[sharbeldam]

By the way, this is not exam or homework, I just study chemistry for my own.

I found this question and I'm really confused how to start, usually to find the hydration number we find the moles of H2O and the moles of the salt alone and we usually do these calculations in the lab (as far as i remember)

but here Im given the total moles of oxalic acid unknown hydrate in 1 Liter water.

Oxalic acid in water should dissociate twice like that:

H2C2O4 <----> HC2O4+ + H+
HC2O4 <-----> C2O4 + H+

The final concentration of H+ should be 10^-2.055.

how would one finish such question?

[AWK]

You can calculate the molar concentration of oxalic acid in the solution from the concentration of H₃O⁺ with sufficient accuracy using the not simplified quadratic equation and K_a1. By the way, solid oxalic acid crystallized from water is a well-defined dihydrate.

[sharbeldam]

Are you suggesting I do
Oxalic acid + H2O <---> 2H3O+ + A-2
initial                               
-x
x                                      2x
and use Pka1 to solve it?

or just use H2C2O4 <----> HC2O4+ + H+
and solve for H2C2O4?

Thanks.

[AWK]

K_a1 concerns the 1st-degree dissociation. Just in the case show your calculations.

[sharbeldam]

 H2C2O4 <----> HC2O4+ + H+
y                                           0
-x                                         +x
y-x                                         x

x=10^-2.055

Ka1 = (10^-2.055)^2/y-(10^-2.055)

I find Y which is the concentration of oxalic acid, I multiply by Volume to find the number of moles and then I substract the number of moles of oxalic acid from the total moles of the hydrate to get the moles of water?

[mjc123]

and then I substract the number of moles of oxalic acid from the total moles of the hydrate to get the moles of water?

No. The number of moles of hydrate is equal to the number of moles of oxalic acid. You have the mass of hydrate, so work out the formula weight and deduce what n must be.

[sharbeldam]

Yesss the mass not moles i meant.
thank you so much both!

[AWK]

Ka1 = (10^-2.055)^2/(y-(10^-2.055))

which is equivalent to
c=[H₃O⁺]²/K_a1 + [H₃O⁺]