Disclaimer: Long answer!
Oxalic acid is a weak acid that dissociates in two steps into Oxonium ions [H_3O^+]. To find how much KOH is needed to neutralize the acid we must first determine the number of moles of Oxonium ions in the solution, as these will react in a 1:1 ratio with the Hydroxide ions to form water.
As it is a weak diprotic acid it has a K_a value for both its acid form and anion form (hydrogen oxalate ion).
K_a (oxalic acid) =5.4 times 10^-2
K_a (hydrogen oxalate ion)= 5.4 times 10^-5
Remember: K_a = ([H_3O^+] times [ Anion])/ ([Acid])
Hence: 5.4 times 10^-2= ([H_3O^+] times [ Anion])/ ([0.333])
0.0179982 = ([H_3O^+] times [ Anion])
sqrt0.01799282=[H_3O^+] = [ Anion]
The concentration of the anion and the oxonium ion in the first dissociation are equal (0.134157 mol dm^-3)
Then in the dissociation of the hydrogen oxalate ion we have:
5.4 times 10^-5= ([H_3O^+] times [ Anion])/ ([0.1341573703])
As the anion formed in the first step is acting as an acid in the second dissociation, we use its concentration of the anion found from the first dissociation.
7.244497996 times 10^-6= ([H_3O^+] times [ Anion])
sqrt(7.244497996 times 10^-6)=[H_3O^+]=[ Anion]
Which is (0.0026915605 mol dm^-3)
All of the dissociations took place in a volume of 25 ml (0.025 dm^3) Hence we can find the number of moles of Oxonium ions formed from both dissociations.
concentration (mol dm^-3)= (Moles)/ (volume) (dm^3)
0.134157 times 0.025=0.0033539343
and
0.0026915695 times 0.025=6.72890125 times 10^-5
Add them up to get: 0.0034212233 mol of [H_3O^+]
This is the same amount of moles we need of OH^- to neutralize the acid.
Hence the volume of KOH we need is:
concentration (mol dm^-3)= (Moles)/ (volume) (dm^3)
0.1=(0.0034212233)/v
v=0.034212233 dm^3 or multiply it by 1000 get it in ml.
v approx 34.2 ml
