# For "1.25 M" of acetic acid (K_a = 1.8 xx 10^(-5)), determine the percent dissociation?

Apr 24, 2017

Since ${K}_{a}$ ~ ${10}^{- 5}$, we expect to be able to use the small x approximation to simplify calculations. However, this is somewhat borderline, unless you already know what average concentration gives you a small percent dissociation.

Write out the dissociation reaction of a weak acid in water and construct its ICE table:

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

$\text{I"" ""1.25 M"" "" "" "-" "" "" ""0 M"" "" "" ""0 M}$
$\text{C"" "-x" "" "" "" "-" "" "" "+x" "" "" } + x$
$\text{E"" ""(1.25 - x) M"" "-" "" "" "x" ""M"" "" "x" ""M}$

where $\text{HA}$ is acetic acid and ${\text{A}}^{-}$ is therefore acetate.

So, its equilibrium expression (its mass action expression) would be:

${K}_{a} = \left(\left[\text{H"_3"O"^(+)]["A"^(-)])/(["HA}\right]\right) = \frac{{x}^{2}}{1.25 - x}$

In the small $x$ approximation, we say that $x$ $\text{<<}$ $1.25$, i.e. that $1.25 - x \approx 1.25$. Therefore:

$1.25 {K}_{a} \approx {x}^{2}$

$\implies x \approx \sqrt{1.25 \stackrel{1.8 \times {10}^{- 5}}{\overbrace{{K}_{a}}}} = 4.74 \times {10}^{- 3}$ "M" = ["H"^(+)] = ["H"_3"O"^(+)]

(In general, under the small $x$ approximation, $\textcolor{g r e e n}{x \approx \sqrt{\left[\text{HA}\right] {K}_{a}}}$.)

The percent dissociation is then

color(blue)(%" dissoc") = (["HA"]_(lost))/(["HA"]_i) = x/(["HA"]_i)

= (4.74 xx 10^(-3) "M")/("1.25 M") xx 100%

~~ color(blue)(0.38%)

Another measure to determine whether the small $x$ approximation is valid is if the percent dissociation is under 5%... and since it obviously is, we don't have to check the true answer (where we don't say $x$ $\text{<<}$ $\left[\text{HA}\right]$).

However, the true $x$ via the quadratic formula would have been $4.73 \times {10}^{- 3} \text{M}$... close enough. The true percent dissociation would then be 0.37_9% ~~ 0.38%.