Question

Find the conditional formation (Kf') constant for Ba(EDTA)2- at pH 11.00 where log Kf is 7.88 and alphaY4- is 0.81

Find the concentration of free Ba2+ in 0.05 M Na2[Ba(EDTA)] at pH 11.00

Answer 1

Solution:

The problem can be solved by following formulla.

Conditional formation constant K_f^'= K_f X Y^4- ------------------(1)

logKf = 7.88 Or Kf = 7.5 X 10⁷

Substituting value of Kf in (1)

K_f^'=  7.5 X 10⁷ X 0.81 = 6.07 X 10⁷

Ba²⁺ + EDTA(Y) <--------------> BaY^2-

initial - - 0.05M

Equilibrium x x 0.05-x

K_f^'= [BaY^2-]/[Ba²⁺][ EDTA] = 0.05-x/x²

^Or K_f^'x² + x -0.05 = 0

6.07 X 10⁷x²   + x + 0.05 = 0

This is quadratic equation its solution will yeild us x = 2.86 X 10^-5 M