In a titration experiment, 22.50 cm3 of an acid solution A containing 10.6 g of NaHSO4 per dm3 reacted with 25.0 cm3 of solution B containing X g of NaOH per dm3. The equation for the reaction is:

(a) From the information given above, calculate the:
(i) concentrate of A in moldm– 3;
(ii) concentrate of B in moldm– 3;
(iii) value of X;
(iv) mass of Na2SO4 formed during the reaction. [ H = 1.00; 0 = 16.0; Na = 23.0; S = 32.0] (16 marks)

(b) (i) Name a suitable indicator for the titration experiment.
(ii) State the apparatus used to measure the volume of solution: I. A; II B; (3 marks)

This question was attempted by majority of the candidates and their performance was fair.

In (a), majority of the candidates lost mark for writing the unit of molar mass in “g” instead of in “g mol”. Although majority of the candidates were able to calculate the concentration of both the acid and base in moldm– 3, a lot of candidate lost marks for not expressing their answers to 3 significant figures.

In (b), candidates were able to name suitable indicators for the titration and they were also able to name the apparatus used to measure the volume of the solution of the acid and base.

The required answers include: (a) (i) Molar mass NaHSO4 = [23.0 + 1.00 + 32.0 + 64.0] = 120gmol-1 (No score for wrong Unit)
Conc. of A = 10.6/120
= 0.0883mol dm-3 (correct evaluation to 3sig.figure) (No score for wrong Unit).

(ii) Reaction: NaHSO4(aq) + NaOH(aq) ? Na2SO4(aq) + H2O(l)
From the reaction equation CAVA / CBVB = 1 /1
CB = CAVA / VB
CB = (0.0883 X 22.5) / 25.0= 0.0795mol dm-3(No score for wrong Unit)

iii) Molar mass of NaOH = (23.0 + 16.0 + 1.00) = 40gmol-1 (No score for wrong Unit)
Mass of NaOH gdm-3 = 40 x 0.0795
= 3.18gdm-3
Value of X = 3.18

(iv) Amount of NaOH in 25.0cm3 of B
= (25.0 x 0.0795)/1000
= 0.00199 mol
From the reaction equation
1 mol NaOH gives 1 mol Na2SO4
0.00199 mol NaOH = 0.00199 mol Na2SO4

Molar mass of Na2SO4 = (2 x 23.0) + 32.0 + (4 x 16.0)
= 142gmol-1
Mass of Na2SO4 = 142 x 0.00199
= 0.283g

Alternative Method 2 to (iv) Mass of NaOH in 25.0cm3 of 0.0795moldm-3
m = MCV
= 40 x 0.0795 x 25 / 1000
= 0.0795g
Molar Mass of Na2SO4 = 142gmol-1
From the balanced equation:
40g NaOH = 142g of Na2SO4
0.0795g NaOH = Xg Na2SO4
mass of Na2SO4 = 0.0795 x 142 g / 40
= 0.282g

(b) (i) Phenolphtalein (Accept methyl orange)
(ii) I - Burette
II - Pipette