At `25^@C` the degree of ionization of water was found to be `1.8 xx 10^(–9)`. Calculate the ionization constant and ionic product of water at this temperature

Explanation

If x is the degree of ionization of water, then
`{:(H_2O, hArr , H^(+), + , OH^(-)),(c(1-alpha), , calpha , , calpha):}`
`c=[H_2O]=1000/18`=55.56 M
`K_(eq)=([H^+][OH^+])/([H_2O])=((calpha)^2)/(c(1-alpha))=calpha^2` (since `alpha` is very much less than 1)
`K_(eq)=55.56xx(1.8xx10^(-9))^2=1.8xx10^(-16)` M
`K_W=[H^+][OH^-]=(c alpha )^2 =(55.56xx1.8xx10^(-9))^2`
`K_W=1.0xx10^(-14) M^2`