Question

A certain solution has pH 3.89 at 0°C. What is the pOH and `[OH^-]`?

Answer 1

`pOH` `=` `14-3.89` `=` `10.11`

Explanation:

Given `pOH`, `[HO^-]` `=` `10^(-pOH)` `=` `10^(-10.11)` `mol*L^-1`.

Note that in water, `pH + pOH = 14` at `298*K`.