Question

Calculate the acid dissociation constant of a weak monoprotic acid if a 0.5M solution of this acid gives a hydrogen-ion concentration of 0.000 1M? Show your work.
Hint: Monoprotic means containing one proton.

Please help thanks a lot!

Answer 1

Let the acid be HA.
The chemical formula for this acid will be the following:

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

The formula for the acid dissociation constant will be the following:

$K_a= \dfrac{[H^+][A^-]}{[HA]}$

We know [H+]=0.0001 (it's given).
However, we must find [A-] and [HA] in order to solve for the constant.

We find that [A-]=[H+] by using a electroneutrality equation.
Also, we can create a concentration equation to find [HA].

$0.5M=[A^-]+[HA]$
$[HA]=0.5M-[A^-]$

Now, we can find the acid dissociation constant.

$K_a= \dfrac{[H^+][A^-]}{0.5M-[A^-]}$

$= \dfrac{0.0001*0.0001}{0.5-0.0001} = 2.0*10^{-8}$