[Lindsay]

Problem 1

This one I'm not too sure about. Actually, I'm not too sure on the balanced equation. I can figure out the rest but the equation got me stuck. Please help

Oxalic acid, H2C2O4•2H2O - 126.07 g/mol is often used as a primary standard for the standardization of a NaOH solution. If 0.147 g of oxalic acid dihydrate is neutralized by 23.64 mL of a NaOH solution, what is the molar concentration of the NaOH solution? Oxalic acid is a diprotic acid. (Hint: what is the balanced equation?)

I think this is the wrong balanced equation
H2C2O4•2H2O + 2NaOH ---- 2NaC2O4•2H2O + H2O

0.147g/126.07g/mole - 0.00117 mole (oxalic acid)

Appreciate any help

[Astrokel]

H2C2O4•2H2O + 2NaOH ---- 2NaC2O4•2H2O + H2O

Oxalic acid is diprotic!

[Lindsay]

Meaning it has two Hydrogen Ions that can completely dissociate, right?

Is the equation like this

H2C2O4•H2O + NaOH ---- NaHC2O4•H2O + H2O

Thanks

Lindsay

[macman104]

No, your base equation was correct, but it was not balanced (count your hydrogens).  Also, I don't know about this, but maybe someone else will, but will the oxalic acid still be a dihydrate after the reaction?

[Arkcon]

Generally, in solution, a hydrate's water molecules are assumed to be, well, not gone, not really dissociated or ionized, but well, not considered.  Of course, the original poster can't just ignore them, they are there, when the dry solid is massed out, but usually, when the reaction is written, we assume they don't participate.

[Astrokel]

No, your base equation was correct, but it was not balanced (count your hydrogens).

I thought it was wrong because it should be Na₂C₂O₄.2H₂O?
.

[macman104]

OH!  I didn't even look at that, lol, I saw the Na and was done, lol.  Fair enough about the hydrate, I wasn't sure if you dissociate them or leave them with the molecule.