# How do you calculate the pH of a weak acid?

Sep 17, 2015

A handy expression to use is $p H = \frac{1}{2} \left(p {K}_{a} - \log a\right)$

#### Explanation:

This is a quick and useful method if you are given the $p {K}_{a}$ and concentration $a$ of the weak acid.

I'll derive it using the usual ICE method:

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

$\text{Initial:}$$\text{ }$a$\text{ "" "0" " " " " } 0$

If $x$ moles dissociate the equilibrium moles $\Rightarrow$

$\left(a - x\right) \text{ " " " " " " " x " " " " " } x$

${K}_{a} = \frac{{x}^{2}}{a - x}$

We assume that $x$ is negligible compared to to $a$ so this approximates to:

${K}_{a} = \frac{{x}^{2}}{a}$

$\frac{{x}^{2}}{a} = {K}_{a}$

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

$x = {\left({K}_{a} \times a\right)}^{\frac{1}{2}}$

Taking -ve logs of both sides $\Rightarrow$

$- \log x = - \frac{1}{2} \log \left({K}_{a} + a\right)$

$p H = \frac{1}{2} \left(p {K}_{a} - \log a\right)$

Here's an example:

$p H$ of a $0.1 \text{M}$ solution of ethanoic acid whose $p {K}_{a} = 4.75$ ?

$p H = \frac{1}{2} \left(4.75 - \left(- 1\right)\right)$

$p H = 5.75$

If you can remember it then fine. If not you need to use the ICE method described in the answer by @Dr Hayek

Sep 18, 2015

$p H = - \log \left[{H}_{3} {O}^{+}\right]$

#### Explanation:

By measuring the concentration of ${H}_{3} {O}^{+}$ ions we get the pH using $p H = - \log \left[{H}_{3} {O}^{+}\right]$

Consider the dissociation of the following weak acid HA with initial concentration of $1.0 M$ and ${K}_{a} = 2.0 \times {10}^{-} 6$:

$\text{ " " " " " " " " " }$ $H A \left(a q\right) + {H}_{2} O \left(a q\right) \to {H}_{3} {O}^{+} \left(a q\right) + {A}^{-} \left(a q\right)$
$\text{ " " " " }$ Initial: $1.0 M \text{ " " " " " " " " " " " 0 M " " " " " } 0 M$
$\text{ " " " }$ Change: $- x M \text{ " " " " " " " " " " +x M" " " " } + x M$
$\text{ " }$ Equilibrium: $\left(1.0 - x\right) M \text{ " " " " " " " "x M" " " " " " " } x M$

The expression of ${K}_{a}$ can be written as follows:

${K}_{a} = \frac{\left[{H}_{3} {O}^{+} \left(a q\right)\right] \cdot \left[{A}^{-} \left(a q\right)\right]}{\left[H A \left(a q\right)\right]} = \frac{x \cdot x}{1.0 - x} = 2.0 \times {10}^{-} 6$

solve for $x$ using calculator, you get

$x \approx 1.4 \times {10}^{-} 3 \implies p H = - \log \left(1.4 \times {10}^{-} 3\right) = 2.895$

that can be rounded to $p H = 2.9$ (2 significant figures).

Here is a video which summarizes this technique which requires use of the quadratic formula. The values are different than Dr. Hayek's example, but the technique is the same.

Video from: Noel Pauller