Answer to Question #158254 in Chemistry for Dexther Villanueva

Question

A sample of pure CaCO3 weighing 0.3677g is dissolved in hydrochloric acid and the solution diluted to 250.0ml in a volumetric flask. A 25.00ml aliquot requires 30.26ml of an EDTA solution for titration. Calculate a) the molarity of the EDTA solution; b) the number of grams Na2H2Y•2H2O (FW = 372.2) required to prepare 500.0ml of the solution. 

Answer 1

Solution:

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⁺

PART A:

Find the molarity of Ca²⁺:

(0.3677 g CaCO₃)×(1 mol CaCO₃/100.09 g CaCO₃)×(1 mol Ca²⁺/1 mol CaCO₃)= 0.003674 mol Ca²⁺ ions

Molarity of Ca²⁺ = Moles of Ca²⁺ / Volume solution = 0.003674 mol / 0.250 L = 0.01469 M Ca²⁺ ions

Molarity of Ca²⁺ = 0.01469 M Ca²⁺ ions

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 molarity of EDTA:

Moles of EDTA = (0.01469 mol Ca²⁺ / L) × (0.025 L) × (1 mol H₂Y^2- / 1 mol Ca²⁺) = 0.0003673 mol EDTA

Molarity of EDTA = Moles of EDTA / Volume of EDTA = 0.0003673 mol / 0.03026 L = 0.01214 M EDTA

Molarity of EDTA = 0.01214 M

PART B:

Molarity of EDTA = 0.01214 M

Molarity of EDTA = Moles of EDTA / Volume solution

Hence,

Moles of EDTA = (0.01214 mol EDTA / L) × (0.500 L / 1) = 0.00607 mol EDTA

Moles of EDTA = Mass of EDTA / Molar mass of EDTA

Hence,

Mass of EDTA = (0.00607 mol EDTA) × (372.2 g EDTA / 1 mol) = 2.2593 g

Mass of EDTA = 2.2593 g

Answer:

(a): The molarity of the EDTA solution is 0.01214 M

(b): 2.2593 grams of EDTA required