A student investigated how the concentration of sodium thiosulfate affects the rate of its reaction with dilute hydrochloric acid, using the disappearing cr...
Question 1 Report
A student investigated how the concentration of sodium thiosulfate affects the rate of its reaction with dilute hydrochloric acid, using the disappearing cross method. Water was used to change the concentration while keeping the total volume the same.
| Experiment | Volume of thiosulfate / cm3 | Volume of water / cm3 | Volume of acid / cm3 | Time for cross to disappear / s |
|---|---|---|---|---|
| 1 | 40 | 0 | 5 | 25 |
| 2 | 30 | 10 | 5 | 34 |
| 3 | 20 | 20 | 5 | 52 |
| 4 | 10 | 30 | 5 | 105 |
(a) Explain why water is added in experiments 2 to 4. [1]
(b) Describe the relationship between the volume of thiosulfate used and the time for the cross to disappear. [1]
(c) Calculate the rate (1/time) for experiment 3. [2]
(d) Explain, in terms of collisions, why a lower concentration of thiosulfate gives a longer time. [3]
(e) State two variables that must be kept constant for a fair test. [2]
(f) Name the observation used to decide that the reaction has finished. [1]
(g) The cloudiness is caused by a solid spread through the liquid. Name the solid and name this type of mixture. [2]
Answer Details
This is a full concentration-versus-rate investigation using the disappearing cross method, testing fair-test design, a rate calculation, and collision theory.
(a) Water is added in experiments 2 to 4 because it lowers (changes) the concentration of the thiosulfate while keeping the total volume the same, so only the concentration is being varied. [1]
(b) Reading the table, as the thiosulfate volume falls from 40 to 10 cm³ the time rises from 25 to 105 s: the smaller the volume of thiosulfate (the lower the concentration), the longer the time for the cross to disappear. [1]
(c) Rate is \(1/\text{time}\). For experiment 3 the time is 52 s:
\[ \text{rate} = \frac{1}{52\ \text{s}} = 0.019\ \text{s}^{-1} \](accept \(0.0192\ \text{s}^{-1}\)). One mark for \(1/52\), one for the value. [2]
(d) A lower concentration of thiosulfate gives a longer time because: a lower concentration has fewer thiosulfate particles in the same volume; so collisions happen less frequently; and the rate is slower, so the cross takes longer to disappear. Three marks. [3]
(e) For a fair test keep any two constant, for example the temperature, the volume (concentration) of acid, the total volume, the size of the cross, or using the same observer. [2]
(f) The reaction is judged finished when the cross can no longer be seen (the mixture has become opaque). [1]
(g) The solid causing the cloudiness is sulfur, and a solid spread through a liquid like this is a suspension (accept precipitate). One mark each. [2]
Exam takeaway: diluting with water while fixing the total volume is the standard way to change concentration alone; it is what keeps the experiment a fair test.
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