In one litre of water `10^(-10)` moles of HCl were added. The pH of the solution will be approximately .

Answer

7

Explanation

Molar conc.of HCl`=10^(-10)M`
`[H_(3)O^(+)]` ion already present in water `=10^(-7)M`
Net `[H_(3)O^(+)]=(10^(-7)+10^(-10))`
`=10^(-7)(1+10^(-3))=1.001xx10^(-7)`
`pH=-log(1.001xx10^(-7))=7-log1.001~~7`