Question

How do you calculate the pH of a .40 solution of `Ba(OH)_2` when 25.0 mL is added to 250 mL of water?

Answer 1

Well, `pH=12-13.`

Explanation:

We first calculate `[Ba(OH)_2]="Moles of stuff"/"Volume of solution"`

`=(0.40*mol*L^-1xx25.0xx10^-3*mol*L^-1)/(275.0xx10^-3*L)`

`=0.0364*mol` with respect to `[Ba(OH)_2]`, i.e. `0.07272*mol*L^-1` with respect to `[HO^-]`.

`pOH=-log_10[HO^-]=-log_10(0.07272)=1.14`

`pH=14-pOH=14-1.14=12-13`.

Note that you might have been a bit careless with the wording of your question. You get a different answer when the solution is diluted to `250*mL` rather than when `25*mL` is added to `250*mL` water......