Question

A `10*mL` volume of oxalic acid was titrated with a `34.40*mL` volume of `NaOH(aq)` whose concentration was `0.250*mol*L^-1`. What was the mass of oxalic acid present in the initial `10*mL` volume?

Answer 1

`[C_2O_4H_2]~=0.5*mol*L^-1`.

Explanation:

We need (i) a stoichiometric equation.....

`"HO(O=)CC(=O)OH + 2NaOH"rarrC_2O_4^(2-)Na_2^+ + 2H_2O`

And (ii) equivalent quantities of sodium hydroxide......

`"Moles of NaOH"=34.40*mLxx10^-3*L*mL^-1xx0.250*mol*L^-1=8.60xx10^-3*mol`.

And HALF this molar quantity was equivalent to the oxalic acid solution contained in `10.00*mL` solution.

`"Concentration of oxalic acid"=(1/2xx8.60xx10^-3*mol)/(10.00xx10^-3*L^-1)`

`=0.430*mol*L^-1`

And thus in a `10*mL` volume, there were `10xx10^-3*Lxx0.430*mol*L^-1xx90.03*g*mol^-1=`

`~=0.4*g` `""`