# Would distilling a sodium fluoride solution generate hydrogen fluoride to any noticeable extent?

If you have a solution of NaF, and you boil it (to get pure water by distillation) can the fluoride ions in water separate from sodium ions and combine with hydrogen ions to form HF? If yes, how much of the molecules vaporized is pure water and how much is HF?

• The solubility of sodium fluoride in water is about 40 g/L. Distill off enough water and sodium fluoride will start to precipitate. That's it. – Klaus-Dieter Warzecha Nov 19 '16 at 20:52

For all intents and purposes, you can consider a solution of sodium fluoride that of a weak base in water. Remember the formula to calculate an acid constant $K_\mathrm{a}$:

$$\begin{gather}\ce{HF + H2O <=> F- + H3O+}\tag{1}\\ K_a = \frac{[\ce{F-}][\ce{H3O+}]}{[\ce{HF}]}\tag{2}\end{gather}$$

For convenience, the acid constant is typically given in logarithmic form as $\mathrm{p}K_\mathrm{a}$; Wikipedia gives us the following value for $\ce{HF}$:

$$\mathrm{p}K_\mathrm{a} (\ce{HF}) = 3.17\tag{3}$$

If we use equations $(2)$ and $(4)$ (the ion product of water) plus a little maths, that allows us to calculate the concentration of $\ce{HF}$ under the assumption of a certain initial amount of $\ce{NaF}$ (e.g. $1~\mathrm{mol\cdot l^{-1}}$). $c_0$ is my notation for the inital concentration (i.e. before any changes).

\begin{align}K_\mathrm{w} = [\ce{H3O+}][\ce{OH-}] &= 10^{-14}~\mathrm{mol^2 \cdot l^{-2}}\tag{4}\\ \Longrightarrow c_0(\ce{H3O+}) &= 10^{-7}~\mathrm{mol\cdot l^{-1}}\tag{4'}\end{align}

From equation $(1)$, we can derive equations $(5)$ and $(6)$, linking together our unknowns to give us the big picture:

\begin{align}[\ce{H3O+}] &= c_0(\ce{H3O+}) - [\ce{HF}]\tag{5}\\ [\ce{HF}] &= c_0(\ce{F-}) - [\ce{F-}]\tag{6}\\ [\ce{H3O+}] &= c_0(\ce{H3O+}) - c_0(\ce{F-}) + [\ce{F-}]\tag{5'}\end{align}

\begin{align}K_\mathrm{a} &= \frac{[\ce{H3O+}][\ce{F-}]}{[\ce{HF}]}\tag{7.1}\\ K_\mathrm{a} [\ce{HF}] &= [\ce{H3O+}][\ce{F-}]\tag{7.2}\\ K_\mathrm{a} \left(c_0(\ce{F-}) - [\ce{F-}]\right) &= \left(c_0(\ce{H3O+}) - c_0(\ce{F-}) + [\ce{F-}]\right) [\ce{F-}]\tag{7.3} \end{align}

Henceforth, any $c_0$ appearing will be $c_0(\ce{F-})$ and $c_0(\ce{H3O+})$ will be replaced by $10^{-7}$ — ignoring the unit.

\begin{align}K_\mathrm{a}c_0 - K_\mathrm{a}[\ce{F-}] &= 10^{-7}[\ce{F-}] - c_0[\ce{F-}] + [\ce{F-}]^2\tag{7.4}\\ 0 &= [\ce{F-}]^2 + \left(10^{-7} - c_0 + K_\mathrm{a}\right)[\ce{F-}] - K_\mathrm{a}c_0\tag{7.5}\end{align}

Thankfully, the equation is quadratic and we can solve it for $[\ce{F-}]_{1/2}$:

\begin{align}[\ce{F-}]_{1/2} &= \frac{- \left(10^{-7} - c_0 + K_\mathrm{a}\right) \pm \sqrt{\left(10^{-7} - c_0 + K_\mathrm{a}\right)^2 + 4 K_\mathrm{a}c_0}}{2}\tag{8.1}\\ [\ce{F-}]_{1/2} &= \frac{1 - 10^{-7} - 10^{-3.17} \pm \sqrt{\left(10^{-7} - 1 + 10^{-3.17}\right)^2 + 4 \times 10^{-3.17}}}{2}\tag{8.2}\\ [\ce{F-}]_{1/2} &= \frac{ 0.9993 \pm \sqrt{0.9986 +0.0027}}{2}\tag{8.3}\\ [\ce{F-}]_{1/2} &= \frac{ 0.9993 \pm \sqrt{1.0013}}{2}\tag{8.4}\\ [\ce{F-}]_{1/2} &= \frac{ 0.9993 \pm 1.0006}{2}\tag{8.5}\end{align}

As always, we need to perform a reality check. Subtracting is obviously not an option, as it would lead to a negative concentration. Thus we must add.

\begin{align}[\ce{F-}] &= \frac{ 0.9993 + 1.0006}{2}\tag{8.6}\\ [\ce{F-}] &= \frac{1.9999998}{2}\tag{8.7}\\ [\ce{F-}] &= 0.9999999\tag{8.8}\end{align}

So of the $1~\mathrm{mol \cdot l^{-1}}$ solution of sodium fluoride we have, a $10^{-7}~\mathrm{mol \cdot l^{-1}}$ solution of $\ce{HF}$ will result. (By the way, the units check out; we do indeed get $\mathrm{mol \cdot l^{-1}}$ at the end.)

You get similar results at other concentrations; if your initial fluoride concentration was $10^{-3}~\mathrm{mol \cdot l^{-1}}$, the resulting $\ce{HF}$ concentration will be $5.966 \times 10^{-8}~\mathrm{mol \cdot l^{-1}}$. There is no need to fear (or hope for) any $\ce{HF}$ development during distillation.