Question 1B
F is a solution of NaOH containing 0.0500 mol of the alkali in 500 cm3 of solution.
G is a solution of a monobasic acid, HY.
(a) Put solution G into the burette.
Titrate 20.0cm3 or 25.0 cm3 portions of F against G using phenolphthalein as indicator.
Repeat the exercise to obtain consistent titres.
Tabulate your burette readings and calculate the average volume of G used.
The equation of reaction is:
NaOH(aq) + HY(aq) 
NaY(aq) + H2O(l)
(b) From your results and the information provided, calculate the:
(i) concerntration of F in moldm-3;
(ii) concerntration of G in moldm-3;
(iii) molar mass of the acid, HY if the concentration of acid in solution G is 3.95 gdm-3
(iv) relative atomic mass of Y;
(v) percentage by mass of Y in HY. [H = 1.00] [21 marks]
Observation
In the part (b), majority of the candidates showed good skills in the calculation of the concentration of F and G in moldm-3 respectively, but failed to give the correct substitution
of 3.95gdm-3 when they were calculating the molar mass of the acid, HY.
Expected answers include:
1(a) Two consistent / concordant titres Averaging
(b)
(i) Conc. of Fin moldm-3
Conc. of F (NaOH) = 0.050 x 1000 OR 0.050 × 2
500
= 0.100moldm-3
(ii) Conc. of G in moldm-3
Conc. of G(HY) = CGVG = 
CFVF
CG = CFVF = 0.100x 20.0/25.0
VG VG
= amol dm-3
Alternative Method
Amount of base in 20/25 cm3 B = 0.100 x V
1000
= 0.001 VBmol
From balance equation of reaction
1mol of B ≡ 1mol of A
0.0001VB of B ≡ 0.0001VBmol of A
VAcm3 ≡ 0.0001VB × 100mol
VA
= 0.1VB mol
VA
Conc. of A = 0.1VB moldm-3
VA
(iii) Molar mass of the acid G(HY)
CG =
MG(HY) = 
= 3.95
CG a
= bgmol-1
(iv) Relative atomic mass of Y in HY
HY = b
1 + Y = b
Y = b – 1
= X say
(v) Percentage by mass of Y in HY
% of Y in HY = X × 100
b