# How can I calculate pH of weak acid?

Dec 6, 2014

Let's call the acid HA.

As it's a weak acid it will dissociate according to :

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

So we can write:

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

You need to know its initial concentration and ${K}_{a}$ value so here's an example:

What is the pH of a 0.01 mol/litre solution of ethanoic acid?

${K}_{a} = 1.7 \times {10}^{- 5} m o l . {\mathrm{dm}}^{- 3}$

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

Initially there are 0.01 mol $C {H}_{3} C O O H$

When equilibrium is established we can say that $x$ moles dissociate so we must be left with $\left(0.01 - x\right)$ moles of $C {H}_{3} C O O H$ and $x$ moles of $C {H}_{3} C O {O}^{-}$ and $x$ moles ${H}^{+}$

So

${K}_{a} = \frac{x . x}{\left(0.01 - x\right)}$

At this point we make an assumption that the amount $x$ is so small compared to $0.01$ that we can assume $\left(0.01 - x\right) \rightarrow 0.01$

So ${x}^{2} = {K}_{a} \times 0.01$

${x}^{2} = 1.7 \times {10}^{- 5} \times 0.01 = 1.7 \times {10}^{- 7}$

$x = 4.12 \times {10}^{- 4} = \left[{H}^{+}\right]$

pH =$- \log \left[{H}^{+}\right]$ = $- \log 4.12 \times {10}^{- 4} = 3.4$

This is at 298K.