# Need help with weak diprotic acids…?

Many household cleaning products contain oxalic acid, H2C2O4, a diprotic acid with the following dissociation constants: Ka1 = 5.9*10^-2, Ka2 = 6.4*10^-5.

A) H2C2O4 + H2O <-----> H3O+ + HC2O4-
5.9 x 10^-2 = x^2 / 0.20-x
0.0118 – 5.9 x 10^-2 x = x^2
x^2 + 5.9 x 10^-2 x – 0.0118=0
x = – 5.9 x 10^-2 + sq.rt ( 0.00348 + 0.0472)/2 = 0.083

HC2O4- + H2O <-----> H3O+ + C2O42-
6.4 x 10^-5 = ( 0.083+x)(x) / 0.083-x

solve for x by quadratic formula
[H3O+]= x + 0.083

part B)

you use part a to set up your next equation ( the template is given to you on the left if you are using mastering chemistry) therefore….

6.4x 10^-5 =(.083+x)x/ (.083-x)

you are left with

6.4x 10^-5 =x because the other terms are so insignificant that they cancel out. and that is your answer

it’s really easy if you use the ICE chart

initial, change, equilibrium.

Source(s): mastering chemistry and also part a was previously answered by another user

Part C:
HC2O4- is equal to H3O+ Initial concentration.
HC2O4- = .083M Initial – The change in C2O4 2- (6.4*10^-5)
.083M – 6.4*10^-4 = .0829 or .083

HC2O4- Equilibrium = .083M