The Elements of Qualitative Chemical Analysis, vol. 1, parts 1 and 2. / With Special Consideration of the Application of the Laws of Equilibrium and of the Modern Theories of Solution.

by Julius Stieglitz

(1)

The condition for equilibrium will be

[I⁻]⁴ × [O₂] / ([I₂]² × [HO⁻]⁴) = K_equil.

(2)[603]

A system in which I⁻ is directly in equilibrium with I₂ (for which [I⁻]₁² : [I₂]₁ = K_{I⁻, Iodine} = 5.6E29, at room temperature (p. [298])) and in which, at the same time, HO⁻ is directly in equilibrium with O₂ (for which at room temperature [HO⁻]₁⁴ : [O₂]₁ = K_{HO⁻, Oxygen} = 8.2E49 (p. [298])) would also represent a condition of equilibrium for the four components. We find thus

K_equil. = K_{I⁻, Iodine}² / K_{HO⁻, Oxygen} = (5.6E29)² / (8.2E49) = 4E9.

(3)

With the aid of this constant and of equation (2) we can obtain, at least, an approximate interpretation of the results of the exposure of hydroiodic acid and of potassium iodide to the influence of atmospheric oxygen.[604] We may [p307] calculate, first, what concentration of free iodine would be required to prevent oxidation of hydroiodic acid, in molar solution, by the oxygen of the air, i.e. to establish equilibrium. We will call x that concentration of I₂. As hydroiodic acid is a very strong acid, ionized to the extent of about 80% in molar solution, we may, with sufficient accuracy for our purpose, consider it completely ionized and put [I⁻] = 1 and [H⁺] = 1. Since at 25° [H⁺] × [HO⁻] = 1.2E−14 (p. [104]), we may put [HO⁻] = 10⁻¹⁴. The concentration of oxygen in the air, at room temperature, may be considered to be approximately [O₂] = (1/5) × (1 / 23.9). Inserting all these given values in equation (2), we have

[I−]4 × [O2]	=	1 × (1/5) × (1 / 23.9)	= 4 × 109.
[I2]2 × [HO−]4		x2 × (10−14)4

(4)