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(\(\text{NaOH}\)) \(= \dfrac{\text{n}}{\text{V}} =
\dfrac{\text{0,016}\text{
mol}}{\text{0,035}\text{ dm$^{3}$}} =\)
\(\text{0,46}\) \(\text{mol·dm$^{-3}$}\)