Calculate the dissociation canstant of water at `25^(@)C` from te followig data specific conductance of `H_(2)O=5.8xx10^(-8)mhocm^(-1),lamda_(H^(+))^(infty)=350.0` and `lamda_(OH)^(infty)=198.0mhocm^(2)mol^(-1)`

Explanation

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")`