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Equilibrium

Question
CBSEENCH11006579

Derive the relations:
straight alpha space equals space square root of fraction numerator straight K subscript straight w over denominator straight K subscript straight a space cross times space straight c end fraction end root

Solution
Derivate space of space straight alpha space equals space square root of fraction numerator straight K subscript straight w over denominator straight K subscript straight a space cross times space straight c end fraction end root
The hydrolysis of the salt (BA) may be represented as:
straight A to the power of minus space plus space straight H subscript 2 straight O space space space space rightwards harpoon over leftwards harpoon space space space space space space OH to the power of minus space plus space HA
Let c be the initial concentration of the salt and a be the degree of hydrolysis. At equilibrium, the concentration of various species can be represented as:
                                straight A to the power of minus space plus space straight H subscript 2 straight O space space space rightwards harpoon over leftwards harpoon space space space space OH to the power of minus space plus space space HA
Let c be the initial concentration of the salt and a be the degree of hydrolysis. At equilibrium, the concentration of various species can be represented as:
                                A-  +  H2O        rightwards harpoon over leftwards harpoon     OH-     +    HA
Initial conc.                c                                0                0
Equilibrium conc.       straight c left parenthesis 1 minus straight alpha right parenthesis                        cα            cα
therefore space space space space space straight K subscript straight h space equals space fraction numerator cα space cross times space cα over denominator straight c left parenthesis 1 minus straight alpha right parenthesis end fraction space equals fraction numerator cα squared over denominator 1 minus straight alpha end fraction
therefore space space space space space straight K subscript straight h space equals space fraction numerator cα squared over denominator 1 minus straight alpha end fraction 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... left parenthesis 1 right parenthesis
Since straight alpha is very small as compared to unity,
 1 minus straight alpha space space almost equal to space 1
space space space straight K subscript straight h space space equals space ca squared
or space space space straight alpha squared space equals space straight K subscript straight h over straight c
or space space space space straight alpha space equals space square root of straight K subscript straight h over straight c end root space straight i. straight e. space space straight alpha space space space straight alpha square root of 1 over straight c end root
But space space space straight K subscript straight h space equals space straight K subscript straight w over straight K subscript straight a
therefore space space space space straight alpha space equals space square root of fraction numerator straight K subscript straight w over denominator straight K subscript straight a straight c end fraction end root space space space space space space... left parenthesis 2 right parenthesis
From equation (2), it is clear that smaller the value of Ka (i.e. weaker the acid), the greater will be the value of a. As Kw increases with temperature, the degree of hydrolysis also increases with the increase in temperature.