[_cheers]

Calculate the number of molecules of water in oxalic acid hydrate, from the following: 5.00 g of the coumpound is made up to 250  ml and 25 ml of the solution requires 15.90 ml of .5 M NaOH to reach neutralization;

What I have so far is the moles of NaOH: .00795
Moles of oxalic acid in 250 ml= .3702 mol and in 25 ml = .009255 mole.

If I subtract the amount of moles of acid (25 ml) from the amound of NaOH, I end up with .001305 mole of acid that is used up as it looks like a one to one ratio? no?

H₂C₂O₄ x H₂0 + NaOH ->  H⁺ + 2 CO₂ + Na⁺ + 2H₂0

Am I on the right track? but then what?

Thanks in advance

[Borek]

Oxalic acid is diprotic, whet you get after neutralization is sodium oxalate - (COONa)₂.

[_cheers]

ok, does that make it 2:1 (NaOH:acid)? and then would I just use those coefficients to find the moles acid used? then find the mass? So, I end up with 0.4293 g of oxalic acid hydrate? it doesnt seem right....am I right off track?

[xiankai]

start from the beginning: how are you going to find the number of moles of water per mole of oxalic acid hydrate? (hint look at the formula, H₂C₂O₄.xH₂O)

What I have so far is the moles of NaOH: .00795
Moles of oxalic acid in 250 ml= .3702 mol and in 25 ml = .009255 mole.

ok, does that make it 2:1 (NaOH:acid)?

yes, so you better change your calculations.