An acid-base reaction is shown below. What is the value of Krxn?

Question

An acid-base reaction is shown below. What is the value of Krxn?

OH1- (aq) + HF(aq) --> F1- (aq) + H2O(l)

A) 2.1 x 10-4

B) 1.0 x 107

C) 7.2 x 1010

D) 7.2 x 10-4

E) 2.5 x 10-11

B is the correct answer but I'm having trouble getting that answer again

I know Krxn is ka(reacting acid)/ Ka(product)

Answer 1

Hi Rachel. Try looking at it this way.

First note that, just as you have written it, an equilibrium constant is always written as products over reactants, by convention in the normally favored direction.

K_rxn = [F^-] * [H₂O] / ( [OH^-] * [HF] ) = [F^-] / ( [OH^-] * [HF] )

Note that in dilute aqueous solution reactions, the amount of water used up or formed (denominator or numerator) is negligible compared to the amount of solvent. Hence the [H₂O] is essentially a constant, and gets included in the K_rxn .

Now since water and OH^- are involved, we use the Kw for water as

Kw = [H⁺] * [OH^-] / [H₂O] = [H⁺] * [OH^-] = 1 * 10^-7

{note that water drops out again for the same reason.)

Rearranging for the hydroxide gives

[OH^-] = 1 * 10^-7 / [H⁺]

Substitute this [OH-] into the K_rxn equation:

K_rxn = [F^-] = [F^-] * [H⁺] = [HF] = 1 * 10⁺⁷

[OH^-] * [HF] [HF] 1 * 10^-7 [HF] * 1*10^-7

The [HF]'s cancel out and we have our answer. ................ ;-)