# Approximately what is the pH of a 0.1 M acetic acid solution?

May 8, 2017

About $2.87$ in aqueous solution.

Well, I assume that you mean in water. I have to assume that, because it matters.

The $\text{pH}$ can be found by finding the $\left[{\text{H}}^{+}\right]$ dissociated from acetic acid into water. Therefore, we must write the dissociation reaction to construct the ICE table.

${\text{HA"(aq) " "+" " "H"_2"O"(l) rightleftharpoons "H"_3"O"^(+)(aq) + "A}}^{-} \left(a q\right)$

$\text{I"" ""0.1 M"" "" "" "" "-" "" "" ""0 M"" "" "" "" ""0 M}$
$\text{C"" "-x" "" "" "" "" "-" "" "+x" "" "" "" } + x$
$\text{E"" "(0.1 - x)"M"" "" "-" "" "" "x" "" "" "" "" } x$

The equilibrium expression is then:

${K}_{a} = {x}^{2} / \left(0.1 - x\right) = 1.8 \times {10}^{- 5}$

We assume that for a ${K}_{a}$ on the order of ${10}^{- 5}$ or less, the small $x$ approximation works. We can verify this by checking that the percent dissociation is less than 5% later.

$1.8 \times {10}^{- 5} \approx {x}^{2} / 0.1$

$\implies x \approx \sqrt{0.1 {K}_{a}}$

$= \left[{\text{H}}^{+}\right] =$ $\text{0.001342 M}$

Therefore, the $\text{pH}$ is:

$\textcolor{b l u e}{{\text{pH") ~~ -log["H}}^{+}} = \textcolor{b l u e}{2.87}$

And to verify that the percent dissociation is sufficiently small, we check that

bb(%"dissoc.") = x/(["HA"]) < 0.05:

$\text{0.001342 M"/"0.1 M} = \boldsymbol{0.01342} < 0.05$ color(blue)(sqrt"")