Question

If the `pH=12` for a solution of potassium hydroxide, what is `[HO^-]`?

Answer 1

`[HO^-]` `=` `0.01*mol*L^-1`.

Explanation:

We know, or should know that (i) `pH=-log_10[H_3O^+]`, and `pOH=-log_10[HO^-]`; and (ii), given that `K_w=[H_3O^+][""^(-)OH]=10^-14` under standard conditions, `pK_w=pH+pOH=14.`

We have `pH=-log_10[H_3O^+]=12`; thus `[H_3O^+]=10^-12*mol*L^-1`, and `pOH=2`, and thus `[HO^-]=10^-2*mol*L^-1`.

What is `[K^+(aq)]`?