Answer to Question #158256 in Chemistry for Dexther Villanueva

Question

A sample of pure CaCO3 weighing 0.2428g is dissolved in HCl and the solution diluted to 250ml in a volumetric flask. A 50.00ml aliquot requires 42.74ml of an EDTA solution for titration. A 200ml sample of water containing Ca+2 is titrated with 16.38ml of the EDTA solution. Calculate the degree of hardness of water in ppm CaCO3. 

Answer 1

Solution:

1)

The reaction between CaCO₃ and hydrochloric acid can be described using the following equation:

CaCO₃(s) + 2HCl(aq) → Ca²⁺(aq) + 2Cl^-(aq) + H₂O(l) + CO₂(g)

According to the equation: Moles of CaCO₃ = Moles of Ca²⁺

Find the Moles of Ca²⁺:

(0.2428 g CaCO₃) × (1 mol CaCO₃/100.09 g CaCO₃) × (1 mol Ca²⁺/1 mol CaCO₃) = 0.002426 mol Ca²⁺

Moles of Ca²⁺ = 0.002426 mol Ca²⁺

Find the Molarity of Ca²⁺:

Molarity of Ca²⁺ = Moles of Ca²⁺ / Volume solution = 0.002426 mol / 0.250 L = 0.0097 M Ca²⁺ions

Molarity of Ca²⁺ = 0.0097 M Ca²⁺ ions

2)

When calcium ions (Ca²⁺) are titrated with EDTA a relatively stable calcium complex is formed:

Ca²⁺ + H₂Y^2- = CaY^2- + 2H⁺

According to the equation: Moles of Ca²⁺ = Moles of H₂Y^2- (EDTA)

Find the Moles of EDTA:

Moles of EDTA = (0.0097 mol Ca²⁺ / L) × (0.050 L) × (1 mol H₂Y^2- / 1 mol Ca²⁺) = 0.000485 mol EDTA

Moles of EDTA = 0.000485 mol EDTA

Find the molarity of EDTA:

Molarity of EDTA = Moles of EDTA / Volume of EDTA = 0.000485 mol / 0.04274 L = 0.01135 M EDTA

Molarity of EDTA = 0.01135 M

3)

The hardness of water is calculated in terms of the weight of CaCO₃ (1 ppm = 1 mg/L).

When calcium ions (Ca²⁺) are titrated with EDTA a relatively stable calcium complex is formed:

Ca²⁺ + H₂Y^2- = CaY^2- + 2H⁺

According to the equation: Moles of Ca²⁺ = Moles of H₂Y^2- (EDTA)

Find the Moles of Ca²⁺:

Moles of Ca²⁺ = (0.01135 mol EDTA/L) × (0.01638 L/1) × (1 mol Ca²⁺/1 mol EDTA) = 0.000186 mol Ca²⁺

Moles of Ca²⁺ = 0.000186 mol Ca²⁺

Find the Mass of CaCO₃:

Moles of CaCO₃ = Moles of Ca²⁺ = 0.000186 mol

(0.000186 mol Ca²⁺) × (1 mol CaCO₃/1 mol Ca²⁺) × (100.09 g CaCO₃/1 mol CaCO₃) = 0.01862 g CaCO₃

Mass of CaCO₃ = 0.01862 g

Find the ppm concentration CaCO₃:

ppm CaCO₃ = mg CaCO₃ / L solution

ppm CaCO₃ = (0.01862 g CaCO₃) × (1000 mg CaCO₃/1g CaCO₃) × (1/0.200 L solution) = 93.1

ppm CaCO₃ = 93.1

Answer: the degree of hardness of water is 93.1 ppm CaCO₃