Question

A solution of potassium fluoride turns litmus paper blue. What are the products of the modified Arrhenius net ionic equation for this compound? Question options: `~H_2O` `~F^-` `~HF` `~H_3O^+` `~HKF^+` `~OH^-`

Answer 1

`F^(-) + H_2O(l) rightleftharpoonsHF(aq) + H_2O(l)`

Explanation:

`HF` is a WEAK acid in aqueous solution.......i.e.

`HF(aq)+H_2O(l)rightleftharpoonsF^(-) + H_3O^+`

`K_a=10^(-3.17)=6.76xx10^-4`.

If we describe something as a weak acid, we might also describe its conjugate base as reasonably strong in that it competes for the proton effectively. Fluoride thus gives rise to a basic solution, and this is in contrast to the solution behaviour of `HCl`, `HBr`, `HI`, for which in aqueous solution, the equilibrium LIES strongly to the right.

`HX(aq)+H_2O(l)rarrX^(-) + H_3O^+`
`X!=F, X=Cl, Br, I`

And thus while `Cl^-`, `Br^-`, and `I^-` ARE WEAK Bronsted bases, `F^-`, being smaller and more charge dense, is a moderately strong Bronsted base.