Conductivity water (prepared by repeated distillation of water containing a small quantity of `NaOH` and `KMnO_(4))` has conductivity `5.50 xx 10^(-6) S m^(-1)`. If `lamda_(H+)6(0) = 3.498 S m^(2) mol^(-1)` and `lamda_(OH-)^(0) = 1.980 xx 10^(9-2) S m^(2) mol^(-1)`, then calculated lthe ionic product of water.
Strategy : Water is known to be slightly ionized as `{:(,H_(2)O(1)hArr,H^(+)(aq.)+,OH^(-)(aq.)),("I.C",C,0,0),("F.C",C(1-alpha),C alpha,C alpha):}`
According to the law of chemical equilibrium
`K_(eq).xx=([H^(+)][OH^(-)])/([H_(2)O])`
Since water ionizes only very slightly, the concentration of `H_(2)O` may be taken as constant. Thus
`K_(eq) xx "constant" = K_(w) = [H^(+)][OH^(-)]`
The product of the concentrationof `H^(+)` and `OH^(-)` ions expressed in mol `L^(-1)` is known as ionic product of water `(K_(w))`:
`(K_(w)) = [H^(+)][OH^(-)]`
`= (Calpha)(Calpha)`
`= C^(2)alpha^(2)` Thus we need to find `C` and `alpha`

Explanation

Step 1: Calculating the concentration `(C) = ("Moles of water")/("Volume of water")`
`= ("Mass of water")/("Molar mass of water") xx (1)/("Volume of water")`
Using the fact that `1 L` (i.e. `10^(-3) m^(3)`) of water has a mass of `1 Kg`, we have
`= (1 Kg)/(0.018 Kg mol^(-1)) xx (1)/(10^(-3) m^(3))`
`= 5.55 xx 10^(4) mol m^(-3)`
Calculating molar conducativity at concentration `C`
`Lambda_(m)=(K)/(C)= (5.50xx10^(-6)SM^(-1))/(5.55xx10^(4)Sm^(2)mol^(-1))`
`= 9.90 xx 10^(-11) S M^(2) mol^(-1)`
Step 3: Calculating limiting molar conductivity.
The limiting molar conductivity of water can be calculated using the Kohlrausch's law of independent migration of ions.
Since water ionizes to a samll extent into `H^(+)` and OH^(-) ions, we can write
`Lambda_(m)^(0)(H_(2)O) = lamda_(H^(+))^(0)+lamda_(OH^(-))^(0)= (3.498 + 1.980) xx 106(-2) S m^(9-2) mol^(-1))`
Step 4: Calculating degree of ionization
`alpha = (Lambda_(m))/(Lambda_(m)6(0))`
`(9.90 xx 10^(-11) S m^(2)mol^(-1))/(5.478 xx 10^(-2) S m^(2)mol^(-1))`
`= 1.81 xx 10^(-9)`
Step 5: Calculating ionic product of water
`K_(w) = [H^(+)][OH^(-)]`
`= (Calpha)(Calpha)`
`= C^(2)alpha^(2)`
Substituting the values of `C` and `alpha`, we get
`K_(w) = (5.55 xx 10^(4) mol m^(-3))^(2) (1.81 xx 10^(-9))^(2)`
`= 1.01 xx 10^(-8) mol^(2)m^(-6)`
`= 1.01 xx 10^(-8) xx 10^(-6) mol^(2) dm^(-6)`
`{:((1m=10 dm),((1m)^(-6)=(10dm)^(-6)),(=10^(-6) dm^(-6)),(1 dm=1L)):}`
`= 1.01 xx 10^(-14) mol^(2)dm^(-6)`