Chemistry Paper 3B WASSCE (SC), 2023

Question 1B

 

           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 ºC
           per dm3 of solution.

  1. Put E into the burette and titrate it against 20.0 cm3 or 25.0 cm3 portion 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

 

  1. From your results and the information provided, calculate the:

(i)            Concentration of HCl in E mol dm-3;
(ii)           concentration of Y(OH)2 in F in mol dm-3-;
(iii)          solubility of the substance Y(OH)2 in mol dm-3;
(iv)          mass of Y(OH)2 that would be deposited if 1 dm3 of saturated solution is 
evaporated to dryness.

                                    [H=1.0; O =16.0; Cl = 35.5; Y(OH)2 = 74.0]

Observation

 

This question was on titration experiment. Majority of the candidates that responded to this question performed above average.

In part (a), majority of the candidates obtained concordant values from the titration experiment;

In part (b), majority of the candidates were able to calculate the concentration of HCl in E in mol dm-3 However, they could not calculate the solubility of the substance Y(OH)2 in mol dm-3.

The expected answers include:

(a) Two concordant titres
Averaging


(b) (i) Molar mass of HCl = 1 + 35.5
= 36.5 g mol--1
Concentration of E = 2.92
36.5 x 1
= 0.080 mol dm-3

  (ii) From the equation
CEVE = 2  
CFVF    1
0.080 x VE = 2
CF x VF            1

CF =   0.080 x VE
                2 x VF
= Say W mol dm-3

(iii) C1V1(Saturated solution) = C2V2 (dilute solution)
C1 x 20 = W x1000

C1 = W x 1000
                20
= Say X mold m-3
Alternative
20cm3 of saturated solution contains W moles
Hence 1000 cm3 of saturated solution = W x 1000
                                                                    20
= Say X mol dm-3
(iv)      Mass of Y(OH)2 deposited
= X x molar mass (Y(OH)2)
= X x 74 g
= Z g say