Calculate the pH of `N/(1000)` sodium hydroxide (NaOH) solution assuming complete ionisation.

Explanation

NaOH completely ionizes as : `NaOH rarr Na^(+) + OH^(-)`
`:. [OH^(-)]=[NaOH]=(N)/(1000)=10^(-3)N=10^(-3)M ` (`:'` Eq. mass = Mol. Mass in case of NaOH)
Now, as `[H_(3)O^(+)][OH^(-)]=10^(-14) :. [H_(3)O^(+)]=(10^(-14))/([OH^(-)])=(10^(-14))/(10^(-3))=10^(-11)`
`:. pH = - log [H_(3)O^(+)]= - log 10^(-11) =11`