Aluminum ions determined by titrating with EDTA:
Al^3+^ + H_2_Y^2-^ >>> AlY^-^ + 2H^+^
1.00 g sample requires 20.5 ml EDTA for titration. the EDTA was standardized by titrating 25.0 mL of a 0.1 M CaCl2 solution, requiring 30.0mL EDTA. Calculate the percent Al2O3 in the sample. I want also to know how we can deduce the relation between Al and EDTA, AL2O3 and EDTA? And why the three Oxygen molecules are not involved in the reaction written above?
Thank you in advance.
EDTA reacts with 1:1 ratio with the Al ion.
Thus the equations are Al+EDTA=Al-EDTA (1:1 RATIO of AL:EDTA)
Al2O3+2EDTA=2 Al-EDTA
1M Al2O3 reacts with 2M of EDTA
Thus 1M Al2O3=0.5M EDTA. The Equilibrium moles of the related Al and EDTA at the equilibrium is shown above .
EDTA was standarised by using 25.0 mL of 0.1M CaCl2 solution.
We also know that V1*S1 (for EDTA) = V2*S2 (for Ca)
30.0 mL* S1=25mL* 0.1000 M
S1( strength of EDTA) = 0.833 M.
Now, 1M EDTA=2M Al2O3.
1M Al2O3=0.5M EDTA.
Let the ampont of Al present is x mg. and the molecular weight of Al2O3 is 101.96 gm/mole.
xmg/101.96 mg/mMole (Moles of Al2O3)=0.5*20.5*0.833 (Moles of EDTA)
x= 87.1 mg.
% of AL in AL2O3=(87.1/1000) 8 100
=8.71 % AL content
The oxygen molecules are not related to the Al-EDTA equilibrium in terms of the no. of moles reacting as shown in the reactions above. Thus they wont effect the reaction.
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