Answer to Question #163869 in Chemistry for clay

Question

A 50.00 mL aliquot solution containing 0.4500 g of magnesium sulfate in 0.500 L required 37.60 mL of EDTA solution for titration. How many milligrams of Calcium carbonate will react with 1.00 mL of the same EDTA solution?

Answer 1

Solution:

Short form of solution:

EDTA forms 1:1 complexes with metals.

M + EDTA → M*EDTA

Hence,

50 mL × (0.450 g MgSO₄ / 0.5 L) × (1 mol MgSO₄ / 120.37 g MgSO₄) × (1 mol Mg²⁺/ 1 mol MgSO₄) ×

× (1 mol EDTA / 1 mol Mg²⁺) = 0.374 mmol EDTA

(0.374 mmol EDTA / 37.60 mL EDTA) × (1 mol Ca²⁺ / 1 mol EDTA) × (1 mol CaCO₃ / 1 mol Ca²⁺) ×

× (100.09 g CaCO₃ / 1 mol CaCO₃) = 0.995 mg CaCO₃ / mL

0.995 mg CaCO₃ / 1.00 mL EDTA

Full form of solution:

EDTA = H₂Y^2-

magnesium sulfate: MgSO₄

The molar mass of MgSO₄ is 120.37 g mol^-1

Moles of MgSO₄ = Mass of MgSO₄/Molar mass of MgSO₄ = 0.4500 g / 120.37 g mol^-1 = 0.0037385 mol

n(MgSO₄) = 0.0037385 mol

MgSO₄ → Mg²⁺ + SO₄^2-

According to the equation above: n(MgSO₄) = n(Mg²⁺)

n(Mg²⁺) = n(MgSO₄) = 0.0037385 mol (in 0.5 L of solution)

n(Mg²⁺) = 0.0037385 mol × (0.050 L / 0.500L) = 0.00037385 (in 50.00 mL of aliquot solution)

Mg²⁺ + H₂Y^2- → MgY^2- + 2H⁺

According to the equation above: n(Mg²⁺) = n(H₂Y^2-)

n(H₂Y^2-) = n(Mg²⁺) = 0.00037385 mol

Moles of H₂Y^2- = Molarity of H₂Y^2- × Volume of H₂Y^2-

Molarity of H₂Y^2- = Moles of H₂Y^2- / Volume of H₂Y^2- = 0.00037385 mol / 0.03760 L = 0.009943 M

Molarity of H₂Y^2- = 0.009943 M

calcium carbonate: CaCO₃

The molar mass of CaCO₃ is 100.09 g mol^-1

CaCO₃ → Ca²⁺ + CO₃^2-

Ca²⁺ + H₂Y^2- → CaY^2- + 2H⁺

According to the equations above: n(H₂Y^2-) = n(Ca²⁺) = n(CaCO₃)

n(H₂Y^2-) = n(CaCO₃)

Molarity of H₂Y^2- × Volume of H₂Y^2- = Mass of CaCO₃ / Molar mass of CaCO₃

0.009943 M × 0.0010 L = Mass of CaCO₃ / 100.09 g mol^-1

Mass of CaCO₃ = 0.009943 M × 0.0010 L × 100.09 g mol^-1 = 0.0009952 g = 0.9952 mg

Mass of CaCO₃ = 0.9952 mg CaCO₃

Answer: 0.9952 milligrams of CaCO₃ will react with 1.00 mL of this EDTA solution.