# How does a pKa value relate to pH?

Mar 25, 2016

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

#### Explanation:

Consider the general acid dissociation reaction in water:

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

Now, this is an equilibrium reaction, and for a given temperature, we can write the equlibrium expression:

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

This is an equation, and like any equation we can add/subtract/multiply it etc. and maintain the equality, SO LONG AS WE DO THE SAME OPERATION TO BOTH SIDES OF THE EQUATION. We can certainly take the ${\log}_{10}$ of both sides give:

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

ON REARRANGEMENT, subtract ${\log}_{10} \left[{H}_{3} {O}^{+}\right]$ AND ${\log}_{10} {K}_{a}$ FROM BOTH SIDES:

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

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

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

So, upon titration of a weak acid, at the point of half-equivalence when $\left[H A\right] = \left[{A}^{-}\right]$, $p H$ $=$ $p {K}_{a}$ because ${\log}_{10} \left\{\frac{\left[{A}^{-}\right]}{\left[H A\right]}\right\}$ $=$ ${\log}_{10} \left\{1\right\}$ $=$ $0$