Question #536b8

The molarity of the sodium hydroxide solution is `"0.1632 M"`.

The balanced chemical equation for this titration, in which you're essentially performing a neutralization reaction, looks like this

`H_2C_2O_(4(Aq)) + 2NaOH_((aq)) -> Na_2C_2O_(4(aq)) + 2H_2O_((l))`

In order for the oxalic acid to solution to completely neutralize the sodium hydroxide solution, you must add exactly 1 mole of oxalic acid for every 2 moles of sodium hydroxide.

This results from the `"1:2"` mole ratio that exists between the two compounds - 1 mole of oxalic acid needs 2 moles of sodium hydroxide for the reaction to takes place.

The number of moles of oxalic acid used can be calculated using the solution's molarity

`C = n/V => n_("oxalic acid") = C * V = 20.34 * 10^(-3)"L" * "0.1003 M"`

`n_("oxalic acid") = = "0.002040 moles oxalic acid"`

According to the aforementioned mole ratio, the number of sodium hydroxide moles must be

`"0.002040 moles oxalic acid" * "2 moles sodium hydroxide"/"1 mole oxalic acid" = "0.004080 moles sodium hydroxide"`

FInally, use the volume of the sodium hydroxide solution to solve for its concentration

`C = n_("sodium hydroxide")/V_("sodium hydroxide") = "0.004080 moles"/(25 * 10^(-3)"L") = "0.1632 M"`