The density of the sample is 0.9977 g/mL, and the molar mass of calcium carbonate is 100.0 g/mol.
Calculate the mass of calcium carbonate present in a 50.00 mL sample of an aqueous calcium carbonate standard, assuming the standard is known to have a hardness of 75.0 ppm. If this wsa titrated with a 0.00501 M EDTA, what volume of EDTA solution would be needed to reach the endpoint?
(using the given density of this problem is throwing me off - how is this applied?)
Volume of solution = 50ml density of the sample =0.9997 g/ml
Mass of sample= 50*0.9997= 49,985 gms
sample contains 75ppm that is 75 gm per 106 gms
106 gms contain 75 gms
49,985 gms contains 49,985*75/106 =0.003749 gms
moles of CaCO3 in the sample= 0.003749/100=3.749*10-5
the reaction between CaCO3 and EDTA is
Ca+2+ EDTA4- ----> Ca(EDTA)2-
1 mole of Ca+2 requires 1 mole of EDTA ( 1 mole of CaCO3---> Ca+2 +CO3-2)
Mole of EDTA =3.749*10-5
volume of EDTA= 3.749*10-5/ 0.00501 L=0.00748 L=7.48 ml
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