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= ([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= ([H^+][A^-])/(0.5M-[A^-])`


`= (0.0001*0.0001)/(0.5-0.0001) = 2.0*10^(-8)`