Question

A 3.7145-g portion of disodium EDTA dihydrate (M.W. 372.24) was dissolved in enough distilled water to prepare 1.000 liter of solution. A 100.0-mL water sample was adjusted to pH 10 and titrated to endpoint with 11.23 mL of the EDTA solution. Calculate the concentration of the EDTA solution and th e hardness of the water sample. Express the hardness of mg caCO3 /L.

Answer 1

Molarity of EDTA solution

Molarity = weight of EDTA*1/molar mass of EDTA *volume of water (lit)

Molarity = 3.7145/372.25

= 0.00997M

We know that

M1V1 = M2V2

M1 = Molarity of water sample = ?M

V1= volume of water sample = 100 ml

M2 = Molarity of EDTA solution = 0.0099M

V2 = volume of EDTA solution for titration = 11.23 ml

M1 = M2V2/V1

= 0.00997*11.23/100 M

= 0.001119 M

Molarity of water sample (M1) = 0.001119 M

Hardness of water sample = M1*100*1000 ppm

= 0.001119*100*1000 ppm

= 111.9 ppm

Hardness of the given water sample = 111.9 ppm (or) 111.9 mg/lit.