Question

A `40.16*mL` volume of sodium hydroxide solution with concentration of `0.1693*g*mol^-1` reached an endpoint with a quantity of oxalic acid. If the oxalic acid was initially dissolved in a `30.00*mL` what was the concentration of oxalic acid?

Answer 1

Concentration of oxalic acid `=` `0.1133` `mol*L^-1`.

Explanation:

`HO(O=)C-C(=O)OH + 2NaOHrarr{C(=O)O}_2^(2-)Na_2^+ + 2H_2O`

Moles of `NaOH` for equivalence `=` `40.16xx10^-3Lxx0.1693` `mol*L^-1 =` `6.80xx10^-3` `mol`.

Thus there were `3.40xx10^-3` `mol` oxalic acid (because each equiv acid contributes 2 equiv `H^+`).

Since this quantity was dissolved in a `30.00` `mL` volume, concentration was `(3.40xx10^-3*mol)/(30.00xx10^(-3)L)` `=` `0.1133` `mol*L^-1` in oxalic acid.