Aluminum ions strong by titrating with EDTA:
Al^3+^ + H_2_Y^2-^ >>> AlY^-^ + 2H^+^
1.00 g pattern requires 20.5 ml EDTA ce titration. the EDTA was standardized by titrating 25.0 mL of a 0.1 M CaCl2 key, requiring 30.0mL EDTA. Calculate the percent Al2O3 in the pattern. I neglect to-boot to distinguish how we can draw the narration betwixt Al and EDTA, AL2O3 and EDTA? And why the three Oxygen molecules are referable confused in the reaction written over?
Thank you in progression.
EDTA reacts with 1:1 pertinency 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 Makeweight moles of the allied Al and EDTA at the makeweight is shown over .
EDTA was standarised by using 25.0 mL of 0.1M CaCl2 key.
We to-boot distinguish that V1*S1 (ce EDTA) = V2*S2 (ce 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 gravity 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 allied to the Al-EDTA makeweight in stipulations of the no. of moles reacting as shown in the reactions over. Thus they custom result the reaction.