The pH of a solution is 5.5 at `25^(@)C`. Calculate its `[OH^(-)]`.

Explanation

pH = 5.5
Since `" "pH+pOH=14`
`pOH=14-pH`
`=14-5.5=8.5`
`[OH^(-)]=10^(-pOH)`
`=10^(-8.5)`
`=3.16xx10^(-9)`M
(OR)
`[OH^(-)]="antilog"[-8.5]`
`=3.16xx10^(-9)`M
`[OH^(-)]=3.16xx10^(-9)`M