Answer to Question #163750 in Organic Chemistry for Wincy Mispa

Question

0.2 g of CaCO3 was dissolved in 500 mL of water. 100 ml of this solution on titration required 20 ml of EDTA. 100 ml of an unknown hard water sample with total hardness 600 mg/L was titrated against same EDTA solution. What is the expected end – point?

Answer 1

During the course of titration against EDTA solution, EDTA combines with M. +2 ... 50 ml of standard hard water = V1 ml of EDTA. ... Solution: 500ml of SHW = 0.5g or 500mg CaCO3 eq.