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

Explanation

In neutral water ` [H_3O^(+) ] = [OH^(-)]`
` " . K_w =[H_3O^(+) ][OH^(-)] =[H_3O^(+)] ^(2)`
`or " " [H_3O^(+) ]^(2) =K_w = 2.7 xx 10 ^(-14)`
`therefore " " [H_3O^(+) ] = (2.7 xx 10 ^(-14) )^(1//2) = 1.64 xx 10^(-7)`
Hence , `" " pH = - log_(10) [H_3O^(+)]`
` " " = - log_(10) (1.64 xx 10 ^(-7)) =6.78`