Buffers

Hi Alex,

For Question 12, I could not figure out exactly what the question was asking.  I think I'm having a problem understanding the concept of buffers.  This statement confuses me:  "A certain buffer solution is 3 M in HF, and 2 M in NaF."  So is the mixture of HF and F- the buffer??  Or is HF/F- the solution being buffered?   And if so, how can the equation Ka=HxF/HF be used?  Wouldn't you have to use the concentrations of the buffer mixture?

Sorry if this is confusing!  Thank you!
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The statement is saying that the buffer is composed of 3M HF and 2 M NaF.  Since NaF is a salt, and will completely dissociate, then the buffer is 3M HF and 2 M F-.

The equation here that you'd want to use is the Henderson-Hasselbalch equation.  Remember that this is the one that relates pH and the pKa of an acid (you can also use the Ka directly, but I think it's more math-heavy).  So, Henderson-Hasselbalch says pH = pKa + log [CB]/[A].  Here that would be
pH = -log(7 x 10^-4) + log [2/3].  The log of a difference is the division of logs:
log (2/3)/(7x10^-4) = log (2/21 x 10^4) = log (2/21) + 4.
log (2/21) is around log (1/10), which is -1.  Thus, the pH = 4-1 or 3.  The actual answer is 2.98, but this is plenty close.