A 3.7145 g portion of disodium EDTA dihydrate (M.W. 372.24) was dissolved in eno

Question

A 3.7145 g portion of disodium EDTA dihydrate (M.W. 372.24) was dissolved in enough distilled water of prepare 1.000 L of solution. A 100.0 mL water sample was adjusted to pH 10 and titrated to the Eriochrome Black T end point with 11.23 mL of the EDTA solution. When the pH of another 100.0 mL water sample was adjusted to pH 12 and titrated with EDTA, 2.11 mL of titrant was required. Calculate the concentration of Ca(II) and Mg(II) in the water. Express the answers in mg/L of CaCO3 and MgCO3, respectively.

Explanation / Answer

Mass of disodium EDTA dihydrate = 3.7145 g

Moles of disodium EDTA dihydrate = 3.7145 g / 372.24 g/mol

= 0.00997 moles

Concentration of disodium EDTA dihydrate solution = 0.00997 moles / 0.100 L

= 0.0997 M

At pH 10:

Volume of disodium EDTA dihydrate solution taken = 11.23 mL

Moles of disodium EDTA dihydrate solution taken =  0.0997 M * 0.01123

= 0.00111 moles

We know that at pH 10 Ca2+ and Mg2+ both are titrated.

Moles of (Ca2+ + Mg2+) at equivalence point = Moles of titrant.

Moles of (Ca2+ + Mg2+) = 0.00111 moles ..........(1)

At pH = 12

Volume of disodium EDTA dihydrate solution taken = 2.11 mL

Moles of disodium EDTA dihydrate solution taken =  0.0997 M * 0.00211

= 0.00021 moles

We know that at pH 12 only Ca2+ is titrated but not Mg2+.

Moles of Ca2+ at equivalence point = Moles of titrant.

Moles of Ca2+ = 0.00021

Thus moles of Mg2+ = 0.00111 - 0.00021

= 0.0009 moles

Mass of CaCO3 = 0.00021 * 100

= 0.021 g = 21 mg

Volume of water = 100 mL = 0.10 L

[Ca2+] = 21/0.10 = 210 mg/L of CaCO3

Mass of MgCO3 = 0.0009 * 84.3139

= 0.0758 g = 75.8 mg

[Mg2+] = 75.8 / 0.10

= 758 mg/L of MgCO3

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