ionic product of water

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

The ionic product of an ionic solid