Ionic product of water at 310 K is 2.7xx10^(-14) .What is the pH of neutral water at this temperature ?

Ionic product of water at 310 K is 2.7xx10^(-14) .What is the pH of ne

1.	Ionic product of water at 310 K is 2.7xx10^(-14) .What is the pH of neutral water at this temperature ?
Answer» Solution :Calculation `[H^+]` : There is FOLLOWING EQUILIBRIUM in water. `H_2O_((l))+ H_2O_((l)) =H_3O_((aq))^(+) + OH_((aq))^(-)` `K_W=[H_3O^+][OH^-]= 2.7xx10^(-14)` but `[H_3O^+]=[OH^-]` `THEREFORE [H_3O^+]^2=2.7xx10^(-14)` `therefore [H_3O^+]= sqrt(2.7xx10^(-14)) =1.6432xx10^(-7)` Calculation of pH : pH=-log `[H^+]` =-log `(1.6432xx10^(-7))` =-(0.2157-7.0)=6.7843