Please help!

Question

If equal volumes of the following are combined, would they form buffered solutions? Why or why not?

.1M HF and .05M NaF

I'm am a bit confused about how I can figure out if these will form buffers or not, so I'd really appreciate it if someone could explain it to me. Thanks.

Answer 1

I answered the similar question below. You want a weak acid and a salt of the weak acid OR a weak base and a salt of the weak base. I see no reason why this would not be a reasonably good buffer. Why don't you try it. I'll get you started.
100 mL x 0.1M HF = 10 mmols
100 mL x 0.05M NaF = 5 mmols.
We will add 5 mL of 0.1M HCl or 5 mL of 0.1M NaOH.
When we start the pH of the original solution will be
pH = 3.17 + log (5/10) = 2.87

For the HCl addition we have
..........F^- + H^+ ==. HF
I.........5.....0........10
add.............0.5...........
C........-0.5..-0.5......+0.5
E........4.5.....0........10.5
pH = 3.17 + log(4.5/10.5) = 2.80 which looks pretty good when compared to 2.87 initially.

Now try the other one; i.e., with the NaOH addition.
..........HF + OH^- ==> F^- + H2O
I.........10....0.......5
add............0.5...........
C......-0.5...-0.5......+0.5
E.......9.5...0.......5.5

pH = 3.17 + log (5.5/9.5) = 2.93 which again looks pretty good when compared to the initial value of 2.87

Hope this helps.

Answer 2

Okay, thank you Dr. Bob. So just to confirm, it does not matter that the weak acid and its conjugate base are added in different amounts?

Answer 3

So the answer is that this will form a buffer?

Answer 4

Yes to both. The weak acid and its salt (or the weak base and its salt) do not have to be the same concentration. It is best if they are because that gives you the best possible buffering capability but any reasonable values will work. Of course a VERY SMALL value for one of them makes the buffering action less useful.
And yes to your next add on,0.1M HF and 0.05M NaF is a buffer solution.