Calculate the mass of oxalic acid which the learner must dissolve to make up the required standard solution.
We need the mass of oxalic acid. However, we don't know the number of moles yet.
V = \(\text{500}\) \(\text{cm$^{3}$}\) \(\times \dfrac{\text{0,001} {\text{ dm}}^{3}}{\text{1} {\text{ cm}}^{3}}\) = \(\text{0,5}\) \(\text{dm$^{3}$}\)
\(\text{C} = \dfrac{\text{n}}{\text{V}}\), therefore n = C x V
n = \(\text{0,2}\) \(\text{mol.dm$^{-3}$}\) x \(\text{0,5}\) \(\text{dm$^{3}$}\) = \(\text{0,1}\) \(\text{mol}\)
M(\(\text{H}_{2}\text{C}_{2}\text{O}_{4}\)) = (\(\text{2}\) x \(\text{1,01}\) + \(\text{2}\) x \(\text{12,0}\) + \(\text{4}\) x \(\text{16,0}\)) \(\text{g.mol$^{-1}$}\) = \(\text{90,02}\) \(\text{g.mol$^{-1}$}\)
\(\text{n} = \dfrac{\text{m}}{\text{M}}\), therefore m = n x M
m = \(\text{0,1}\) \(\text{mol}\) x \(\text{90,02}\) \(\text{g.mol$^{-1}$}\) = \(\text{9,00}\) \(\text{g}\)
The learner titrates this \(\text{0,2}\) \(\text{mol.dm$^{-3}$}\) oxalic acid solution against a solution of sodium hydroxide. He finds that \(\text{40}\) \(\text{cm$^{3}$}\) of the oxalic acid solution completely neutralises \(\text{35}\) \(\text{cm$^{3}$}\) of the sodium hydroxide solution.
Calculate the concentration of the sodium hydroxide solution.
The balanced equations is:
\(\text{H}_{2}\text{C}_{2}\text{O}_{4} + 2\text{NaOH}\) \(\to\) \(\text{Na}_{2}\text{C}_{2}\text{O}_{4} + 2\text{H}_{2}\text{O}\)
The number of moles of oxalic acid used is the number of moles in \(\text{40}\) \(\text{cm$^{3}$}\) of the standard solution:
V = \(\text{40}\) \(\text{cm$^{3}$}\) \(\times \dfrac{\text{0,001} {\text{ dm}}^{3}}{\text{1} {\text{ cm}}^{3}}\) = \(\text{0,04}\) \(\text{dm$^{3}$}\)
n = C x V = \(\text{0,2}\) \(\text{mol.dm$^{-3}$}\) x \(\text{0,04}\) \(\text{dm$^{3}$}\) = \(\text{0,008}\) \(\text{mol}\)
The molar ratio of oxalic acid to sodium hydroxide is \(\text{1}\):\(\text{2}\). For every one mole of oxalic acid there are two moles of sodium hydroxide.
n(\(\text{NaOH}\)) = \(\text{2}\) x \(\text{0,008}\) \(\text{mol}\) = \(\text{0,016}\) \(\text{mol}\)
V(\(\text{NaOH}\)) = \(\text{35}\) \(\text{cm$^{3}$}\) \(\times \dfrac{\text{0,001} {\text{ dm}}^{3}}{\text{1} {\text{ cm}}^{3}}\) = \(\text{0,035}\) \(\text{dm$^{3}$}\)
C(\(\text{NaOH}\)) \(= \dfrac{\text{n}}{\text{V}} = \dfrac{\text{0,016}\text{ mol}}{\text{0,035}\text{ dm$^{3}$}} =\) \(\text{0,46}\) \(\text{mol.dm$^{-3}$}\)