Titration with Diprotic Acid!!!

[delco714]

The question is:

 

Oxalic acid, H2C2O4•2H2O (molar mass = 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?)

 

My calculations (which the lab site said were wrong) are as follows:

 

0.147g/126.07g/mole - 0.001 mole (oxalic acid)

 

M (moles/Liter) = (0.001 mole / 0.02364 L) X (2 mole NaOH / 1 mole acid) = 0.085 mole/Liter NaOH.

 

is this right and I should disregard the site, or where did I mistake?

[Fuzzwood]

How far off are you from the answer from the site? (Which you failed to mention btw).

[hermanntrude]

the usual procedure for a volumetric analysis is to use any method to find the number of moles of one reactant (often this is just a simple n=m/M calculation, as in this example, sometimes it involves c=n/v). The next step is to find the number of moles of the other reactant. You can do this using the stoichiometry of the balanced equation. the final step involves using c=n/v to find the concentration of the other reactant (NaOH in this case)

 

let me know which bit is giving you trouble

[John Cuthber]

"0.147g/126.07g/mole - 0.001 "

Pardon?