Calculating the pH of a solution containg a weak acid and a weak base.?

Question

A buffer contains 0.309M of HF and 0.275M NaF.
The Ka of HF is 6.8*10^-4
What is the pH of the buffer?

How do I set this up? I know how to calculate the pH of a solution containing a weak acid and its conjugate base, how is this done differently or is it?

So I could then use
pH = -log(6.8*10^-4)-log(.309/.275) = 3.21

Is that right?

Answer 1

1) The equilibrium equation in buffer solution is:

HF + H2O + NaF <<===>> H3O(+) + Na(+) + 2 F(-)

2) So, you can use the Henderson - Hasselbach expression to calculate pH:

pH = pKa + log ([Base] / [Acid])

in this case, Base is F(-) and the Acid is HF

pH = pKa + log( [F-] / [HF])

we know all the data:

pKa = -log (Ka) = -log (6.8 x 10^-4) = 3.16
[F-] = 0.275M
[HF] = 0.309M

3) Substitute the given values in Henderson -Hasselbach expression:

pH = 3.16 + log{ (0.275 M)/(0.309M) }

pH = 3.16 - 0.0506 = 3.109

pH = 3.109

NOTE: It is important to note that this buffer is quite "acid" because HF is not really a weak acid (although NaF is a weak base). Hendersson - Hasselbach just give you an approximation (a good one) and could have a bias when pH of solution is a couple of units away of neutrality.

That´s It!

Good luck!

Answer 2

yup. Your answer is correct.

Answer 3

no it is not done any different