Ionic product of water at 310 K is `2.7` X `10^(-14)` . What is the pH of neutral water at this temperature ?

Explanation

Initial `pH:alpha=sqrt((K_(a))/(C))=0.06(lt0.1)`
So `[H^(+)]=Calpha=0.06xx0.05=0.003M`
`:.pH=2.52`
final `pH=-log(0.01)/(0.1)=1` :.Change =`1-2.52=-1.52`.
Now `K_(a)=([H^(+)][HCOO^(-)])/([HCOOH])rArr 1.8xx10^(-4)=(0.1xx[HCOO^(-)])/(0.05)`
`rArr [HCOOH^(-)]=9xx10^(-5)M`