Question

a) A metal ion buffer was prepared from 0.022 M ML and 0.026 M L, where ML is a metal-ligand complex and L is free ligand. M + L equilibrium reaction arrow ML Kf = 3.7 ✕ 108 Calculate the concentration of free metal ion, M, in this buffer.

b) By how many volts will the potential of an ideal Mg²⁺ ion-selective electrode change if the electrode is removed from 2.55 ✕ 10⁻⁴M MgCl₂ and placed in 2.05 ✕ 10⁻³M MgCl₂at 25°C? (Assume that the Nernst potential is 0.05916 V.)

Answer 1

1. Given: Concentration of ML = 0.022 M

Concentration of L= 0.026 M

Kf value (given) 3.7x10⁸   

To find the concentration of free ion M.

Solution:

Kf = ML / [M⁺] [L]

3.7 x 10⁸ = 0.022 / [M⁺] [0.026]

[M⁺] = 0.022/3.7 x 10⁸ x 0.026

[M⁺] = 0.022/9620000

[M⁺] = 2.28 x 10^-9

Therefore, concentration of free ion M in the buffer is 2.28 x 10^-9.

2. Given : concentration of Mg⁺² (c₁) = 2.55 x 10^-4 M

concentration of Mg⁺² (c₂) = 2.05 x 10^-3 M

Nernst Potential = 0.05916 V

To find = no. of volts change

Solution:

delta E = E₂ - E₁

= -0.05916/n ( log C₂ - log C₁ )

= -0.05916/2 ( log 2.05x 10^-3 - log 2.55 x 10^-4 )

= - 0.02958 x (-2.6882 - (-3.5934))

= -0.02958 x (-2.6882 + 3.5934)

= -0.02958 x 0.9052

= - 0.02677 Volts

Therefore, by -0.02677 volts the potential of an ideal Mg⁺² ion selecttive electrode will change.