Question 1A
A is 0.100 mol dm-3 of HCl.
B is 0.030 mol dm-3 of a trioxocarbonate (IV) salt.
(a) Put A into the burette and titrate it with 20.0 cm3 or 25.0 cm3 portion of B using methyl orange as indicator.
Repeat the titration to obtain concordant titre values.
Tabulate your results and calculate the average volume of A used.
Put A into the burette and titrate it with 20.0 cm3 or 25.0 cm3 portion of B using methyl orange as Indicator.
Repeat the titration to obtain concordant titre values.
Tabulate your results and calculate the average volume of A used.
(b) From your results and the information provided, calculate the mole ratio of the acid to the trioxocarbonate (IV) in the reaction.
(c) Given that B contains 5.0 g dm-3 of the hydrated trioxocarbonate (IV) salt, calculate the:
(i) concentration of anhydrous salt in B in g dm-3,
(ii) percentage of water hydration in B;
(iii) number of moles of hydrogen ions in the average titre value.
[ Molar mass of an anhydrous salt in B = 106 g mol-1]
Observation
This question was based on acid-base titration, and was attempted by the majority of the candidates.
In part (a), majority of the candidates correctly calculated the average volume of A used.
In part (b), majority of the candidates were unable to calculate the mole ratio of the acid to the trioxocarbonate (IV) in the reaction.
In part (c), few candidates were able to calculate the concentration of anhydrous salt in B in g dm-3. Also, percentage of water of hydration in B was correctly calculated by few candidates.
The expected answers include:
(a) Two concordant titres
Averaging
(b) Amount of acid in A = 0.1 x VA
1000
= say a mol
Amount of CO32- in B =0.030 x VB
1000
=Say b mol
Mole ratio = a: b
Simpler whole number ratio = A : B
(c) (i) Conc. of anhydrous salt in g dm-3
= molar conc. x molar mass
= 0.030 x 106
= 3.18 g dm-3
(correct to 3 sig. fig to score, wrong unit no score)
(ii) % of water of hydration
mass of water of hydration in dm3 x 100
mass of hydrated salt in dm3
Mass of water of hydration = 5.0 g dm-3 – 3.18 g dm-3
= 1.82 g
% of water of hydration = 1.82 x 100
5.0
= 36.4%
(iii) 1 mole of HCl contain 1 mole of H+
1000 cm3 of A contain 0.1 moles H+
VAcm3 will contain VA x 0.1
1000
No. of moles of H+ = 0.0001 VA
Say Q mol