What volume of a 0.3300-M solution of sodium hydroxide would be required to titrate 15.00 mL of 0.1500 M oxalic acid?

\({{\rm{C}}_2}{{\rm{O}}_4}{{\rm{H}}_{2({\rm{aq}})}}{\rm{ + 2NaO}}{{\rm{H}}_{({\rm{aq}})}}{\rm{ }} \to {\rm{ N}}{{\rm{a}}_2}{{\rm{C}}_2}{{\rm{O}}_{4({\rm{aq}})}}{\rm{ + 2}}{{\rm{H}}_2}{{\rm{O}}_{({\rm{l}})}}\)

According to the mole concept, the number of moles of a compound is the product of its concentration and volume. It is given that the concentration of oxalic acid is 0.1500 M and the volume of oxalic acid is 15.0 ml or 0.150 L.

Therefore, the number of moles = \(0.150{\rm{ mL }} \times {\rm{ 0}}{\rm{.1500 moles/L = 2}}{\rm{.25 }} \times {\rm{ }}{10^{ - 3}}\)

Now, as 1 moles of oxalic acid is required to react with 2 moles of \({\rm{NaOH}}\), thus the number of moles of sodium hydroxide will be \({\rm{2}}{\rm{.25 }} \times {\rm{ }}{10^{ - 3}}{\rm{ }} \times {\rm{ 2 = 4}}{\rm{.5 }} \times {\rm{ }}{10^{ - 3}}{\rm{ moles}}\)

It is given that the concentration of \({\rm{NaOH}}\) is 0.3300 M and the moles of \({\rm{NaOH}}\)from above calculation is \({\rm{4}}{\rm{.5 }} \times {\rm{ }}{10^{ - 3}}{\rm{ moles}}\), then the volume of \({\rm{NaOH}}\) will be calculated as

\({\rm{Volume = }}\frac{{{\rm{moles}}}}{{{\rm{concentration}}}}\)

\({\rm{Volume = }}\frac{{{\rm{4}}{\rm{.5 }} \times {\rm{ 1}}{{\rm{0}}^{ - 3}}}}{{{\rm{0}}{\rm{.330}}}}{\rm{ = 13}}{\rm{.64 mL}}\)