Question

Zn²⁺ and Ni²⁺ both react quantitatively with EDTA (Y^4-)

Zn²⁺ + Y^4- → ZnY^2-
Ni²⁺ + Y^4- → NiY^2-

The reagent 2,3-dimercapto-1-propanol (Z) reacts with the zinc-EDTA complex to liberate EDTA but it does not react with the nickel-EDTA complex.

ZnY^2- + Z → ZnZ²⁺ + Y^4-
NiY^2- + Z → N.R.

50.00 mL of a sample solution that contains Ni²⁺ and Zn²⁺ are mixed with 50.00 mL of 0.01981 M EDTA. The excess EDTA is titrated with 0.02010 M Mg²⁺, requiring 14.78 mL. Then, 15 mL of 0.10 M 2,3-dimercapto-1-propanol are added. The resulting solution is titrated again with 0.02010 M Mg²⁺, requiring 18.92 mL. Calculate the molarity of Zn²⁺ and Ni²⁺ in the sample solution.
[Zn²⁺] =    mol/L
[Ni²⁺] =   mol/L

Answer 1

[Zn2+] = 0.007606 mol/L

[Ni2+] = 0.006262 mol/L

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Total EDTA added,

No. of moles of EDTA in 50.00 mL, 0.01981 M EDTA = 0.01981 M x 50 mL/1000 = 0.0009905 mol.

Unreacted EDTA,

No. of moles of Excess EDTA, 14.78 mL, 0.02010 M = 0.02010 M x 14.78 mL/1000 = 0.0002971 mol

Reacted EDTA = total Zn2+ and Ni2+,

No. of moles of EDTA reacted with Ni2+ and Zn2+ = 0.0009905 - 0.0002971 = 0.0006934 mol

Total No. of moles of Ni2+ and Zn2+ ions = 0.0006934 mol

Liberated Zn2+ ion on adding 2,3-dimercapto-1-propanol,

No. of moles of Zn2+ ion (after adding 2,3-dimercapto-1-propanol) = 18.92 mL. 0.02010 M Mg2+ ions.

No. of moles of Zn2+ ion = 0.02010 M x 18.92 mL/1000 = 0.0003803 mol

No. of moles of Ni2+ ions = 0.0006934 - 0.0003803 = 0.0003131 mol.

Volume of sample solution = 50.00 mL = 0.050 L

Molarity of Zn2+ ions in sample solution = 0.0003803 mol/0.050 L = 0.007606 M

Molarity of Ni2+ ions in sample solution = 0.0003131 mol/0.050 L = 0.006262 M