Question

Hi,
I have been doing review for my upcoming Chem exam and was going through sample exams for practice when I stumbled upon this question:

In the equilibrium below, which is true if NaOH(aq) is added to the system?

2 HF(aq) + Ba(s) ⇌ BaF2(aq) + H2(g)

Equilibrium Shift | pH
(A) left | decrease
(B) left | increase
(C) right | decrease
(D) right | increase

I picked D) but apparently the answer was B); I cannot understand why that is the right answer?
I'll explain below this post as to why I chose D):

Answer 1

(Continued)

I picked D) as the answer because I thought that since a base is being added to what I assume to be the reactant side, wouldn't the equilibrium shift to the right to balance it out? Or is this a buffer solution? I already knew the pH would increase since a strong base is being added to a weak acid (again I am assuming its being added to the reactant side of the equation).

The sample exam does not explain the answers for multiple choice type questions so any help as to why B) is the answer would be greatly appreciated!

Thanks!

Answer 2

Sorry, I'm not entirely sure how to answer this - what you said would have been my first thought as well. However, I wouldn't think about NaOH being added to the reactant side. It's being added to the entire system, meaning it will release NA+ and OH- ions. Perhaps the correct answer has to do with how these ions react with each of the species present?

Answer 3

First of all, you can't think of the NaOH being added "to the reactant side." You have an equilbrium, which means both forward and reverse reactions are possible, so that means EITHER side of that equation can be regarded as "the reactant" side. Best regard the NaOH as being added to the mix, and able to react with anything that's present, whichever side of the reaction it appears on.

Second, it sounds like you are thinking Le Chatelier, and in general that's a good guide. Since adding NaOH would immediately act to raise the pH, it's reasonable to assume the equilibrium will shift in such a way as to moderate that shift -- to lower the pH again. Well, which way is that? Clearly if it shifts to the LEFT you replace a salt (BaF2) with an acid, so you lower the pH. So Le Chatelier already points to B.

Another way to look at is that there are two equilibria actually going on here:

H+(aq) + Ba(s) <-> Ba^2+(aq) + H2(g)

\[K = \frac{[Ba^{2+}]}{[H^{+}]}\]
and
H+(aq) + OH-(aq) <-> H2O(l)

with K = K_w, the usual autodissociation of water reaction.

When you add the OH-, it acts only through the second equilibrium, since it doesn't participate in the first. What does it do? Clearly, since K_w has to stay constant, it is necessary for [H+] to decrease if [OH-] goes up. Now turn to the first equilibrium: if [H+] goes down, then to keep that K constant, [Ba^2+] has to drop as well -- that is, the equilibrium must shift left.

Still another way to look at it is that because the added [OH-] acts through the water autoionization equilbrium, what you are effectively doing is taking H+ out of the system. If you look at the original equilibrium and ask yourself what Le Chatelier would predict if you pull H+ out, what would you say? That the equilibrium shifts left, right?

I think the key realization you need here is that the added OH- acts through the water autodissociation equilibrium, since it doesn't directly participate in the reaction you're given. And it's better to work this out by considering the two equilibria in balance, each satisfying at all times their K equations, than to try to take the "shortcut" of guessing how the pH changes and trying to infer what that means for the reaction. Focussing on the pH obscures the underlying reality of the second equilibrium (the water autodissociation) that's important here.

Answer 4

In trying to understand, is it essentially the base actually decreasing the concentration of the acid and the equilibrium trying to balance this?

Answer 5

Something like that, yes. The base decreases [H+] because the water autoionization equilibrium must be maintained ([H+][OH-] = K_w always). The decrease in [H+] then shifts the first equilibrium to the left, because in that equilibrium [Ba^2+]/[H+] must equal a constant always.

Answer 6

I see, well that clears it up a great deal :)

Thank you very much for the help, all answers were very informative!