`pH=-log[H_(3)O^(+)] :. Log[H_(3)O^(+)]=-pH=-12`
`:. [H_(3)O^(+)]`= antilog (-12) = antilog `(bar(12))=10^(-12)` g ions/litre
But we know that `[H_(3)O^(+)][OH^(-)]=K_(w)=10^(-14) :. [OH^(-)]=(K_(w))/([H_(3)O^(+)])=(10^(-14))/(10^(-12))=10^(-2)` g ions/litre
Since NaOH is a strong electrolyte, it undergoes complete ionization as : `NaOH rarr Na^(+)+OH^(-)`
`:. [OH^(-)]=[NaOH]=10^(-2)M = 10^(-2) "mol" L^(-1)`
Mol. mass of NaOH = 40 `:.` Amount of NaOH dissolved per litre `=10^(-2) xx 40 = 0.4 g`
