$ ${\log}_{10} \left[{H}_{3} {O}^{+}\right] + {\log}_{10} \left(\frac{\left[{A}^{-}\right]}{\left[H A\right]}\right)$

On rearrangement:

$- {\log}_{10} \left[{H}_{3} {O}^{+}\right]$ $=$ $- {\log}_{10} {K}_{a} + {\log}_{10} \left(\frac{\left[{A}^{-}\right]}{\left[H A\right]}\right)$

But by definition, $- {\log}_{10} \left[{H}_{3} {O}^{+}\right]$ $=$ $p H$; and $- {\log}_{10} {K}_{a}$ $=$ $p {K}_{a}$.

(The $p$ function just means take the negative logarithm to the base 10.)

Thus $p H = p {K}_{a} + {\log}_{10} \left(\frac{\left[{A}^{-}\right]}{\left[H A\right]}\right)$

When $\left[{A}^{-}\right] = \left[H A\right]$, ${\log}_{10} \left(\frac{\left[{A}^{-}\right]}{\left[H A\right]}\right)$ $=$ $0$, because ${\log}_{10} 1 = 0$ i.e.${10}^{0} = 1$). So during an acid base titration, at the point of half equivalence, $p H$ $=$ $p {K}_{a}$.

For stronger acids, i.e. $H C l , H C l {O}_{4} , {H}_{2} S {O}_{4}$, the $p {K}_{a}$ values will tend to be negative, because the equilibrium lies strongly to the right in the equation as given. In water, their dissociation would be almost complete. If a more acidic solvent were used (e.g. acetic acid or hydrogen fluoride), the $p {K}_{a}$ values could be quantitatively measured.

Look up the $p {K}_{a}$ values of acetic acid, phenol, $H F$, and phosphoric acid, ${H}_{3} P {O}_{4}$, and arrange them in order of acid strength.