# Answer to Question #151144 in Inorganic Chemistry for SANTHOSH

Question #151144
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.
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2020-12-15T07:51:51-0500

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

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