Question

Why do fluoride salts give basic solutions in water?

Answer 1

Because `F^-` is an effective base.

Explanation:

We interrogate the equilibrium:

`BaX_2(s) stackrel(H_2O)rightleftharpoonsBa^(+) +2 X^-`

Consider the basicity of the counterion, `X^-`. The fluoride is a reasonable base, and addition of `H_3O^+` to solution will protonate the fluoride to give `HF(aq)`, which drives the solubility reaction to the right as written. On the other hand, `Cl^-` is not an effective base, and does not contribute to solution `pH` by hydrolysis,

For both salts we examine the solubility product:

`K_"sp"=[Ba^(2+)][X^-]^2`.

While I have few data available, the fluoride salt is soluble to the extent of `0.16*"g per 100 mL"` at `20` `""^@C`.

Answer 2

Both `"BaF"_2` and `"BaCl"_2` are soluble in water, but `"F"^(-)` is a weak base while `"Cl"^(-)` is essentially `"pH"`-neutral.

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If you decrease the `"pH"` of the solution, you increase `["H"^(+)]`. Recall that the dissolution for `"BaF"_2` is written as:

`"BaF"_2(s) rightleftharpoons "Ba"^(2+)(aq) + 2"F"^(-)(aq)`

with `K_(sp) = ["Ba"^(2+)]["F"^(-)]^2 ~~ 3.04 xx 10^(-6)` (the `K_(sp)` was derived from the `"g/100 mL"` solubility on Wikipedia).

But `"F"^(-)` has its own equilibrium:

`"F"^(-)(aq) + "H"^(+)(aq) rightleftharpoons "HF"(aq)`

with `K_c = (["HF"])/(["H"^(+)]["F"^(-)])`.

Thus, if the `"pH"` is lower, this equilibrium is skewed towards `"HF"`, and `["F"^(-)]` has decreased. We could then say that the reaction quotient for `"BaF"_2`, `Q_(sp)`, is now less than `K_(sp)`.

As a result, by Le Chatelier's principle, `"BaF"_2` dissociates more, since the equilibrium wishes to shift towards the products. Since `"BaF"_2` dissociates more at lower `"pH"`, a more acidic solution increases the solubility of `bb("BaF"_2)`.

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The same cannot be said about `"BaCl"_2`, because `"HCl"` is a strong acid, and thus, lowering the `"pH"` does not significantly decrease `["Cl"^(-)]`. That means `"BaCl"_2` essentially does not dissociate any more, and to a good approximation, its solubility does not change.