HELP!!!! I\'LL GIVE ALL POINTS!! PLEASE SHOW ALL STEPS PreLab Questions: A Ca^2+
ID: 808776 • Letter: H
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HELP!!!! I'LL GIVE ALL POINTS!! PLEASE SHOW ALL STEPS
PreLab Questions: A Ca^2+ solution is prepared by dissolving 0.2085 g of CaCO3 in a 250.00 volumetric flask. A 1.00 mL aliquot of this solution is pipetted into a flask and titrated using 0.75 mL of an EDTA solution. What is the molarity of the EDTA solution? A 0.0985 g sample of unknown calcium carbonate is dissolved and transferred to a 100.00 mL volumetric flask. A 1.00 mL aliquot of this solution is titrated using 0.77 mL of 0.01015 M EDTA. Calculate the percent CaO in the sample.Explanation / Answer
1. Strength of EDTA
no. of moles of CaCO3 = 0,2085g/100.0869g/mol = 0.002083 mole ( using the formula to calculate the number of moles) which is dissolved in 250ml . Thus the molarity of CaCO3 solution is= 0.02083mol/0.25L=0.00833 M.
Now, the solution is titrated using EDTA solution we have to calcualte the molarity of EDTA solution. Using the formula V1S1=V2S2, where V1= volume of aliquot taken= 1ml , V2= Vol of EDTA= 0.75ml, S1= Molarity of CaCO3 solution =0.00833M, S2= Molarity of EDTA solution which is asked. Thus, S2= (1*0.00833/0.75)M = 0.011M
2. %of CaO in the sample = (Vol. of EDTA X 56.08 X Molarity of EDTA X100X 100)/( 1000 X weight of the aliquot taken) where vol of EDTA is in ml and weight of the aliquot take is in g.
thus, substituting the respective value we get.
% of Cao in sample= (0.77 X 56.08 X 0.01015 X100X 100)/( 1000X 0.0985) = 44%
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