Question

How many `"mL"` of `"H"_2"C"_2"O"_4` can be neutralized by `"23.65 mL"` of `"1.115 M"` `"NaOH"`?

Answer 1

The volume of oxalic acid required is `6.75 mL`.

First, start with the balanced chemical equation for this particular neutralization reaction

`H_2C_2O_4(aq) + 2NaOH(aq) -> Na_2C_2O_4(aq) + 2H_2O(l)`

From the chemical reaction we can see that `1` mole of `H_2C_2O_4` needs `2` moles of `NaOH` in order to produce `1` mole of `NaC_2O_4` and `2` moles of `H_2O`.

We can determine the number of moles of `NaOH` by using

`n_(NaOH) = C * V = 1.115 (mol es)/(L) *0.02365 L = 0.026 mo l es`

This means that we need

`n_(H_2C_2O_4) = (n_(NaOH))/2 = 0.026/2 = 0.013 m o l es` of oxalic acid.

Therefore, the volume of oxalic acid can be calculated from

`C_(H_2C_2O_4) = n_(H_2C_2O_4)/V_(H_2C_2O_4)`

`V_(H_2C_2O_4) = (0.013 mo l es)/(1.978 (mo l es)/L) = 6.57 mL`