Question 1
A is a solution of tetraoxosulphate (VI) acid.
B contains 3.95 g of sodium trioxocarbonate (IV) in 750 cm3 of solution. If 22.70 cm3 of the tetraoxosulphate (VI) acid neutralized 25.0 cm3 of sodium trioxocarbonate (IV).
- Calculate the concentration of:
(i) B in g dm-3;
(ii) B in mol dm-3;
- A in mol dm-3;
- A in g dm-3.
- Calculate the number of hydroxonium ions which could be obtained from the result in 1 (a)(iii).
- Suggest a possible indicator for the reaction.
Give a reason for your answer.
The equation of the reaction is:
H2SO4 (aq) + Na2SO4 (aq) Na2SO4 (aq) + H2O(l) + CO2 (g)
[H = 1.00; C = 12.0; O = 16.0; Na = 23.0; S = 32.0,
Avogadro’s constant = 6.02x 10 23] [16 marks]
Observation
In part (a), candidates lost marks to approximation error; many of them failed to have their answers rounded up to 3 significant figures. Some of them lost marks for wrong units.
In part (b), majority of the candidates could not interpret the ionization of H2SO4(aq), using the mole concepts. Consequently, candidates failed to obtain 2 moles of the hydroxonium H3O+ from 1 mole of H2SO4, to get the correct number of ions using Avogadro’s constant.
In part (c), majority of the candidates gave the correct answer and a reason.
The expected answers include:
- To calculate the concentration ofB in gdm ̄ 3
= 
=
= 5.27gdms-3
ALTERNATIVE METHOD
750cm3 of B contain 3.97g
1000cm3 (1dm3) contain 3.97 x 1000
750
= 5.27gdm-3
- To calculate the concentration ofB in moldm ̄ 3
M (Na2CO3) = 46 +12+48=106gmol ̄ 1
Cʙ=
= 
=0.0497moldm̵ ³
- To calculate the concentration ofA in moldm ̄ ³
= 
CA=
=0.0547mol dm ̄ ³
ALTERNATIVE METHOD
1000cm3 of B contain 0.0497 mol
25cm3 of B contain 0.0497 x 25
1000
= 1.24 x 10-3/0. 00124 mol
From Equation 1mol Na2CO3 (B) ≡1mol H2SO4 (A)
mol H2SO4 (A) = 0.00124 /1.24 x 10-3mol.
22.70cm3A contain 0.00124 mol
10003 (1dm3) contain 0.00124 x 1000
22.70
= 0.0547 moldm-3
To calculate the concentration of A in gdm-3
M (H2SO4) = 2 + 32 + 64 = 98g mol ̵ ¹
= CM = 0.0547×98
= 5.36gdm ̵ ³
(b)Number of hydroxonium ions obtained from 1(a) (iii)
H2SO4 + 2H2O 2H3O+ + SO42-
1mol H2SO4 produces 2 mol H3O+
0.0547 mol H2SO4 produces 2 × 0.0547 mol = 0.1094mol
1mol H3O+ = 6.02 x 1023 ions
0.1094 mol H30+ = 6.02 x 0.1094 x1023
= 6.59 x 1022 ions.
(c) Possible indicator = methy1orange
Reason: Strong acid verse weak base.