Question

Help me please?? Thanks!

Answer 1

`[HO^-]=1.78xx10^-11*mol*L^-1`

Explanation:

In aqueous solution, water undergoes the famous autoprotolysis reaction:

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

And under standard condition of `298*K`, and near `1*atm`...

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

Taking `"log_10` of BOTH sides........

`log_10K_w=log_10[HO^-]+log_10[H_3O^+]`.....on rearrangment...

`underbrace(-log_10[HO^-])_(pH)underbrace(-log_10[H_3O^+])_(pH)=underbrace(-log_10(10^-14))_14`

And thus our working relationship.......which you will have to commit to memory:

`pH+pOH=14`

We have `pH=3.25`, thus `pOH=10.75`, and .........

`[HO^-]=10^(-10.75)=1.78xx10^-11*mol*L^-1`......

For the use of `pH` in buffer equations, see here.