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Equilibrium

Question
CBSEENCH11006577

Show that for an aqueous solution of a salt of weak base and strong acid,

pH space equals space minus log space square root of straight K subscript wc over straight K subscript straight b end root.

Solution

Let the salt of strong acid and weak base be XY and hydrolysis reaction be represented as:
straight X to the power of plus space plus space straight H subscript 2 straight O space space space rightwards harpoon over leftwards harpoon space space space space space stack XOH space plus space straight H to the power of plus with open square brackets Cationic space hydrolysis close square brackets below
Let c be the initial concentration of the salt and α be the degree of hydrolysis. At equilibrium, the concentration of various species is represented as:
  X+    +      H2O       rightwards harpoon over leftwards harpoon    XOH      +     H+
Initial conc.
  c                 c                0                  0
Equilibrium conc.            
        straight c left parenthesis 1 minus straight alpha right parenthesis                   cα               cα
                         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 space fraction numerator cα squared over denominator left parenthesis 1 minus straight alpha right parenthesis end fraction
Since straight alpha is very small as compared to unity, 


therefore space space space space 1 minus straight alpha space equals space 1 space
therefore space space space space space space space space space straight K subscript straight h space equals space cα squared
or space space space space space space space straight alpha squared space equals space straight K subscript straight h over straight c
or space space space space space space space space straight alpha space equals space square root of straight K subscript straight h over straight c end root space
But space space space space space straight K subscript straight h space space equals space straight K subscript straight w over straight K subscript straight b
therefore space space space space 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 b straight c end fraction end root space space
straight i. straight e. space space space open square brackets straight H to the power of plus close square brackets space equals space cα
Substituting space the space value space of space straight alpha comma
space space space space space space space space open square brackets straight H to the power of plus close square brackets space equals space straight c square root of fraction numerator straight K subscript straight w over denominator straight K subscript straight b cross times straight c end fraction end root space equals space square root of fraction numerator straight K subscript straight w straight c over denominator straight K subscript straight b end fraction end root
But space pH space equals negative log space open square brackets straight H to the power of plus close square brackets
space space space space space space space space space space equals space minus log space square root of fraction numerator straight K subscript straight w straight c over denominator straight K subscript straight b end fraction end root
space space space space space space space space space space equals negative log space open parentheses fraction numerator straight K subscript straight w straight c over denominator straight K subscript straight b end fraction close parentheses to the power of 1 divided by 2 end exponent
or space pH space equals space minus 1 half open square brackets log space straight K subscript straight w space minus space log space straight K subscript straight b space plus space log space straight c close square brackets