Calculate the pH value of 0.0001 M `HNO_(3)`

Answer

`pH =4`

Explanation

`HNO_(3)` is a strong acid and it dissociates completely in aqueous solution. The concentration of the ions is the same as that of the acid
`HNO_(3) overset((aq))(to) underset(0.0001 M)(H_(3)O^(+)) +underset(0.0001M)(NO_(3)^(-)(aq))`
`[H_(3)O^(+)] =0.0001 M =10^(-4) M`
`pH =- log [H_(3)O^(+)] =-log [10^(-4)] =(-) (-4) log 10=4`