Answer to Question #286976 in Chemistry for Garry

Question

Calculate the volume of 0.0500 M EDTA needed to titrate the Ca and Mg in a 0.1612-g sample that is 92.5% dolomite, CaCO 3 ·MgCO 3

(184.4 g/mol)

Answer 1

Solution:

Calculate the mass of CaCO₃ ·MgCO₃ in a sample:

Mass of CaCO₃ ·MgCO₃ = Mass of a sample × %CaCO₃ ·MgCO₃

Mass of CaCO₃ ·MgCO₃ = (0.1612 g × 0.925) = 0.14911 g

Calculate the moles of CaCO₃ ·MgCO₃:

The molar mass of CaCO₃ ·MgCO₃ is 184.4 g/mol

Hence,

Moles of CaCO₃ ·MgCO₃ = (0.14911 g) × (1 mol / 184.4 g) = 0.0008086 mol

CaCO₃ ·MgCO₃ → Ca²⁺ + Mg²⁺ + CO₃^2-

According to stoichiometry:

1 mol of CaCO₃ ·MgCO₃ contain 1 mol of Ca²⁺ and 1 mol of Mg²⁺

Hence,

0.0008086 mol of CaCO₃ ·MgCO₃ contain 0.0008086 mol of Ca²⁺ and 0.0008086 mol of Mg²⁺

When calcium ions (Ca²⁺) are titrated with EDTA (H₂Y^2-) a relatively stable calcium complex is formed:

Ca²⁺ + H₂Y^2- = CaY^2- + 2H⁺

According to stoichiometry:

1 mol of Ca²⁺ reacts with 1 mol EDTA (H₂Y^2-)

Hence,

0.0008086 mol of Ca²⁺ reacts with 0.0008086 mol EDTA (H₂Y^2-)

Moles 1 of EDTA = 0.0008086 mol

When calcium ions (Mg²⁺) are titrated with EDTA (H₂Y^2-) a relatively stable magnesium complex is formed:

Mg²⁺ + H₂Y^2- = MgY^2- + 2H⁺

According to stoichiometry:

1 mol of Mg²⁺ reacts with 1 mol EDTA (H₂Y^2-)

Hence,

0.0008086 mol of Mg²⁺ reacts with 0.0008086 mol EDTA (H₂Y^2-)

Moles 2 of EDTA = 0.0008086 mol

Calculate the moles of EDTA needed to titrate the Ca²⁺ and Mg²⁺:

Moles of EDTA = Moles 1 of EDTA + Moles 2 of EDTA

Moles of EDTA = 0.0008086 mol + 0.0008086 mol = 0.0016172 mol

Calculate the volume of EDTA solution:

Volume of EDTA = Moles of EDTA / Molarity of EDTA

Volume of EDTA = (0.0016172 mol) / (0.0500 M) = 0.032344 L = 32.34 mL

Volume of EDTA = 32.34 mL

Answer: The volume of 0.0500 M EDTA needed to titrate the Ca²⁺ and Mg²⁺ is 32.34 mL