# How can I calculate the pH of a weak acid with an example?

Dec 6, 2014

Let's use acetic acid, $C {H}_{3} C O O H$, as an example.

Weak acids, unlike strong ones, do not dissociate completely in an aqueous solution, which means that the final concentrations of ${H}_{3}^{+} O$ and ,in this case, $C {H}_{3} C O {O}^{-}$ , the weak acid's conjugate base, must be calculated.

The chemical equation can be set up like this:

$C {H}_{3} C O O H \left(a q\right) \iff {H}_{3}^{+} O \left(a q\right) + C {H}_{3} C O {O}^{-} \left(a q\right)$

When dealing with acids in general, the value of the acid dissociation constant, ${K}_{a}$, is usually given. In this case, ${K}_{a}$ is equal to $1.76 \cdot {10}^{- 5} \frac{m o l e}{L}$.

Even before doing any calculations, judging by the value of ${K}_{a}$, which in this case is <<1, one can see that the reaction will favor the reactants, with very little product being formed.

Now, the ICE method (I've added a wikipedia link) is the most appropriate method for this sort of calculations.

We will start with a $0.1 M$ concentration for the acetic acid.

$C {H}_{3} C O O H \left(a q\right) \iff {H}_{3}^{+} O \left(a q\right) + O {H}^{-} \left(a q\right)$

I..........$0.1 M$..........................$0$...............$0$
C............$- x$........................$+ x$.........$+ x$
E......$\left(0.1 - x\right)$........................$x$............$x$

This sums up to:

$I$ - initial - the concentrations of the substances present in the solution before the reaction; notice that ${H}_{3}^{+} O$ and $C {H}_{3} C O {O}^{-}$ are both zero, since the reaction has not yet taken place.
$C$ - consumed- this shows how the concentrations will change before an equilibrium is reached; notice that the acid's concentration should decrease by the same amount the products' concentrations increase;
$E$ - equilibrium - this shows the concentrations after an equilibrium is reached.

Now, we know that

${K}_{a} = \frac{\left[{H}_{3}^{+} O\right] \cdot \left[C {H}_{3} C O {O}^{-}\right]}{\left[C {H}_{3} C O O H\right]} = \frac{x \cdot x}{0.1 - x} = {x}^{2} / \left(0.1 - x\right)$

SInce ${K}_{a}$ is <<1, we can approximate $0.1 - x$ as 0.1, which leads to ${x}^{2} = 1.76 \cdot {10}^{- 6}$ -> $x = 1.3 \cdot {10}^{- 3} M$.

This is equal to the concentration of ${H}_{3}^{+} O$, therefore

$p H = - \log \left(\left[{H}_{3}^{+} O\right]\right) = - \log \left(1.3 \cdot {10}^{- 3}\right) = 2.89$