EXPT. NO.
:2 DATE:
SOLID-LIQUID EXRACTION
Aim: - To determine the percentage recovery of solute by solid liquid leaching operation
(multistage cross-current).
Apparatus: - Beaker 500m, pipette, measuring cylinder, conical flask, stirrer, weighing machine etc.
Chemicals: - Oxalic acid powder, 1 N NaOH, phenolphthalein indicator, sand.
Theory: - The extraction of soluble constituted from a solid by means of solvent is generally
referred as leaching. The process may be employed either for production of concentrate solution of a
valuable solid material with which it is contaminated or if the solute is uniformly dispersed The
material close to the surface will get dissolved fist leaving a porous structure in solid residue. The
solvent will have to penetrate this outer layer before it can reach further solute and process will
progressively more difficult and extraction rate will become less. Leaching is done in the following
way.
1. Single stage leaching
2. Multistage cross-current leaching
3. Multistage countercurrent leaching
Procedure: - 1) Take 75 gms of clean sand and 25 gms of oxalic acid.
2) Add 100 gms of water in it and mix the same for 5 min.
3) Allow to settle down. Take the extract 1 in another beaker by pouring carefully.
4) Add 100 gms of water in the residue of the first beaker.
5) Repeat the above procedure for extract 3.
6) Titrate above extract against 0.1 NaOH in order to find out % Recovery.
Observations Table:
Sr. Volume of NaOH Conc. Salt extracted(gm) Recovery Cumulative
No Extract(ml) Required( gms/lit % Recovery
ml)
1
Calculations
NaOH v/s Oxalic Aid (Extract)
N1 V1 = N2 V2
1) 1*23.5= N2*10= 2.35
Calculate normality of oxalic acid (N2)
Conc. in gms/lit = N × Eq wt.
Oxalic acid recovered in 100ml.
% Recovery = Oxalic acid recovered × 100
Oxalic acid in feed
Result:-
% recovery of solute by multistage cross current Extraction is found to be :
1st Stage _ _ _ %
2nd Stage _ _ _ _ _ %
3rd Stage _ _ _ _ _ %
R1 R2 R3 R4 Total Signature
(5 Marks) (5 Marks) (5 Marks) (5 Marks) (20Marks)
