Question

A 45.20 mL aliquot from a 0.500 L solution that contains 0.420 g of MnSO 4 ( MW = 151.00 g/mol) required 35.3 mL of an EDTA solution to reach the end point in a titration. What mass, in milligrams, of CaCO 3 ( MW = 100.09 g/mol) will react with 1.40 mL of the EDTA solution? in mg

Answer 1

mass CaCO₃ = 0.998 mg

Explanation

Mass MnSO₄ = 0.420 g

moles MnSO₄ = (mass MnSO₄) / (molar mass MnSO₄)

moles MnSO₄ = (0.420 g) / (151.00 g/mol)

moles MnSO₄ = 2.78 x 10^-3 mol

concentration MnSO₄ = (moles MnSO₄) / (total volume in Liter)

concentration MnSO₄ = (2.78 x 10^-3 mol) / (0.500 L)

concentration MnSO₄ = 5.56 x 10^-3 M

volume of aliquot = 45.20 mL = 0.04520 L

moles MnSO₄ in aliquot = (concentration MnSO₄) * (volume of aliquot in Liter)

moles MnSO₄ in aliquot = (5.56 x 10^-3 M) * (0.04520 L)

moles MnSO₄ in aliquot = 2.51 x 10^-4 mol

moles EDTA used = moles MnSO₄ in aliquot

moles EDTA used = 2.51 x 10^-4 mol

volume EDTA used = 35.3 mL = 0.0353 L

concentration EDTA = (moles EDTA used) / (volume EDTA used in Liter)

concentration EDTA = (2.51 x 10^-4 mol) / (0.0353 L)

concentration EDTA = 7.12 x 10^-3 M

Volume EDTA reacted with CaCO₃ = 1.40 mL = 0.00140 L

moles EDTA reacted with CaCO₃ = (concentration EDTA) * (Volume EDTA reacted with CaCO₃)

moles EDTA reacted with CaCO₃ = (7.12 x 10^-3 M) * (0.00140 L)

moles EDTA reacted with CaCO₃ = 9.97 x 10^-6 mol

moles CaCO₃ = moles EDTA reacted with CaCO₃

moles CaCO₃ = 9.97 x 10^-6 mol

mass CaCO₃ = (moles CaCO₃) * (molar mass CaCO₃)

mass CaCO₃ = (9.97 x 10^-6 mol) * (100.09 g/mol)

mass CaCO₃ = 9.98 x 10^-4 g

mass CaCO₃ = 0.998 mg