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

Nov 21, 2016

You have the measurements, but you must use the definitions of $p H$ and $p O H$. On the basis of the data given, you cannot measure $p H$ for the basic sodium fluoride solution.

#### Explanation:

Water dissociates according to the following equilibrium:

$2 {H}_{2} O r i g h t \le f t h a r p \infty n s {H}_{3} {O}^{+} + H {O}^{-}$

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

${K}_{w} = \frac{\left[{H}_{3} {O}^{+}\right] \left[H {O}^{-}\right]}{{\left[{H}_{2} O\right]}^{2}}$

And we can simplify this to
${K}_{w} = \left[{H}_{3} {O}^{+}\right] \left[H {O}^{-}\right]$ because $\left[{H}_{2} O\right]$ is effectively constant.

By careful measurement we know that ${K}_{w} = {10}^{- 14}$ at $298 \cdot K$

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:

${\log}_{10} {K}_{w}$ $=$ log_10[H_3O^+] +log_10[""^(-)OH]

But by definition $- {\log}_{10} \left[{H}_{3} {O}^{+}\right] = p H$ and $- {\log}_{10} \left[H {O}^{-}\right] = p O H$.

And also by definition $p {K}_{w} = - {\log}_{10} {K}_{w}$ $= - {\log}_{10} {10}^{- 14} = 14.$

So we get our defining relationship:

$p H + p O H = 14$, 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, $= {10}^{- 3.17}$.