Question

You are given 10.00 mL of a solution of an unknown acid. The pH of this solution is exactly 2.18. You determine that the concentration of the unknown acid was 0.2230 M. You also determined that the acid was monoprotic (HA). What is the pKa of your unknown acid

Answer 1

Answer:

$pKa=3.70$

Explanation:

Hello,

In this case, given the information, we can compute the concentration of hydronium given the pH:

$pH=-log([H^+])$

$[H^+]=10^{-pH}=10^{-2.18}=6.61x10^{-3}M$

Next, given the concentration of the acid and due to the fact it is monoprotic, its dissociation should be:

$HA\rightleftharpoons H^++A^-$

We can write the law of mass action for equilibrium:

$Ka=\frac{[H^+][A^-]}{[HA]}$

Thus, due to the stoichiometry, the concentration of hydronium and A⁻ are the same at equilibrium and the concentration of acid is:

$[HA]=0.2230M-6.61x10^{-3}M=0.2164M$

As the concentration of hydronium also equals the reaction extent ($x$). Thereby, the acid dissociation constant turns out:

$Ka=\frac{(6.61x10^{-3})^2}{0.2164}\\ \\Ka=2.02x10^{-4}$

And the pKa:

$pKa=-log(Ka)=-log(2.02x10^{-4})\\\\pKa=3.70$

Regards.