How do you calculate the acidity constant?

Question

a student investigating an equilibrium mixture produced by reacting methanoic acid (HCOOH) and H2O finds that the pH of the solution is 2.4 and the [HCOOH]=0.10M. Calculate the acidity constant of HCOOH.

Answer 1

Ka = [H+]*[CHCOO-]/[CH3COOH]

You must have [H+]=[CH3COO-]

Ka= [H+]^2/[CH3COO-]

[H+]^2= 10^(-2.4*2)=10^-4.8/0.1=10^-3.8

pKa=3.8

Answer 2

methanoic acid is monoprotic so it dissociates as follows :

HCOOH + H2O ------  H3O+ + HCOO-

if one starts with 0.1 M metahnoic acid a small amount x will ionize to give X amounts of H+  and the same X amount of HCOO-

since the pH of the solution is 2.4 that means that -log[H+] = 2.4

so [H+ ] = 3.998 X 10^-3   since[ HCOO-1]=[H+]  then HCOO-1 = 3.998 X 10^-3 as well.

The Ka for the weak acid is   Ka = {[X]X]}/0.1 -X    X is small compared to 0.1 and neglected so

Ka = (3.998 x 10^-3)^2 / 0.1     = 1.598 X 10^-4