[skibum143]

A buffer that is 0.324 M in acid, HA, and 0.152 M in the potassium salt of its conjugate base, KA, has a pH of 5.13. What is the pH of the buffer after 160. mL of 0.255 M LiOH is added to 1.150 L of this buffer? Assume that the volumes are additive.

I tried to solve this problem by first solving for the pKa using:
pKa = pH - log(base/acid)
pKa = 5.13 - log(.152/.324) = 5.4587

Then I used M1V1 = M2V2 to find the moles of the LiOH
.255*160 = 1310*M2
M2 = .0311
moles = M2*volume(1.31) = .0407

Then I added the moles of the LiOH to the moles of the base, and subtracted the moles of the LiOH from the moles of the acid, and then used the Henderson-Hasselbach equation to solve for pH:
(moles base = .152M*1.15L = .19912 moles) + .0407 = .23982 moles
(moles acid = .324M*1.15L = .4244 moles) - .0407 = .3837 moles

Then tried to solve for pH
pH = 5.4587 + log (.23982/.3837) = 5.25

But this is wrong, I'm not sure where I'm going wrong?

[Borek]

Check your math, you are on the right track.

Note that you don't have to calculate concentration of LiOH after dilution - number of moles is the same.

[skibum143]

I see, bad math. Thanks!