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)

Calculate the dissociation constant of water at 25^(@)C from the follo

1.	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» Solution :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")`.