OSTWALD’S DILUTION LAW
According to Arrhenius theory of electrolyte dissociation, the molecules of an electrolyte in
solution are constantly splitting up into ions and the ions are constantly reuniting to form
unionized molecules. Therefore, a dynamic equilibrium exists between ions and unionized
molecules of the electrolyte in solution. It was pointed out by Ostwald that like chemical
equilibrium, law of mass action van be applied to such systems also.
Consider a binary electrolyte AB which dissociates into A+ and B- ions and the equilibrium
state is represented by the equation:
AB ↔ A+ + B-
Initially t = o C 0 0
At equilibrium C(1-α) Cα Cα
So, dissociation constant may be given as
K = [A+][B-]/[AB] = (Cα * Cα)/C(1-α)
= Cα2 /(1-α) ....... (i)
For very weak electrolytes,
α <<< 1, (1 - α ) = 1
.·. K = Cα2
α = √K/C ....... (ii)
Concentration of any ion = Cα = √CK .
From equation (ii) it is a clear that degree of ionization increases on dilution.
Thus, degree of dissociation of a weak electrolyte is proportional to the square root of
dilution.
Limitations of Ostwald's dilution law:
The law holds good only for weak electrolytes and fails completely in the case of strong
electrolytes. The value of 'α' is determined by conductivity measurements by applying the
formula Λ/Λ ∞ . The value of 'α' determined at various dilutions of an electrolyte when
substituted in Eq. (i) gives a constant value of K only in the case of weak electrolytes like
CH 3 COOH, NH 4 OH, etc. the cause of failure of Ostwald's dilution law in the case of strong
electrolytes is due to the following factors"
(i) The law is based on the fact that only a portion of the electrolyte is dissociated into ions
at ordinary dilution and completely at infinite dilution. Strong electrolytes are almost
completely ionized at all dilutions and Λ/Λ ∞ does not give accurate value of 'α'.
(ii) When concentration of the ions is very high, the presence of charges on the ions
appreciably effects the equilibrium. Hence, law of mass action its simple form cannot be
strictly applied in the case of string electrolytes.
SOME SOLVED EXAMPLES
Example 1: A 0.01 M solution of acetic is 5% ionized at 25o C. Calculate itsdissociation
constant.
Solution: According to Ostwald's dilution law
K α = α2/(1-α)V
α = 0.05, V = 1/0.01 = 100 litres
Hence, K a = 0.05 * 0.05/(1-0.05)100 = 2.63 * 10-5
Example 2: Calculate the H+ ion concentration of a 0.02 N weak monobasic acid. The
value of dissociation constant is 4.0 × 10-10.
Solution: HA ↔ H+ + A-
Applying Ostwald's dilution law of a weak acid,
α = √k a V
Ka= 4.0 ×10-10, V = 1/0.01 = 100 litres
α = √(4 * 10-10 * 102) = 2 * 10-4
Concentration of hydrogen ions
a/√V = (2*10-4)/100 = 2*10-6 mol L-1
or Concentration of hydrogen ions
= √(CK) = √(0.01 * 4 *10-10) = 2 * 10-6mol L-1
Source : http://ciseche10.files.wordpress.com/2013/12/ionic-equilibrium.pdf
