Calculate the degree of ionization of pure water at `25^@C`.

Explanation

Strategy: The degree of ionization of pure water `(alpha)` is the ratio of ionized water to that of unionized water.
Solution: The density of pure water is `1g mL^(-1)` or `1000gL^(-1)`. The molar mass of water is `18gmol^(-1)`. Thus, the molarity of pure water can be given as
`C_(H_2O)=(d_(H_2O))/("Molar mass"_(H_2O))=(1000gL^(-1))/(18gmol^(-1))`
Now, we can calculate the degree of ionization of pure water as
`alpha=(C_(H^+))/(C_(H_2O))` or `(C_(OH^-))/(C_(H_2O))=(1.0xx10^-7)/(55.5)`
`=1.8xx10^-9`