Question

equilibrium Calculate the concentration of Ni2+ in a solution with a formal Niy2-concentration of 0.015 M at pH = 3.0. NiY2-: logK; = 18.4; Aya- = 2.1x10-11

Answer 1

Consider a complex formation reaction, Ni ²⁺ +Y ^4-$\rightleftharpoons$NiY ^2-

For above reaction, K'f = [Ni Y ^2- ] / [Ni ²⁺][Y ^4- ]

At pH = 3.0 , $\alpha$_Y 4- is 2.1 $\times$ 10 ^-11

We have, pK f = - log K f

$\therefore$ K f = 10 ^18.4

$\therefore$ The conditional formation constant K'f = ($\alpha$_Y 4- ) K f = (2.1 $\times$ 10 ^-11 ) ( 10 ^18.4) =5.275 $\times$ 10 ⁷

Let's use ICE table.

Concentration (M)	Ni 2+	Y 4-	Ni Y 2-
Initial	-	-	0.015
Change	+X	+ X	- X
Equilibrium	X	X	0.015 - X

Therefore, K'f = ( 0.015 - X) / (X )(X) = 5.275 $\times$ 10 ⁷

Value of K'f is very large, hence at equilibrium X is very small in comparison with 0.015. Therefore, we can write 0.015 - X $\simeq$ 0.015

$\therefore$ 0.015 / X ² = 5.275 $\times$ 10 ⁷

0.015 / 5.275 $\times$ 10 ⁷ = X ²

X ² = 2.844 $\times$ 10 ^-10

X = 1.69 $\times$ 10 ^-05 M = [Ni ²⁺]

We can check assumption , ( 1.69 $\times$ 10 ^-05 / 0.015 ) 100 = 0.11

This value is less than 5 % , hence our assumption is correct.

ANSWER : [Ni ²⁺] = 1.69 $\times$ 10 ^-05 M