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Equilibrium

Question
CBSEENCH11006493

Discuss the ionisation of weak electrolytes (Ostwald’s dilution law).

Solution
As the weak electrolytes are ionised to a small extent, there is a state of equilibrium between the unionised electrolyte and the ions formed in solution. To illustrate.this, let us consider a weak electrolyte HA to be dissolved in water, a be the degree of ionisation and C be the molar concentration of HA.
                                  HA rightwards harpoon over leftwards harpoon    H+(aq) + A-(aq)
Initial  conc.                C            0               0
Equilibrium conc.     straight C minus Cα      Cα              Cα
Applying law of chemical equilibrium,
                 straight K subscript straight a space equals space fraction numerator open square brackets straight H to the power of plus left parenthesis aq right parenthesis close square brackets space space open square brackets straight A to the power of minus left parenthesis aq right parenthesis close square brackets over denominator open square brackets HA left parenthesis aq right parenthesis close square brackets end fraction
where Ka is ionisation (or dissociation) constant of the weak acid HA.
or space space space space space straight K subscript straight a space equals space fraction numerator Cα space cross times space Cα over denominator straight C minus Cα end fraction space equals space fraction numerator straight C squared straight alpha squared over denominator straight C left parenthesis 1 minus straight alpha right parenthesis end fraction space equals space fraction numerator Cα squared over denominator 1 minus straight alpha end fraction
This equation is often referred to as Ostwald’ dilution law equation since the degree of dissociation of a weak electrolyte is very small as compared to unity.
therefore space space space space left parenthesis 1 minus straight alpha right parenthesis space can space be space taken space as space one space open square brackets because space straight a less than less than 1 close square brackets
or space space space space straight K subscript straight a space equals space Cα squared
or space space space space space straight alpha squared space equals space straight K subscript straight a over straight C
therefore space space space space straight alpha space equals space square root of straight K subscript straight a over straight C end root space space or space space straight alpha space proportional to space square root of 1 over straight C end root
space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space open square brackets because space space straight K subscript straight a space is space constant close square brackets

Thus, the degree of ionisation of a weak electrolyte is inversely proportional to the square root of the molar concentration of the solution of the electrolyte.
Hence the degree of ionisation of weak electrolyte goes on increasing with the decrease in molar concentration i.e. with dilution and it reaches the maximum value (unity) in very dilute solution. Thus, all weak electrolytes undergo almost complete ionisation at infinite dilution.
For a weak base BOH:
Let us consider a weak base BOH dissolved in water, a be the degree of ionisation and C be the molar concentration of BOH.
                         BOH space space space rightwards harpoon over leftwards harpoon space space space space straight B to the power of plus left parenthesis aq right parenthesis space space space plus space space space space OH to the power of minus left parenthesis aq right parenthesis
Initial conc.           C              0                 0
Equilibrium        straight C minus Cα         Cα                 Cα
conc. 
Applying law of chemical equilibrium,
straight K subscript straight b space equals space fraction numerator open square brackets straight B to the power of plus left parenthesis aq right parenthesis close square brackets space open square brackets OH to the power of minus left parenthesis aq right parenthesis close square brackets over denominator open square brackets BOH space left parenthesis aq right parenthesis close square brackets end fraction space equals space fraction numerator Cα space cross times space Cα over denominator straight C space minus space Cα end fraction
space space space space space space equals space fraction numerator straight C squared straight alpha squared over denominator straight C left parenthesis 1 minus straight alpha right parenthesis end fraction space equals space fraction numerator Cα squared over denominator 1 minus straight alpha end fraction
straight K subscript straight b space equals space Cα squared space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space open square brackets because space space straight a less than less than less than 1 close square brackets
or space space space straight alpha squared space equals space straight K subscript straight b over straight C
or space space space space straight alpha space equals space square root of straight K subscript straight b over straight C end root