Question

How do I find `pH` and `pOH` for solutions of aqueous hydrogen fluoride or sodium fluoride?

Answer 1

You have the measurements, but you must use the definitions of `pH` and `pOH`. On the basis of the data given, you cannot measure `pH` for the basic sodium fluoride solution.

Explanation:

Water dissociates according to the following equilibrium:

`2H_2O rightleftharpoonsH_3O^+ +HO^-`

Now it is well known that for a given temperature, the extent of this autoprotolysis reaction is a constant:

`K_w=([H_3O^+][HO^-])/([H_2O]^2)`

And we can simplify this to
`K_w=[H_3O^+][HO^-]` because `[H_2O]` is effectively constant.

By careful measurement we know that `K_w=10^(-14)` at `298*K`

Now this is a mathematical expression, which we are free to manipulate, provided that we do it to BOTH sides of the equation. Now we can take `log_10` of both sides of the equation:

`log_10K_w` `=` `log_10[H_3O^+] +log_10[""^(-)OH]`

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

And also by definition `pK_w=-log_10K_w` `=-log_(10)10^(-14)=14.`

So we get our defining relationship:

`pH+pOH=14`, under the given conditions of your question. You should commit this relationship to memory. You won't be asked to derive it at A level.

So get an electronic calculator, and have a go at your table. You will need `K_a` for hydrofluoric acid, `=10^(-3.17)`.