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