# Why do fluoride salts give basic solutions in water?

May 16, 2017

Because ${F}^{-}$ is an effective base.

#### Explanation:

We interrogate the equilibrium:

$B a {X}_{2} \left(s\right) \stackrel{{H}_{2} O}{r} i g h t \le f t h a r p \infty n s B {a}^{+} + 2 {X}^{-}$

Consider the basicity of the counterion, ${X}^{-}$. The fluoride is a reasonable base, and addition of ${H}_{3} {O}^{+}$ to solution will protonate the fluoride to give $H F \left(a q\right)$, which drives the solubility reaction to the right as written. On the other hand, $C {l}^{-}$ is not an effective base, and does not contribute to solution $p H$ by hydrolysis,

For both salts we examine the solubility product:

${K}_{\text{sp}} = \left[B {a}^{2 +}\right] {\left[{X}^{-}\right]}^{2}$.

While I have few data available, the fluoride salt is soluble to the extent of $0.16 \cdot \text{g per 100 mL}$ at $20$ ""^@C.

May 16, 2017

Both ${\text{BaF}}_{2}$ and ${\text{BaCl}}_{2}$ are soluble in water, but ${\text{F}}^{-}$ is a weak base while ${\text{Cl}}^{-}$ is essentially $\text{pH}$-neutral.

If you decrease the $\text{pH}$ of the solution, you increase $\left[{\text{H}}^{+}\right]$. Recall that the dissolution for ${\text{BaF}}_{2}$ is written as:

${\text{BaF"_2(s) rightleftharpoons "Ba"^(2+)(aq) + 2"F}}^{-} \left(a q\right)$

with ${K}_{s p} = {\left[{\text{Ba"^(2+)]["F}}^{-}\right]}^{2} \approx 3.04 \times {10}^{- 6}$ (the ${K}_{s p}$ was derived from the $\text{g/100 mL}$ solubility on Wikipedia).

But ${\text{F}}^{-}$ has its own equilibrium:

$\text{F"^(-)(aq) + "H"^(+)(aq) rightleftharpoons "HF} \left(a q\right)$

with ${K}_{c} = \left(\left[{\text{HF"])/(["H"^(+)]["F}}^{-}\right]\right)$.

Thus, if the $\text{pH}$ is lower, this equilibrium is skewed towards $\text{HF}$, and $\left[{\text{F}}^{-}\right]$ has decreased. We could then say that the reaction quotient for ${\text{BaF}}_{2}$, ${Q}_{s p}$, is now less than ${K}_{s p}$.

As a result, by Le Chatelier's principle, ${\text{BaF}}_{2}$ dissociates more, since the equilibrium wishes to shift towards the products. Since ${\text{BaF}}_{2}$ dissociates more at lower $\text{pH}$, a more acidic solution increases the solubility of $\boldsymbol{{\text{BaF}}_{2}}$.

The same cannot be said about ${\text{BaCl}}_{2}$, because $\text{HCl}$ is a strong acid, and thus, lowering the $\text{pH}$ does not significantly decrease $\left[{\text{Cl}}^{-}\right]$. That means ${\text{BaCl}}_{2}$ essentially does not dissociate any more, and to a good approximation, its solubility does not change.