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^-#

1 Answer
May 28, 2017

Answer:

#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.