Question

?. C3) A 100.00 ml. water sample was titrated 20.50 mL of 0.150 M EDTA after pH adjustment and using Eriochrome black T indicator. Knowing that density of water sample equals to 1.10 g/mL. (Caco 00.09 amu). Calculate the total hardness of water as Cacos and accordingly classify

Answer 1

Step1: Calculate the moles of EDTA required To complex with a and ions in water

Concentration of EDTA = 0.15 M

Volume of EDTA Solution = 20.50 ml = 20.50 × L

Moles of EDTA = Concentration of EDTA × Volume of EDTA Solution

= 0.15 × 20.50× =0.003075 mol

Step2: Equation for reaction between EDTA and Ca^2+ and Mg^2+

   +

Step3: Calculate the moles of ( = Ca^2+ + Ma^2+) present in the 100 ml of water sample

From the above equation 1 moles of M^2+ reacts with 1 moles of EDTA

So 0.003075 mol of EDTA reacts with 0.003075 moles M^2+.

Step4: Calculate the concentration of M^2+

Concentration of M^2+ = moles of M^2+ ÷ Volume

Moles of M^2+ = 0.003075 mol

Volume = 100 ml = 100 x10^-3 L

Concentration of M^2+ = 0.003075 ÷ 100 × 10^-3

= 0.03075 M

= 0.03075×100 g/ L of Caco3

= 0.03075× 100× 1000 ppm of Caco3

= 3075 ppm . Caco3

The total hardness of water as caco3 and Mgco3 is 0.03075 M