Question

Co. A 50.0 mL aliquot of water (assume contains only Ca?") required 4.08 ml. of o.0100 M EDTA to titrate all of the Cr" cation present what is the ppm concentration of Ca2+ (molar mass: 400 g/mol).

Answer 1

Molarity of EDTA = 0.0100 M

Volume of EDTA = 4.08 mL

Moles of EDTA = Molarity * Volume = 0.0100M * 4.08 mL = 0.0408 mmol

1 mol of EDTA reacts with 1 mol of Ca²⁺

So, moles of Ca²⁺ = 0.0408 mmol

Volume of water = 50.0 mL

Molarity of Ca²⁺ = moles/ Volume = 0.0408 mmol/ 50 mL = 0.000816 M

i.e. 0.000816 mol of Ca²⁺ are present in 1L of solution

0.000816 M = (0.000816 mol/1L)*(40g/mol)* (10³ mg/g) = 32.64 ppm

So, ppm concentration of Ca²⁺ = 32.64