Q4. 25.0 mL of a NaOH solution is titrated with 0.050 M oxalic acid; 21.6 mL of acid is needed. Find the molarity and strength of the NaOH. (This time the acid is in the burette.)
2 × 0.050 × 21.6 = 1 × M × 25.0, so M = 2.16/25.0 = 0.0864 M; strength = 0.0864 × 40 = 3.4560 g L−1. With phenolphthalein the end point is now pink to colourless, which is harder to see; that is why the manual puts the alkali in the burette.
Q5. Why is sodium hydroxide a secondary standard and not a primary one?
Solid NaOH absorbs water from the air (it is hygroscopic) and reacts with carbon dioxide to form sodium carbonate, so a weighed sample is not pure NaOH. Its solution must be standardised by titration against a primary standard such as oxalic acid or potassium hydrogenphthalate.
Q6. Explain why aqueous sodium carbonate is weakly alkaline though it is a salt (a Laboratory Manual discussion question).
It is the salt of a weak acid (carbonic acid) and a strong base. The carbonate ion hydrolyses, CO32− + H2O ⇌ HCO3− + OH−, so the solution contains more OH− than H+.
Q7. What is the difference between the end point and the equivalence point? (A Laboratory Manual discussion question, page 78.)
The equivalence point is where the acid and base have reacted in exactly the stoichiometric ratio; the end point is where the indicator actually changes colour. A well-chosen indicator makes them almost the same; the small difference is the titration error.
Q8. To what final volume must 50.0 mL of 0.20 M HCl be diluted to make it 0.050 M?
M1V1 = M2V2: 0.20 × 50.0 = 0.050 × V2, so V2 = 200 mL. Dilute with water to a final volume of 200 mL and mix. The water added is approximately 150 mL only if volumes are treated as additive; accurate preparation uses the final-volume mark.