Question

1. Instructions: Answer all questions. All your burette readings (initial and final) as well as the size of your pipette must be recorded, but no account of experimental procedure is required. All calculations must be done in this booklet.

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 portions 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 salt, calculate the:
 (i) concentration of the anhydrous salt in B, in g dm⁻³;
 (ii) percentage of water of hydration in B;
 (iii) number of moles of hydrogen ions in the average titre value. [Molar mass of anhydrous salt in B = 106 g mol⁻¹]

Accepted Answer

(a) Size of pipette used = 25.0 cm3.
Burette readings (cm3):
 Titre 1: Final 24.00, Initial 0.00, Volume used 24.00
 Titre 2: Final 48.30, Initial 24.20, Volume used 24.10
 Titre 3: Final 24.20, Initial 0.00, Volume used 24.20
Average titre = (24.00 + 24.10 + 24.20)/3 = 24.10 cm3

(b) Amount of acid (HCl) in A = Concentration × Volume = 0.1 mol dm⁻³ × (24.10/1000) dm3 = 0.00240 mol
Amount of CO3²⁻ in B = Concentration × Volume = 0.03 mol dm⁻³ × (25.00/1000) dm3 = 0.00075 mol
Mole ratio (acid : carbonate) = 0.00240/0.00075 = 3.2 : 1 ≈ 3 : 1

(c)(i) Concentration of anhydrous salt (g dm⁻³) = molar concentration × molar mass = 0.03 mol dm⁻³ × 106 g mol⁻¹ = 3.18 g dm⁻³
(ii) Mass of water of hydration = 5.0 − 3.18 = 1.82 g dm⁻³. % water of hydration = (1.82/5.0) × 100% = 36.4%
(iii) 1 mole of HCl contains 1 mole of H⁺, so 1000 cm3 of A contains 0.1 mole of H⁺. ∴ 24.10 cm3 of A contains (24.10 × 0.1)/1000 = 0.00241 mol of H⁺.

WAEC Chemistry 2023 Theory — Question 8

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