In: Chemistry
1) A soil sample of 2.0134g was extracted, transferred to a 100.00ml volumetric flask. A 10.00mL aliquot was titrated with EDTA of molarity 0.002052M and the volume of EDTANa2 was found to be 23.82mL. Calculate a) the number of moles of EDTANa2 titrated
b) the corresponding total number of moles of (calcium and magnesium combined) in the titrec
c) the concentration of (calcium and magnesium) combined) in the soil sample of units of moles/g
d) the above concentration in units of moles/kg
Thank you!
Solution :-
a) the number of moles of EDTANa2 titrated
Solution :-
Moles of EDTA = molarity * volume in liter
= 0.002052 mol per L * 0.02382 L
= 4.89*10^-5 mol EDTA
b) the corresponding total number of moles of (calcium and magnesium combined) in the titrec
Solution :- 1 mol EDTA = 1 mol Ca2+ or Mg2+
So the moles of combined Ca2+ or Mg2+ are same as moles of EDTA
Therefore combined moles of Ca2+ and Mg2+ = 4.89*10^-5 mol
c) the concentration of (calcium and magnesium) combined) in the soil sample of units of moles/g
Solution :- concentration of calcium and magnesium combined = moles / mass of soil
= 4.89*10^-5 mol /2.0134 g
= 2.43*10^-5 mol /g
d) the above concentration in units of moles/kg
Solution :-
Lets find concentration is moles per kg
(2.43*10^-5 mol /g )* (1000 g / 1 kg) = 2.43*10^-2 mol per kg
So the concetration is 2.43*10^-2 mol / kg