WAEC GCE 2019 Chemistry Practical Answer – Jan/Feb Expo

(1bi)
Volume of Acid Va = 21.60cm³
Volume of Alkali Vb = 25.0cm³
Mass of NaoH
2.0g = 500cm³
Xg = 1000cm³
X = 1000×2.0/500
X = 2×2 = 4.0g/dm³

Then mass conc = molar conc × molar mass
.:. molar conc of B = mass conc/molar mass

Where the molar mass of NaOH
=23+16+1 = 40gmol^-1
.:. Conc of NaOH in moldm^-³ = 4.0g/dm³/40g/mol
= 0.100mol/dm^-³

(1bii)
CaVa/CbVb = na/nb
Ca×21.60/0.1×25 = 1/2
Ca = 0.1×25/2×21.6
CA = 0.058mol/dm^-³
Concentration of H2SO4 in mol/dm^-³ = 0.058mol/dm-³

(1biii)
Recall that mass conc of H2SO4
= Molar conc × Molar mass
Where molar mass of H2SO4
=2(1)+32+4(16)
= 2 + 32 + 64
= 98g/mol
Mass conc of H2SO4 = 0.058× 98
= 5.684g/dm^-³
= 5.68g/dm^-³

(1c)
(i) I would ensure the air bubble present is removed before taking the initial reading.
(ii) I would ensure I keep my eye in level with the liquid surface while taking the burrette reading.
(iii) I would ensure there is no leakage from the burette during titration.