Answer to Question #163869 in Chemistry for clay

Question #163869

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?


1
Expert's answer
2021-02-15T05:26:05-0500

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 = H2Y2-

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


MgSO→ Mg2+ + SO42-

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+ + H2Y2- → MgY2- + 2H+

According to the equation above: n(Mg2+) = n(H2Y2-)

n(H2Y2-) = n(Mg2+) = 0.00037385 mol

Moles of H2Y2- = Molarity of H2Y2- × Volume of H2Y2-

Molarity of H2Y2- = Moles of H2Y2- / Volume of H2Y2- = 0.00037385 mol / 0.03760 L = 0.009943 M

Molarity of H2Y2- = 0.009943 M


calcium carbonate: CaCO3

The molar mass of CaCO3 is 100.09 g mol-1

CaCO3 → Ca2+ + CO32-

Ca2+ + H2Y2- → CaY2- + 2H+

According to the equations above: n(H2Y2-) = n(Ca2+) = n(CaCO3)

n(H2Y2-) = n(CaCO3)

Molarity of H2Y2- × Volume of H2Y2- = 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 CaCOwill react with 1.00 mL of this EDTA solution.

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