Question

1. A 3.7145g of disodium EDTA dehydrate (MW 372.24) was dissolved in enough distilled water to prepare 1.000-liter of solution. A 100 mL Ca sample was adjusted to pH 10 and titrated to the Eriochrome Black T end point with 11.23 mL of the EDTA solution. Calculate the concentration of the EDTA solution and the mmol Ca2+ in the sample.

Answer 1

Mass of disodium EDTA dihydrate = 3.7145 g

Moles of disodium EDTA dihydrate = 3.7145 g / 372.24 g/mol

= 0.00997 moles

Concentration of disodium EDTA dihydrate solution = 0.00997 moles / 0.100 L

= 0.0997 M

At pH 10:

Volume of disodium EDTA dihydrate solution taken = 11.23 mL

Moles of disodium EDTA dihydrate solution taken =  0.0997 M * 0.01123

= 0.00111 moles

We know that at pH 10 Ca2+ are titrated.

Moles of (Ca2+) at equivalence point = Moles of titrant.

Moles of (Ca2+) = 0.00111 mole or 1.11 mmol