A solution of a mineral acid containing 0.10 mol of the acid per dm of solution. B IS a solution Containing 1.325g of anhydrous sodium trioxocarbonate (IV) per dm of solution.
(a) Put A into the burette and titrate with 20cm\(^3\) or 25cm\(^3\) portions of B using methyl orange as an indicator. Record the volume of your pipette
(b) From your results and the information is given, calculate;
(i) The number of mole of acid in the average titre
(ii) The number of mole of sodium trioxocarbonate (IV) in the volume of B pipette,
(iii) The mole ratio of acid to base in the reaction [H = 1, C = 12, O = 16, Na 23]
(c) Suggest what the acid in (a) Could be giving reasons for your answer. Hence, write the equation for the reaction.
(a) Volume of pipette used 25.00cm\(^3\)
Indicator used: Methyl orange
Colour change at endpoint from orange to pink/red
| Burette readings | Rough | 1st | 2nd | 3rd |
| Final readings in cm\(^3\) | 6.50 | 9.70 | 16.10 | 6.15 |
| Initial reading in cm\(^3\) | 0.00 | 3.50 | 10.00 | 0.00 |
| Volume of A (acid) used | 6.50 | 6.20 | 6.10 | 6.15 |
Average vol of acid used = \(\frac{6.20 + 6.10 + 6.15}{3}\) = 6.15cm\(^3\)
(b)(i) Number of moles of acid = \(\frac{vol. \times mol}{1000} = \frac{6.15 \times 0.1}{1000}\) = 0.000615 moles
(ii) The number of moles of Na\(_2\)CO\(_3\) in the 25.00cm\(^3\) of solution B pipette; conc. of Na\(_2\)CO\(_3\) = 1.325g (given). Relative molecular mass (RMM) of Na\(_2\)CO\(_3\) 106
Number of moles of Na\(_2\)CO\(_3\) = \(\frac{\text{conc. of Na\(_2\)CO\(_3\)}}{\text{RMM of N\(_2\)CO\(_3\)}}\) = \(\frac{1.325}{106}\) = 0.125 moles
The number of moles of Na\(_2\)CO\(_3\) in 25cm\(^3\) of solution B = \(\frac{25 \times 0.0125}{1000}\) = 0.0125 moles
(iii) The ratio of the number of moles of acid to alkali 0.000615; 0.0003125
Dividing all through by the smallest value to obtain smallest ratio
\(\frac{0.000615}{0.0003125}\) : \(\frac{0.0003125}{0.0003125}\)
1.968 : 1 or approximately 2 : 1
(c) Since the ratio of the material acid to the alkali is 2:1, therefore the acid is monobasic. e.g. HCl, HNO\(_3\)
Equation for the reaction is written thus; 2HCl + Na\(_2\)CO\(_3\) = 2 NaCl + CO\(_2\) + H\(_2\)O or 2HNO\(_3\) + NaCO\(_3\) \(\to\) 2NaNO\(_3\) + CO\(_2\) + H\(_2\)O
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