Question

What is the pOH of a solution with a pH of .7?

Answer 1

And it's an aqueous solution?

Explanation:

We know that in water the following equilibrium operates:

`2H_2O(l) rightleftharpoonsH_3O^+ + HO^-` `K_"eq"=10^-14`

`[H_3O^+][HO^-]=10^-14`

We take `log_10` of both sides:

`log_10[H_3O^+]+log_10[HO^-]=log_10[10^-14]`

`log_10[H_3O^+]+log_10[HO^-]=-14`, and multiply both sides by `-1`,

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

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

And thus `pH+pOH=14` (for water under standard conditions). This is something that you have to know.

You have `pH=0.7`, and thus `pOH=13.3`. What is `[H_3O^+]`?