Label the acid-base conjugate pairs in the following equation:
\(\text{HCO}_{3}^{-}(\text{aq}) + \text{H}_{2}\text{O}(\text{l})\) \(\rightleftharpoons\) \(\text{CO}_{3}^{2-}(\text{aq}) + \text{H}_{3}\text{O}^{+}(\text{aq})\)
A certain antacid tablet contains \(\text{22,0}\) \(\text{g}\) of baking soda \((\text{NaHCO}_{3})\). It is used to neutralise the excess hydrochloric acid in the stomach. The balanced equation for the reaction is:
\(\text{NaHCO}_{3}(\text{s}) + \text{HCl}(\text{aq})\) \(\to\) \(\text{NaCl}(\text{aq}) + \text{H}_{2}\text{O}(\text{l}) + \text{CO}_{2}(\text{g})\)
The hydrochloric acid in the stomach has a concentration of \(\text{1,0}\) \(\text{mol.dm$^{-3}$}\). Calculate the volume of the hydrochloric acid that can be neutralised by the antacid tablet.
(DoE Grade 11 Paper 2, 2007)
M(\(\text{NaHCO}_{3}\)) = \(\text{23,0}\) + \(\text{1,01}\) + \(\text{12,0}\) + (\(\text{3}\) x \(\text{16,0}\)) = \(\text{84,01}\) \(\text{g.mol$^{-1}$}\)
\(\text{n(NaHCO}_{3}{\text{)}} = \dfrac{\text{m}}{\text{M}}\)
\(\text{n} = \dfrac{\text{22,0}\text{ g}}{\text{84,01}\text{ g.mol$^{-1}$}}\) = \(\text{0,26}\) \(\text{mol}\)
From the balanced equation we see that the molar ratio of \(\text{NaHCO}_{3}\) to \(\text{HCl}\) is \(\text{1}\):\(\text{1}\).
Therefore, n(\(\text{HCl}\)) = n(\(\text{NaHCO}_{3}\)) = \(\text{0,26}\) \(\text{mol}\)
\(\text{C} = \dfrac{\text{n}}{\text{V}}\) therefore \(\text{V} = \dfrac{\text{n}}{\text{C}}\)
\(\text{V(HCl)} = \dfrac{\text{0,26}\text{ mol}}{\text{1,0}\text{ mol.dm$^{-3}$}}\) = \(\text{0,26}\) \(\text{dm$^{3}$}\)