Calculate the pH of solution with `H_3O^+` concentration in mol `dm^(-3)`.
(iii) `6.8xx10^(-3)`

Explanation

`pH=-log[H_3O^+]`
(iii) `pH=-log[H_3 O^+]`
`pH=-log [6.8xx10^(-3)]`
`pH=log 1-log 6.8 -log 10^(-3)`
`=3-0.8325`
`pH=2.17 (or) 2.2`