Question

The Kf for the formation of the complex ion between Pb2+ and EDTA4− is 1.0 × 1018 at 25° C. Pb2+ + EDTA4− ⇌ Pb(EDTA)2− Calculate the [ Pb2+ ] at equilibrium in a solution containing 2.50 × 10−3 M Pb2+ and 3.50 × 10−3 M EDTA4

Answer 1

Initial concentrations are given

Initial [Pb²⁺] = 2.50 × 10⁻³ M

Initial [EDTA⁴⁻] = 3.50 × 10⁻³ M

given Kf = 1.0 × 10¹⁸

Since formation constant (k_f) is very large

Reaction goes to completion

	Pb2+	EDTA4−	Pb(EDTA)2−
I	2.50 × 10−3	3.50 × 10−3	0
C	-2.50× 10−3	-2.50× 10−3	+2.50× 10−3
F	0	1.00 × 10−3	2.50× 10−3

Since the reaction at equilibrium again Pb(EDTA)²⁻ will undergo dissociation (back reaction takes place)

At equilibrium

E

x

1.00 × 10−3 + x

2.50× 10−3 - x

Since dissociation is very small, we can neglect x

x = 2.5 x 10^-18

[Pb²⁺] = x = 2.5 x 10^-18 M