To estimate the amount of Al as Al2O3 in the given solution of aluminium sulphate https://youtu.be/wETFNg2a-Y0

1. SHREE MALLIKARJUN COLLEGE TYBSC
SAFETY: Refer to MSDS of aluminium sulphate, ammonium chloride, ammonium hydroxide
-Dr. Mithil S. Fal Desai
Aim: To estimate the amount of Al as Al2O3 in the given solution of Aluminium sulphate.
Chemicals: aluminium sulphate, ammonium chloride, ammonium hydroxide.
Apparatus: Beaker, funnel, Whatman 41, glass rod with policemen, water bath, silica crucible.
Theory: In the gravimetric estimation, Al is precipitated as Al(III) hydroxide which is then
sintered to get Al(III) oxide (Al2O3). In concentrated alkali Al(III) oxide is soluble as it forms tetra
hydroxy aluminate ions. Al2O3 is an amphoteric oxide as it reacts with both, acid and bases.
An alternative and better procedure involves separating and weighing the precipitate as the
aluminum 8-hydroxyquinolate, Al(C9H6NO)3.
Reactions:
Al3+
(aq) + 3OH-
(aq)  Al(OH)3 (s) ↓
2Al(OH)3(s)) + Heat  Al2O3 + 3H2O
Procedure:
1) Pipette out 25 mL of the solution in a 250 mL beaker and add 2 drops of methyl red indicator.
2) Add 2 g of NH4Cl and warm the solution.
3) Add 6M NH4OH dropwise till solution turns yellow (white gelatinous precipitate) and digest
the precipitate on water bath for 30 min.
4) After 30 min, cool the solution and add 2-3 drops of of NH4OH solution along the sides of the
beaker to check complete precipitation of Al.
5) Collect the precipitate on a Whatman 41 filter paper and wash with 1% NH4NO3 distilled water
until the filtrate is free from SO4
2- ions.
6) Carefully transfer the precipitate with the filter paper in a previously weighed silica crucible
and heat it on a low flame.
7) Once the paper is charred, ignite the precipitate to get Al2O3.
8) Cool and weigh the crucible containing the precipitate.
(To get the constant weights, repeat the step 7 and 8)
9) The weight of precipitate quantitatively corresponds to the weight of Al2O3.
Observation
Table 1.
The constant
weight of silica
crucible without
precipitate
“W1’ (g)
The constant
weight of silica
crucible with
precipitate
“W2’ (g)
Weight of
Al2O3
‘A’=W2 – W1
(g)
Molecular
weight of Al2O3
‘B’ (g)
Atomic weight
of Al
‘C’(g)
Amount of Al in
25 mL
‘x' = A*C*2/B
(g)
Result: i) Amount of Al in 500 mL is ______g.
ii) Amount of aluminium sulphate in 500 mL is ______g.