Question

The metal ion Mn+ was titrated with 0.0500 M EDTA. The initial solution contained 100.0 mL of 0.0500 M metal ion (Mn⁺) buffered at a pH of 9.00.

a) Calculate the equivalence volume (V_eq).

b) What is the concentration of the free metal ion at volume V = V_eq/2.

c) If the conditional formation constant, K’_f = 5.4 X 1010, calculate the concentration of free metal ion at volume V = V_eq.

Answer 1

a. The equivalence volume (Ve is 100.00 mL b. The concentration of free metal ion at V = Veq/2 .10.025)/.150 = .01667M 0.025 = .05/2) c. If the conditional formation constant, K't = 5.4 X 1010, calculate the concentration of free metal ion at volume V = Vec. Kf = [EDTA@V]/[V ][EDTA] [EDTA.V ] = .10.05)/.2 = .025 [V ] = [EDTA]=> Kf = 0.025-X)/X .025-X^.025 X2 = .025/Kf X= sqrt (.025/ 5.4 X 1010) = 3.67 X 10-10 X =1.916 X 105