Question

The Kf for the formation of the complex ion between Pb2+ and EDTA4− is 1.0 ×...

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

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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 (kf) 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

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