Why do fluoride salts give basic solutions in water?

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#.

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

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)#.

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.