# 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^-

May 28, 2017

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

#### Explanation:

$H F$ is a WEAK acid in aqueous solution.......i.e.

$H F \left(a q\right) + {H}_{2} O \left(l\right) r i g h t \le f t h a r p \infty n s {F}^{-} + {H}_{3} {O}^{+}$

${K}_{a} = {10}^{- 3.17} = 6.76 \times {10}^{-} 4$.

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 $H C l$, $H B r$, $H I$, for which in aqueous solution, the equilibrium LIES strongly to the right.

$H X \left(a q\right) + {H}_{2} O \left(l\right) \rightarrow {X}^{-} + {H}_{3} {O}^{+}$
$X \ne F , X = C l , B r , I$

And thus while $C {l}^{-}$, $B {r}^{-}$, and ${I}^{-}$ ARE WEAK Bronsted bases, ${F}^{-}$, being smaller and more charge dense, is a moderately strong Bronsted base.