Question

What is `pH` of a solution `1.2xx10^-2*mol*L^-1` with respect to `KOH`?

Answer 1

`pH=12.08`

Explanation:

Well, first off we calculates `pOH`

And `pOH=log_(10)[HO^-]=-log_(10)1.2xx10^-2=1.92`.

Now it is a given that `pOH+pH=14`.

Why? Because `K_w=[HO^-][H_3O^+]=10^(-14)`, and when we take `log_10` of both sides, we come up with that relationship.

And so `pH=14-pOH=14-11.92=....??`