Question

If `[H_3O^+]=5.0xx10^-3*mol*L^-1`, what is `pH` and what is `pOH` of this solution?

Answer 1

`pH=2.30`............

Explanation:

.......and `pOH=11.70`

These acid base reactions are governed by the equilibrium:

`2H_2O(l) rightleftharpoons H_3O^(+) + HO^-`.

And at `298*K`, `K_w=[H_3O^+][HO^-]`

`log_10K_w=log_10[H_3O^+]+log_10[HO^-]`

And since we know that `K_w=10^-14`, then..........

`-14=log_10[H_3O^+]+log_10[HO^-]`

OR `14=-log_10[H_3O^+]-log_10[HO^-]`

But, by definition, `-log_10[H_3O^+]=pH`, and `-log_10[H_3O^+]=pOH`

And thus, BY DEFINITON, `pH+pOH=14` under standard conditions. If you can remember this relationship, then these sorts of problems become almost trivial.

And thus, for your problem, `pH=-log_10(0.0050)=-(-2.30)=2.30`.

`pOH=14-2.30=11.70`.

And can you tell me what is `[HO^-]` here? It should be small......