# Can you calculate the acid dissociation constant (Ka) of a weak acid from experimental data?

Sep 6, 2017

Yes, we can, but first we need some experimental data.......and someone must first do the measurement....

#### Explanation:

We interrogate the equilibrium....

$H A \left(a q\right) + {H}_{2} O \left(a q\right) r i g h t \le f t h a r p \infty n s {A}^{-} + {H}_{3} {O}^{+}$

Now ${K}_{a} = \frac{\left[{H}_{3} {O}^{+}\right] \left[{A}^{-}\right]}{\left[H A\right]}$

Now it is a fact, that $\left[{H}_{3} {O}^{+}\right]$ for a $1.0 \cdot m o l \cdot {L}^{-} 1$ solution of $\text{acetic acid}$ is $4.24 \times {10}^{-} 3 \cdot m o l \cdot {L}^{-} 1$. And I got these data from the interwebz; of course someone had to measure these data.....

${H}_{3} C - C {O}_{2} H \left(a q\right) + {H}_{2} O \left(a q\right) r i g h t \le f t h a r p \infty n s {H}_{3} C - C {O}_{2}^{-} + {H}_{3} {O}^{+}$

And ${K}_{a} = \frac{\left[{H}_{3} {O}^{+}\right] \left[{H}_{3} C - C {O}_{2}^{-}\right]}{\left[{H}_{3} C - C {O}_{2} H\right]}$

$= \frac{4.24 \times {10}^{-} 3 \cdot m o l \cdot {L}^{-} 1 \times 4.24 \times {10}^{-} 3 \cdot m o l \cdot {L}^{-} 1}{1.0 \cdot m o l \cdot {L}^{-} 1 - 4.24 \times {10}^{-} 3 \cdot m o l \cdot {L}^{-} 1}$

$= 1.81 \times {10}^{-} 5$

pK_a=-log_10(1.81xx10^-5)=??

Capisce?