# Calculate the "pH" for "0.270 M" benzoic acid? K_a = 6.5 xx 10^(-5)

Apr 24, 2017

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

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

$\text{I"" ""0.270 M"" "" "" "-" "" "" ""0 M"" "" "" ""0 M}$
$\text{C"" "-x" "" "" "" "-" "" "" "+x" "" "" } + x$
$\text{E"" ""(0.270 - x) M"" "-" "" "" "x" ""M"" "" "x" ""M}$

where $\text{HA}$ is benzoic acid and ${\text{A}}^{-}$ is therefore benzoate. So, its equilibrium expression (its mass action expression) would be:

${K}_{a} = \left(\left[\text{H"_3"O"^(+)]["A"^(-)])/(["HA}\right]\right) = \frac{{x}^{2}}{0.270 - x}$

In the small $x$ approximation, we say that $x$ $\text{<<}$ $0.270$, i.e. that $0.270 - x \approx 0.270$. Therefore:

$0.270 {K}_{a} \approx {x}^{2}$

$\implies x \approx \sqrt{0.270 \stackrel{6.5 \times {10}^{- 5}}{\overbrace{{K}_{a}}}} = 4.19 \times {10}^{- 3}$ "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

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

= (4.19 xx 10^(-3) "M")/("0.270 M") xx 100%

~~ color(blue)(1.55%)

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 $4.16 \times {10}^{- 3} \text{M}$... close enough. The true percent dissociation would then be 1.53_96% ~~ 1.54%.

Therefore, the $\text{pH}$ to a good approximation is

$\textcolor{b l u e}{\text{pH") = -log(4.19 xx 10^(-3) "M}} \approx {2.377}_{9} \approx \textcolor{b l u e}{2.38}$