Calculate the pH of solution with `H_3O^+` concentration in mol `dm^(-3)`.
(iv) `3.2xx10^(-5)`

Explanation

`pH=-log[H_3O^+]`
(iv) `pH=-log [3.2xx10^9-5)]`
`pH=log 1-log 3.2 -log 10^(-5)`
`=5-0.5051`
`pH=4.49 (or) 4.5`