[billsauce]

You have 500 ml of an unknown buffer of concentration 0.4 M. The pH is 6.0. You
add 0.05 moles (no volume change) of NaOH to the mixture and discover there is twice
as much salt as acid in the mixture. What is the pKa of the unknown buffer?

I know that there is .2 mols total in solution, and salt + acid = .2

ph = pka + log(salt/acid)  the only thing i dont understand is when there is .05 moles NaOH added, it says there is a 2:1 salt to acid ratio. does this simply mean I plug like this:

6.0 = pka + log(2/1)?

[magician4]

I know that there is .2 mols total in solution, and salt + acid = .2

no , this is not correct: neither is the initial ratio [A^-] : [HA] given to you ( and therefore esp. not the information that those two should be equal!), nor is the sum of the two of them 0.2.
Quite the contrary: it's 0.4 mole/L  that's exactly what "buffer capacity" means, the sum of both

what I would do:

let's call [A^-] "x"
let's call [HA] "y"

when you add NaOH ( a stong base) to a buffer, HA will be consumed and A^- will be formed

therefore, we can say that the original ratio of x :y will change as follows:

[tex] \frac {x}{y}  \xrightarrow {0.05 \ mole \ NaOH}  \frac {x + 0.05 \ mole}{y - 0.05 \ mole} [/tex]

being told that by now you'll have twice as much A^- than HA, you can expand this result to the equation

[tex]  \frac {x + 0.05  \ mole}{y - 0.05 \ mole}  = \frac {2}{1}[/tex]
(this is eq. 1)

however, even with adding base (or acid), your given buffer capacity doesn't change, so at every moment another eq. ist valid, too :  x + y = 0.4 mole   (in one liter, or, 0.4 mole/L in general)
(this is eq. 2)

two equations, two unknowns: you can calculate x and y

after that, use the HH equation, put in x, y and the pH given to you, and rearrange / calculate for pKs


regards

Ingo