Question

1. Instructions: Answer all questions.

E is a solution containing 2.92 g of HCl per dm3. F is a solution obtained by diluting 20.0 cm3 of a saturated solution of Y(OH)2 at 25°C to 1 dm3 of solution.
(a) Put E into the burette and titrate it against 20.0 cm3 or 25.0 cm3 portions of F using phenolphthalein as indicator. Repeat the titration to obtain concordant titre values. Tabulate your results and calculate the average volume of acid used.
The equation for the reaction is: Y(OH)2 + 2HCl → YCl2 + 2H2O
(b) From your results and the information provided, calculate the:
 (i) concentration of HCl in E, in mol dm⁻³;
 (ii) concentration of Y(OH)2 in F, in mol dm⁻³;
 (iii) solubility of Y(OH)2, in mol dm⁻³;
 (iv) mass of Y(OH)2 that would be deposited, given its molar mass is 74 g mol⁻¹.

Accepted Answer

(a) Burette readings (cm3):
 Titre 1: Final 8.40, Initial 0.00, Volume used 8.40
 Titre 2: Final 16.90, Initial 8.40, Volume used 8.50
 Titre 3: Final 8.60, Initial 0.00, Volume used 8.60
Average titre = (8.40 + 8.50 + 8.60)/3 = 8.50 cm3

(b)(i) Molar mass of HCl = 1 + 35.5 = 36.5 g/mol. Concentration of HCl in E = (2.92 g dm⁻³) ÷ (36.5 g mol⁻¹) = 0.080 mol dm⁻³.

(ii) From Y(OH)2 + 2HCl → YCl2 + 2H2O: ηHCl = 2, ηY(OH)2 = 1. Using CE·VE/CF·VF = ηE/ηF:
CF = (CE × VE × ηF)/(VF × ηE) = (0.080 × 8.50 × 1)/(25 × 2) = 0.68/50 = 0.0136 mol dm⁻³.

(iii) The 20 cm3 of saturated Y(OH)2 solution was diluted to 1 dm3 (1000 cm3) to make F, which has a concentration of 0.0136 mol dm⁻³. So the original saturated solution's concentration (solubility) = (0.0136 × 1000)/20 = 0.68 mol dm⁻³.

(iv) Mass of Y(OH)2 = moles × molar mass = 0.68 mol × 74 g/mol = 50.32 g.

WAEC Chemistry 2023 Theory — Question 12

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