The table shows titres from an analysis of sodium carbonate solution used to neutralise acidic wastewater. A student places 25.0 cm3 of the sodium carbonate...
Question 1 Report
The table shows titres from an analysis of sodium carbonate solution used to neutralise acidic wastewater. A student places 25.0 cm3 of the sodium carbonate solution in a flask and adds hydrochloric acid from a burette. The acid concentration is 0.100 mol dm-3. The first result is a rough trial and is not used in the mean.
| trial | volume of hydrochloric acid used / cm3 |
|---|---|
| rough | 26.80 |
| 1 | 25.15 |
| 2 | 25.25 |
| 3 | 25.20 |
(a) Name the gas made when sodium carbonate reacts with hydrochloric acid. [2]
(b) Complete the balanced equation: Na2CO3 + ...HCl → ...NaCl + H2O + CO2. [3]
(c) Calculate the mean titre from trials 1, 2 and 3. [3]
(d) Calculate the amount, in mol, of hydrochloric acid in the mean titre. [2]
(e) Calculate the concentration, in mol dm-3, of the sodium carbonate solution. [4]
Answer Details
(a) The gas is carbon dioxide. It can be confirmed because it turns limewater cloudy. [2]
(b)
\[\mathrm{Na_2CO_3+2HCl\rightarrow2NaCl+H_2O+CO_2}\]
Two moles of hydrochloric acid react with one mole of sodium carbonate. [3]
(c) Do not include the rough titre:
\[\frac{25.15+25.25+25.20}{3}=25.20\text{ cm}^3\]
The mean titre is 25.20 cm3. [3]
(d)
\[25.20\text{ cm}^3=0.02520\text{ dm}^3\]
\[n=cV=0.100\times0.02520=0.00252\text{ mol}\]
The amount of hydrochloric acid is 0.00252 mol. [2]
(e) From the equation, \(2\) mol HCl react with \(1\) mol \(\mathrm{Na_2CO_3}\):
\[n(\mathrm{Na_2CO_3})=\frac{0.00252}{2}=0.00126\text{ mol}\]
\[25.0\text{ cm}^3=0.0250\text{ dm}^3\]
\[c=\frac{0.00126}{0.0250}=0.0504\text{ mol dm}^{-3}\]
The sodium carbonate concentration is 0.0504 mol dm-3. [4]
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