Question

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^(+))^(oo) = 350.0` and `lambda_(OH^(-))^(oo) = 198.0 mho cm^(2)`

Answer 1

Correct Answer - `1.8 xx 10^(-16)` mole/litre
Suppose water contains X moles per L (or X. eq/L) of `H^(+)` ions (or `OH^(-)` ions)
`because X` equivalents of `H^(+)` ions are produces form X eq. of water
`:.` volume (cc) containin 1 eq. of water which dissociated into its ions `= (1000)/(X)`
`:.` eq. conductance of water = sp. cond . xV
`= 5.8 xx 10^(-8) xx (1000)/(X)`
Since water dissociates feebly, i.e., water may be considered to be a dilute solution of `H^(+)` and `OH^(-)` ions.
`Lamda_(H_(2)O)=Lamda_(H_(2)O)^(@) =lamda_(H)^(@) +lamda_(OH^(-))^(@)`
`:. 5.8 xx 10^(-8) xx (1000)/(X)=350 +198 =548`
`:. X = 1.9 xx 10^(-7)`
`:. [H^(+)]=[OH]=1 xx 10^(-7)` ltbtgt For the equilibrium, `H_(2)O = H^(+) + OH^(-)`
Equilibrium constant `(K)=([H^(+)][OH^(-)])/([H_(2)O])`
`K_(w)=K xx [H_(2)O]=[H^(+)O]==[H^(+)][OH]`
`=1.0 xx 10^(-7) xx 1.0 xx 10^(-7) = 1 xx 10^(-14)`
`:. K=(K_(w))/([H_(2)O]) =(1 xx 10^(-14))/(55.5) =1.8 xx 10^(-16) "mole/L"`
`([H_(2)O]=(1000)/(18) =55.5 "mole/L")`.