All your burette readings (initial and final), as well as the size of your pipette, must be recorded but no account of experimental procedure is required. All calculations must be done in your answer book.
A is a solution containing 6.3g dm\(^{-3}\) of impure ethanoic acid. B is 0.10mol. dm\(^{-3}\) sodium hydroxide sodium hydroxide solution.
(a) Put A into the burette and titrate with 20.0cm\(^3\) or 25.0cm\(^3\) portions of B using phenolphthalein as indicator. Record the volume of your pipette Record the volume of your pipette. Tabulate your burette readings and calculate the volume of A used.
(b) From your result and the information provided, calculate the
(i) concentration of solution A in mol.dm\(^{-3}\)
(ii) Concentration of solution A in dm\(^{-3}\) and hence the percentage purity of the ethanoic acid,
(iii) Volume of solution A that would neutralize a solution containing 0.005 mole of sodium hydroxide. The equation for the reaction is;
H\(_2\)C\(_2\)O\(_{4(aq)}\) + 2NaOH\(_{(aq)}\) \(\to\) Na\(_2\)C\(_2\)O\(_{(aq)}\) + 2H\(_2\)O\(_{(l)}\) [H = 1; C = 12. O = 16]
(c)(i) What would be the colour of methylorange indicator in solution B?
(ii) Give the reason why methylorange is not a suitable indicator for titration.
Explanation
Volume of pipette/base used = 25cm\(^3\)
Indicator - phenolphthalein
| Burette readings | Rough | 1st | 2nd | 3rd |
| Final burette readings | 20.80 | 41.45 | 20.70 | 25.60 |
| Initial burette readings | 0.00 | 20.80 | 0.00. | 5.00 |
| volume of acid used | 20.80 | 20.65 | 20.70 | 20.60 |
Average litre value = \(\frac{20.65 + 20.70 + 20.60}{3}\) = 20.65 cm\(^{3}\)
(b)(i) Equation for the reaction; H\(_2\)C\(_2\)O\(_{4(aq)}\) + 2NaOH\(_{(aq)}\) + 2NaOH\(_{(aq)}\) \(\to\) Na\(_2\)C\(_2\)O\(_{(aq)}\) + 2H\(_2\)O
From the equation, mole ratio of acid to base is 1 : 2
Amount of NaOH in 25cm\(^3\) of B = \(\frac{25 \times 0.10}{1000}\) = 0.0025 mol.dm\(^{-3}\)
Amount of acid in 20.65 cm\(^3\) of A = \(\frac{25 \times 0.10}{2 \times 20.65}\)
= 0.60 mol.dm\(^{-3}\)
(ii) Molar concentration in gm/dm\(^{-3}\) of H\(_2\)C\(_2\)O\(_4\) = (2 x 1) + (2 x 1) + (2 x 12) + (4 x 16) + 90g
mass of H\(_2\) C\(_2\)O\(_4\) = (2 x 1) + (2 x 12) + (4 x 16) + 90g
mass of H\(_2\) C\(_2\)O\(_4\) in 1dm\(^3\) of A = \(\frac{90 \times 25 \times 0.1}{2 \times 20.65}\)
= \(\frac{225}{41.3} = 5.45g\)
mass of H\(_2\) C\(_2\)O\(_4\) in 1dm\(^3\) of A = 6.3g
% purity of H\(_2\) C\(_2\)O\(_4\) = \(\frac{5.45}{6.3} \times \frac{100}{1}\)
= 86.51%
(iii) 25dm\(^3\) of 0.10 mol. dn\(^3\) NaOH contains = \(\frac{25 \times 010}{1000}\) = 0.0025
20.65 cm\(^3\) of a neutralised 0.0025 mole NaOH
0.005 mole NaOH will be neutralised by \(\frac{20.65 \times 0.005}{0.0025}\) = 41.3cm\(^3\)
(c)(i) Methyl orange turns yellow
(ii) Ethanoic acid is a weak acid while sodium hydroxide is a strong base, Methyl orange is not suitable for the titration of a weak acid with strong bases because the endpoint will not coincide with the pH at which the indicator exhibits its colour change.