# When "0.2 M" of propionic acid dissolves in water, the "pH" was found to be 4.88. Find the K_a of propionic acid?

Jun 17, 2017

This is an unrealistic answer, but I got $8.69 \times {10}^{- 10}$. The actual ${K}_{a}$ is very different, at $1.34 \times {10}^{- 5}$!! The $\text{pH}$ given was unrealistic, so the ${K}_{a}$ calculated was unrealistic.

By knowing the $\text{pH}$, you know the equilibrium concentration of ${\text{H}}^{+}$, which means you know what $x$ is.

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

$\text{I"" ""0.2 M"" "" "" "" "-" "" "" ""0 M"" "" "" "" "" ""0 M}$
$\text{C"" " - x" "" "" "" "-" "" "" "+x" "" "" "" "" } + x$
$\text{E"" "(0.2 - x) "M"" "-" "" "" "" "x" M"" "" "" "" "x" M}$

Since "pH" = -log["H"_3"O"^(+)]:

${10}^{- {\text{pH") = 10^(-4.88) = ["H"_3"O}}^{+}} = x = 1.318 \times {10}^{- 5}$ $\text{M}$

This gives you a ${K}_{a}$ expression of:

$\textcolor{b l u e}{{K}_{a}} = \left(\left[\text{PropCOO"^(-)]["H"_3"O"^(+)])/(["PropCOOH}\right]\right)$

$= {x}^{2} / \left(0.2 - x\right)$

$= {\left(1.318 \times {10}^{- 5}\right)}^{2} / \left(0.2 - 1.318 \times {10}^{- 5}\right)$

$= \textcolor{b l u e}{8.69 \times {10}^{- 10}}$

By the way, this ${K}_{a}$, although calculated correctly, is completely unrealistic. The actual ${K}_{a}$ of propionic acid is around $1.34 \times {10}^{- 5}$... so the $\text{pH}$ is unrealistic.

Let's calculate the $\text{pH}$ using the correct ${K}_{a}$ and compare. Since ${K}_{a}$ is small (on the order of ${10}^{- 5}$ or less),

${K}_{a} \approx {x}^{2} / \left(0.2\right)$

$\implies x \approx \sqrt{0.2 {K}_{a}} = \sqrt{0.2 \cdot 1.34 \times {10}^{- 5}} = \text{0.001637 M}$

As a result, a more realistic $\text{pH}$ would have been:

$\textcolor{red}{\text{pH"_"realistic}} = - \log \left(0.001637\right) = \textcolor{red}{2.79}$,

which is quite a bit less than $4.88$...