# How do you determine pH from pKa?

Feb 1, 2015

I'll discuss how to determine pH given $\text{pKa}$ for a monoprotic acid, which is an acid that only donates one proton per molecule when placed in aqueous solution.

The general equation for a monoprotic acid in aqueous solution is

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

If you're dealing with a buffer, then you are dealing with a weak acid. In this case, the Henderson-Hasselbalch equation can take you directly from $\text{pKa}$ to the solution's pH (assuming you know the concentrations of the weak acid and its conjugate base)

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

If you're not dealing with a buffer, then you must use the acid dissociation constant, ${\text{K}}_{a}$, to help you determine the pH of the solution. In this case, you need to determine $\left[{H}^{+}\right]$ in order to determine pH, since

$p H = - \log \left(\left[{H}^{+}\right]\right)$

The value of the acid dissociation constant can be derived from $\text{pKa}$

${K}_{a} = {10}^{\text{-pKa}}$

For a strong acid, $\text{pKa} < 1$ and ${\text{K}}_{a} > 1$ ; strong acids dissociate completely in aqueous solution, so $\left[{H}^{+}\right]$ = $\left[H A\right]$, which means

$p H = - \log \left(\left[H A\right]\right)$

If you're dealing with a weak acid, you have to use the ICE table method (more here: http://en.wikipedia.org/wiki/RICE_chart). The initial concentration of the acid is $C$, so

......$H A r i g h t \le f t h a r p \infty n s {H}^{+} + {A}^{-}$
I:......C.........0.........0
C:...(-x).........(+x)......(+x)
E:..(C-x).......(x).......(x)

Remember that ${\text{K}}_{a}$ is defined as

${K}_{a} = \frac{\left[{H}^{+}\right] \cdot \left[{A}^{-}\right]}{\left[H A\right]}$, which means that you'll get

${K}_{a} = \frac{x \cdot x}{C - x} = {x}^{2} / \left(C - x\right)$

At this point, the only unknown you'll usually have is $x$; solve for $x$ and pick the positive solution (remember that $x$ represents the concentration of ${H}^{+}$ and ${A}^{-}$, so it must be a positive number).

As a result, $\left[{H}^{+}\right] = x \implies p H = - \log \left(\left[{H}^{+}\right]\right)$