Question

Find the conditional formation constant for (MgEDTA)^2- at pH 9.00. Then find the concentration of free Mg^2+ in 0.050 M Na2[Mg(EDTA)] at pH 9.00. Please show all the steps because I don't know how to start it. If you do, I'll give it a thumbs up! Thank you!

Answer 1

we know that formation constant for (MgEDTA)^2- is K_f= 10^8.79

and α_Y4- = 0.054 at pH = 9.00

therefore conditional formation constant is

K_f l = α_Y4- x  K_f

= 0.054 x 10^8.79

= 3.3 x 10^7

and the concentration of free Mg2+ can be find as

Mg2+ + EDTA <-----------> Mg(EDTA)2-

initial conc. - - 0.050 M

equilibrium conc. x x ( 0.050 - x) M

at pH 9.00

K_f l = [Mg(EDTA)2-] / [Mg2+] [EDTA]

or

3.3 x 10^7 = ( 0.050 - x ) / x^2

3.3 x 10^7 x^2 + x -0.050 = 0

on solving the quadratic equation , we get

x = 3.9 x 10 -5 M

[Mg2+] = 3.9 x 10 -5 M

therefore the concentration of free Mg^2+ in 0.050 M Na2[Mg(EDTA)] at pH 9.00 is 3.9 x 10 -5 M.