Question

A 0.3205g sample of CaCO3 was dissolved in HCl and the resulting solution diluted to 250.ml in a volumetric flask. A 25 mL sample of the solution required 18.75mL of EDTA solution for titration to the Eriochrome Black T end point. What is the concentration of Ca2+ (mol/L) in the 250.0ml of CaCl2 solution? How many mole of Ca2+ are contained in a 250mL sample? How many mole of EDTA are contained in the 18.75mL used for titration? What is the concentration (mol/L) of the EDTA solution?

Answer 1

i) Let us calculate the moles of CaCO3 first

Moles = Mass / molecular weight = 0.3205 / 100 = 3.205 X 10^-3 moles

These moles were dissolved in 250mL (0.25 L) of water

ii) So molarity of CaCO3 = Moles / Volume of water = 3.205 X10^-3 / 0.25 = 12.82 X10^-3 M

iii) Molarity of Ca+2 = 12.82 X10^-3 M
iv) Moles of Ca+2 in 250mL = Moles of CaCO3 = 3.205 X10^-3 moles

v) the moles of Ca+2 in 25mL = Volume X initital concentration = 25 X 3.205 X10^-3 / 250 = 3.205 X10^-4 moles

vi) moles of EDTa used = moles of Ca+2 present = 3.205 X10^-4

vii) concentration = Moles / Volume = 3.205 X10^-4 X 1000 / 18.75 = 0.01709 Molar