Question

What is `[HO^-]` in an aqueous solution for which `[H_3O^+]=3.2xx10^-3*mol*L^-1`?

Answer 1

`[HO^-]=3.16xx10^-12*mol*L^-1`

Explanation:

We know that water undergoes autoprotolysis according to the following equation......

`2H_2O(l)rightleftharpoonsH_3O^+ + HO^-`

WE know by precise measurement at `298*K` under standard conditions, the ion product......

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

Now this is an equation, that I may divide, subtract, multiply, PROVIDED that I does it to both sides. One thing I can do is to take `log_10` of BOTH SIDES........

`log_10{K_w}=log_10{[H_3O^+][HO^-]}=log_10(10^-14)`

And thus `log_10[H_3O^+]+log_10[HO-]=-14`, and on rearrangement.....

`14=underbrace(-log_10[H_3O^+])_(pH)underbrace(-log_10[HO^-])_(pOH)`

And so (FINALLY) we gets our working relationship, which you should commit to memory.....

`pH+pOH=14`

We have `[H_3O^+]=3.2xx10^-3*mol*L^-1`. `pH=-log_10(3.2xx10^-3)=2.50`. And thus `pOH=11.50`.

And now we take antilogs.....`[HO^-]=10^(-11.50)*mol*L^-1`

`=3.16xx10^-12*mol*L^-1` as required............