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Calculate the dissociation constant of water at 25^(@)C from the following data : Specific conductance of H_(2)O = 5.8 xx 10^(-8) "mho cm"^(-1) lambda_(H^(+))^(@) = 350.0 and lambda_(OH^(-))^(@) = 198.0 "mho cm"^(2) |
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Answer» Solution :SUPPOSE WATER contains X moles per litre (or X eq./L) of `H^(+)` IONS (or `OH^(-)` ions). `because` X equivalents of `H^(+)` ions are produced from X eq. of water `therefore` volume (cc) containing 1 eq. of water which dissociated into its ions `= (100)/(X)`. `therefore` eq. conductance of water = sp. cond. `xx` V `= 5.8 xx 10^(-8) xx (1000)/(X)` Since water dissociated feebly, i.e, water may be considered to be a diliute solution of `H^(+)` and `OH^(-)` ions. `Lambda_(H_(2)O) = Lambda_(H_(2)O)^(@) = lambda_(H)^(@) + lambda_(OH^(-))^(@)` `therefore 5.8 xx 10^(-8) xx (1000)/(X) = 350 + 198 = 548`. `therefore X = 1.0 xx 10^(-7)` `therefore [H^(+)] = [OH^(-)] = 1 xx 10^(-7)` For the equilibrium, `H_(2)O = H^(+) + OH^(-)` Equilibrium constant (K) `= ([H^(+)][OH^(-)])/([H_(2)O])` `K_(w) = K xx [H_(2)O] = [H^(+)][OH^(-)]` `= 1.0 xx 10^(-7) xx 1.0 xx 10^(-7) = 1 xx 10^(-14)` `therefore K = (K_(w))/([H_(2)O]) = (1xx 10^(-14))/(55.5) = 1.8 xx 10^(-16)` mole/litre. `([H_(2)O] = (1000)/(18) = 55.5 " mole/litre")` |
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