Chemistry Practical 

(1a)
Volume of burrete = 50.00cm³
Volume of pipette = 25.00cm³
Indicator = Methyl orange

TABULATE

Final burette readings (cm³); 37.50 | 22.50 | 37.50 | 44.50

Initial burette readings(cm³); 0.00 | 0.00 | 0.00 | 0.00

Volume f Acid used (cm³); 37.50 | 22.50 | 22.50 | 22.50

Volume of A used = 1st + 2nd + 3rd/3
= 22.50+22.50+22.50/3
Average volume of A used = 22.50cm³

(1bi)
Concentration of A ( CA )?
2.03g in 500cm³ of solution

Volume = 500cm³/1000 = 0.500dm³
In g/dm³ = 2.03/0.5 = 4.06g/dm³

To find concentration in mol/dm³
Conc. of A in mol/dm³ = Conc. of A in g/dm³/Molar mass

Conc. of A in mol/dm³ = 4.06/36.5
= 0.1112mol/dm³
Molar mass of A (HCL) = 1+35.5 = 36.5g/mol

(1bii)
Number of mole of the Acid in average titre
Na = CaVa
= 0.1112 × 22.50
= 2.52moles

(1biii)
Nb= ? Na= 3, Ca= 0.1112mol/dm³, Cb = 0.12mol/dm³ , Va = 22.50cm³, Vb = 25.00cm³

CaVa/CbVb= nA/nB

0.1112×22.50/0.12×25 = 3/nB

nB × 2.502 = 9
nB = 9/2.502 =
nB = 3.597moles

(1biv)
Mole ratio of Acid to base
nA:nB
= 3:3
= 1:1

(3ai)
The value will increase

(3aii)
The occur as a result of the decrease in the concentration of base due to the added volume of water

(3bi)
There will be no visible reaction because copper is less than Zinc in the electrochemical series

(3bii)
It absorb water and become sticky because it is hygroscopic

(3biii)
The solution turns pink

(3c)

When NaOH is added to the solution of zn³+, a white precipitate is formed which later dissolve in excess NaOH due to the formation of zinc hydroxide