Question

If `[HO^-]=6.0xx10^-10*mol*L^-1`, what is `pH` for this solution?

Answer 1

`[H^+] ("or "[H_3O^+])=10^(-4.78)*mol*L^-1`, `pH=4.78`, `pOH=9.22`

Explanation:

In aqueous solution at `298*K`, water undergoes the following equilibrium reaction:

`H_2O(l) rightleftharpoons H^(+) + HO^(-)`

Alternatively, and perhaps now more commonly we would write:

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

This equilibrium reaction has been carefully and quantitatively measured, and under standard conditions:

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

If we take negative logarithms to the base 10 of both sides we get:

`pK_w=14=pOH+pH`, where the `pH` functions means `-log_10[H_3O^+]`, etc.

We are given `[HO^-]=6.0xx10^-10`, so `pOH=-log_10(6.0xx10^-10)=-(-9.22)=9.22`, and thus `pH=14-9.22=4.78`.

Is this solution acidic or basic?

The given reaction occurs at `298*K`. Given that this is a bond-breaking reaction, how do you think `pK_w` would evolve at `323*K` or `373*K`?