Question 1
In a titration experiment, 10.0 cm3 portions of 0.15 mol dm-3 solution of potassium trioxocarbonate (IV) were titrated against tetraoxosulphate (VI) acid, using methyl orange as indicator. The results are recorded in the table.
Burette readings/cm3 |
1 |
2 |
3 |
Final reading |
14.45 |
|
38.65 |
Initial reading |
1.80 |
14.45 |
|
Titre value |
|
12.15 |
12.05 |
- (i) Complete the table.
(ii) Calculate the average tire value.
The equation for the reaction is:
K2CO3(aq) + H2SO4(aq) K2SO4(aq) + CO2(g) + H2O(l)
- From the information provided and your results, calculate the:
- concentration of the acid in mol dm-3;
- mass of potassium tetraoxosulphate (VI) that would be formed in solution.
(O=16.0, S=32.0, K = 39.0)
- List two techniques that could be used to recover the potassium tetraoxosulphate (VI) from the solution.
- A is a solution of HCl containing 0.025 mole in 500 cm3 B is 0.100 mol dm-3 NaOH solution. A was titrated against 20.0 cm3 portion of B using a suitable indicator.
- Name an indicator that is suitable for the reaction.
- State the colour change for the reaction at the end-point.
- Write a balanced chemical equation for the reaction.
- Calculate the concentration of A in mol dm-3.
[22 marks]
Observation
This was a popular question among the candidates as majority of them responded to it and their performance was above average.
In part (a), majority of the candidates were able to complete the titration table.
In part (b), majority of the candidates were able to calculate the concentration of the acid in mol dm-3. However, few of them calculated the mass of potassium tetraoxosulphate (VI) that would be formed in solution.
In part (c), majority of the candidates listed two techniques that could be used to recover the potassium tetraoxosulphate (VI) form the solution.
In part (d), majority of the candidates named an indicator that is suitable for the reaction and stated the colour change for the reaction at the end-point.
The expected answers include:
(a) (i)
Burette readings / cm3 |
1 |
2 |
3 |
Final reading |
14.45 |
26.60 |
38.65 |
Initial reading |
1.80 |
14.45 |
26.60 |
Titre Value |
12.65 |
12.15 |
12.05 |
(ii) Average titre = 12.15 + 12.05
2
= 24.20 cm3
2
= 12.10 cm3
(b) (i) CaVa = na
CbVb nb
Ca x 12.10 = 1
0.15 x 10 1
Ca = 0.15 x 10
12.10
= 0.124 mol dm-3
(ii) Molar mass of K2SO4 = (2x39)+32+(4x16)
= 174 g mol-1
With respect to K2CO3,
Amount in moles of K2CO3, used, n = CV
= 0.15 x 10
1000
= 0.0015 mol
From the equation, 1 mole of K2SO4 1 mole of K2CO3,
Amount of K2SO4 = 0.0015 mole
Hence mass of K2SO4 = 0.0015 x 174
= 0.261 g
(c) - Evaporation and crystallization
(d)
(i) - methyl orange
- methyl red
- phenolphthalein
(ii) methyl orange - yellow to orange
methyl red - yellow to red
phenolphthalein - pink to colourless
(iii) HCl(aq) + NaOH(aq) → NaCl(aq) + H2O(l)
(iv) 500 cm3 contains 0.025 mol
1000 cm3 will contain 0.025 x 1000
500