Question

A 50.0-mL solution containing Ni2+ and Zn2+ was treated with 25.0 mL of 0.0452 M EDTA to bind all the metal. The excess unreated EDTA required 12.4 mL of 0.0123 M Mg2+ for complete reaction. An excess of the reagent 2,3-dimercapto-1-propanol was then added to displace the EDTA from zinc. Another 29.2 mL of Mg2+ was required for reaction with the liberated EDTA. Calculate the molarity of Ni2+ and Zn2+ in the original solution

Answer 1

Molarity of EDTA added = 0.0452M = 0.0452 mol/L

Volume added= 25 mL = 0.025 L

Moles of EDTA =Molarity * Volume = 0.0452 M*0.025 L =0.00113 moles

EDTA in excess = moles of Mg²⁺ required

As 1 mol of EDTA reacts with 1mol of Mg²⁺

Moles of Mg²⁺ = 0.0123mol/L * 0.0124 = 0.00015 mol = EDTA in excess

Thus, Moles of EDTA that reacts with Zn²⁺ and Ni²⁺ = Total moles of EDTA - EDTA in excess

= 0.00113 - 0.00015 = 0.00098 mol

Now, 1 mol of EDTA reacts with 1 mol of either of the metal cation . i.e. Zn²⁺ or Ni²⁺

Now, EDTA displaced from Zn²⁺ = moles of Zn2+ present = EDTA that reacts with Mg²⁺

= 0.0292L *0.0123mol/L = 0.00035916 mol

Thus, Moles of Ni²⁺ in solution = Total moles of EDTA - moles of Zn²⁺

= 0.00098 mol -0.00035916 mol = 0.00062084moles

Total volume of mixture = 50 mL = 0.050 L

Thus, Molarity of Ni²⁺ = moles/ Volume = 0.00062084 mol Ni+2 /0.050 L = 0.0124 M

Molarity of Zn²⁺ = 0.00035916 molZn+2 / 0.050 L = 0.0072 M