Answer to Question #151144 in Inorganic Chemistry for SANTHOSH

Question

250 mg of caco3 was dissolved in hcl and the solution was made 250 ml with distilled water. 50 ml of the above solution required 20 ml of edta solution [standardization part]. 50 ml of hard water sample consumed 25 ml of edta with ebt as indicator [total hardness]. calculate the total hardness of water sample in ppm.

Answer 1

1. Standard calcium solution.

(0.250 g CaCO₃) (1 mol CaCO₃ / 100.09 g CaCO₃)(1 mol Ca²⁺ / 1 mol CaCO₃) = 0.00250 mol Ca²⁺ ions

Molarity of Ca²⁺ = 0.00250 mol Ca²⁺ / 0.250 L = 0.01 M Ca²⁺ ions

2. Standardization of EDTA.

(0.01 mol Ca²⁺ / L)(0.050 L / 1)(1 mol EDTA / 1 mol Ca²⁺) = 0.0005 mol EDTA

Molarity of EDTA = 0.0005 mol EDTA / 0.020 L = 0.025 M EDTA

3. Ca²⁺ concentration in an unknown solution CaCO₃ reported as ppm CaCO₃.

(0.025 mol EDTA / L)(0.025 L / 1)(1 mol Ca²⁺ / 1 mol EDTA) = 0.000625 mol Ca²⁺

0.000625 mol Ca²⁺ (1 mol CaCO₃ / 1 mol Ca²⁺)(100.09 g CaCO₃ / 1 mol CaCO₃) = 0.062556 g CaCO₃

ppm CaCO₃ = mg CaCO₃ / L Solution = 0.062556 g CaCO₃ (1000 mg CaCO₃ / 1.00 g CaCO₃) (1 / 0.050 L Solution) = 1251

1251 ppm CaCO₃