Answer to Question #171732 in Physical Chemistry for Akshat

Question

0.28 g of CaCO3 was dissolved in dil. HCl and the solution was made up to 1000 mL with distilled water. 100 mL of the above solution required 28 mL of EDTA solution on titration. 100 mL of the hard water sample required 33 mL of the same EDTA solution on titration. After boiling, 100 mL the sample required 10 mL of EDTA solution. Solve each type of hardness in ppm

Answer 1

M1V1= M2V2.

1×100= M×28

M= 3.57M

M2V2= M3V3

3.57×33= 10×M3

M3= 11.7

Mass= molarity×volume/molar mass

Mass= 11.7×10/100

Mass= 117gm