Question

A metal ion M forms a stable complex with the ligand X:

M + (aq) + 4 X (aq) ⇌ M (X) 4+ (aq) Kf = 1.0 ∙ 10^20

In a solution where [M +] = 0.100 M and [X] = 0.800 M before reaction, what is [X] at equilibrium?

Answer 1

Answer

[X] = 0.400M

Explanation

First consider complete formation of M(X)₄⁺

M⁺(aq) + 4X(aq) ------> M(X)4⁺( aq) .

stoichiometrically, 1mole of M⁺ reacts with 4moles of X to give 1mole of M(X)4⁺

0.100M of M⁺ reacts 0.400M to give 0.100M M(X)4⁺

After completion of reaction

[X] = 0.800M - 0.400M = 0.400M

[M⁺] = 0

[M(X)4⁺] = 0.100M

Now, consider the dissociation equilibrium of M(X)4⁺

M(X)4⁺(aq) <-------> M⁺(aq) + 4X(aq)

K_d = [M⁺][X]⁴/[M(X)4⁺]

K_d = 1/K_f = 1/1.0 ×10²⁰ = 1.0 ×10^-20

Initial concentration

[M(X)4⁺] = 0.100

[M⁺] = 0

[X] = 0.400

chamge in concentration

[M(X)4⁺] = - x

[M⁺] = +x

[X] = +4x

Equilibrium concentration

[M(X)4⁺] = 0.100 - x

[ M⁺] = x

[X] = 0.400 + 4x

so,

x(0.400 +4x)⁴ / 0.100 - x = 1.0 ×10^-20

We can assume 0.400 + 4x = 0.400 and 0.100 - x= 0.100 because x is small value

x(0.400)⁴/0.100 = 1.0 ×10^-20

x 0.256 = 1.0 × 10^-20

x = 3.91 × 10^-20

Therefore,

[X] = 0.400 + 4( 3.91 ×10^-20₎ = 0.400M