Question

Calculate the pZn of a solution prepared by mixing 25.0 mL of 0.0100 M EDTA with 50.0 mL of 0.00500 M Zn2+. Assume that both the Zn2+ and EDTA solutions are buffered with 0.100 M NH3 and 0.176 M NH4Cl.

Answer 1

Given 25 mL of 0.0100 M EDTA with 50 mL of 0.00500 M Zn²⁺ and Buffered with 0.100 M NH₃ and 0.176M NH₄Cl

The analytical concentrations of NH₃ and NH₄Cl then become

C_NH3 = (25*0.100 + 25*0.0100 + 50*0.00500)/100 = 3/100 = 0.03

C_NH4Cl = (25*0.176 + 25*0.0100 + 50*0.00500)/100 = 4.9/100 = 0.049

[Zn²⁺] = 5.70*10^-10 * (0.049/0.03) = 9.31*10^-10

pZn of the solution = - log(9.31*10^-10) = 9.03

Therefore,pZn of the solution prepared by mixing 25.0 mL of 0.0100 M EDTA with 50.0 mL of 0.00500 M Zn²⁺ is equal to 9.03