Question

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?)

Answer 1

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+ EDTA^4- ----> 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