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

Dec 27, 2016

We take the weak acid, hypochlorous acid, $H O C l$, for which $p {K}_{a} = 7.53$, and that $\left[H O C l\right]$ $=$ $0.100 \cdot m o l \cdot {L}^{-} 1$

#### Explanation:

We needed details of the acid dissociation behaviour of this acid; that's why I included the $p {K}_{a}$. Without this value we could not do the problem. But what does $p {K}_{a}$ mean? It is measure of the extent of the following, so-called protonolysis reaction:

HOCl(aq) + H_2O rightleftharpoons H_3O^+ + ""^(-)OCl

For which we could write the equilibrium equation:

K_a=([H_3O^+][""^(-)OCl])/([HOCl])

Now $p {K}_{a} = - {\log}_{10} {K}_{a}$, and thus ${K}_{a} = {10}^{- 7.53}$.

And now we put numbers into our equation. Clearly, [H_3O^+]=[""^(-)OCl]=x, and $\left[H O C l\right] = 0.100 - x$

${K}_{a} = {10}^{- 7.53} = \frac{{x}^{2}}{0.100 - x}$

Now this is a quadratic in $x$, which we could solve exactly. Because chemists are simple souls, and this is a weak acid, we could make the approximation that $0.100 - x \cong 0.100$. We have to justify this approx. later.

And thus ${x}_{1} = \sqrt{{10}^{-} 7.53 \times 0.100} = 5.43 \times {10}^{-} 5$.

Now we have a first approx. we can recycle this into the first equation:

${x}_{2} = 5.43 \times {10}^{-} 5$

And thus $\left[{H}_{3} {O}^{+}\right] = 5.43 \times {10}^{-} 5 \cdot m o l \cdot {L}^{-} 1$.

$p H = - {\log}_{10} \left[{H}_{3} {O}^{+}\right] = - {\log}_{10} \left(5.43 \times {10}^{-} 5\right) = 4.27$.

This is a weak acid, but $\left[{H}_{3} {O}^{+}\right]$ is about 1000 times more concentrated than pure water.