Aluminum ions solid by titrating with EDTA:
Al^3+^ + H_2_Y^2-^ >>> AlY^-^ + 2H^+^
1.00 g illustration requires 20.5 ml EDTA control titration. the EDTA was standardized by titrating 25.0 mL of a 0.1 M CaCl2 disentanglement, requiring 30.0mL EDTA. Calculate the percent Al2O3 in the illustration. I failure too to perceive how we can infer the homogeneity among Al and EDTA, AL2O3 and EDTA? And why the three Oxygen molecules are referable attributable attributable attributable confused in the reaction written over?
Thank you in pace.
EDTA reacts with 1:1 connection with the Al ion.
Thus the equations are Al+EDTA=Al-EDTA (1:1 RATIO of AL:EDTA)
1M Al2O3 reacts with 2M of EDTA
Thus 1M Al2O3=0.5M EDTA. The Makeconsequence moles of the cognate Al and EDTA at the makeconsequence is shown over .
EDTA was standarised by using 25.0 mL of 0.1M CaCl2 disentanglement.
We too perceive that V1*S1 (control EDTA) = V2*S2 (control Ca)
30.0 mL* S1=25mL* 0.1000 M
S1( power of EDTA) = 0.833 M.
Now, 1M EDTA=2M Al2O3.
1M Al2O3=0.5M EDTA.
Let the ampont of Al give is x mg. and the molecular consequence 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 referable attributable attributable attributable cognate to the Al-EDTA makeconsequence in stipulations of the no. of moles reacting as shown in the reactions over. Thus they rule pi the reaction.