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>> Ostwald s Dilution Law-Part1
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 CH3COOH, NH4OH, 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
Kα = α2/(1-α)V
α = 0.05, V = 1/0.01 = 100 litres
Hence, Ka = 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 = √kaV
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
Example 3: The concentration of H+ ions in 0.10 M solution of a weak acid is 1.0 × 10-5 mol L-1. Calculate the dissociation constant of the acid.
Solution: HA ↔ H+ + A-
Initial concentration 0.1 0 0
(mol L-1) 0.1-1.0×10-5 1.0×10-5 1.0×10-5
[HA] can be taken as 0.1 M as 1.0 × 10-5 is very small.
Applying law of mass action.
Kα = [H+][A-]/[HA] = (1.0*10-5 * 1.0 *10-5)/0.10
= 1 × 10-9
Example 4: What will be the dissociation constant of 0.1 N aqueous ammonia solution in terms of degree of dissociation 'α'? What will be the value if the concentration is 0.01 N?
Solution: NH4OH ↔ NH-4 + OH-
At equilibrium (1-α) α α
Since the solution is 0.1 N,
V = 1/0.1 = 10litre
[NH4OH]=(1-α)/10, [NH-4] = α/10 and [OH-] = α/10
Applying law of mass action,
For 0.01 N Solution, Kb remains the same at the same temperature but degree of dissociation value becomes different.
Example 5: A 0.0128 N solution of acetic acid has A = 14 mho equiv-1 and A∞ = 391 mho equiv-1 at 25oC. Calculate the dissociation constant of the acid.
Solution: Degree of dissociation,
α = Λ/Λ∞ = 14/391 = 3.58*10-2
Now applying Ostwald's dilution law,
Kα = α2/(1-α)V
α = 3.58 * 10-2 and V = 1/0.0128 litre
So, Kα = 3.58 * 10-2 * 3.58 * 10-2 * 0.0128 = 1.64 * 10-5