[noppawit]

You have 500.0 mL of a buffer solution containing 0.20M acetic acid (CH₃COOH) and 0.30M sodium acetate (CH₃COONa). What will the pH of this solution be after the addition of 20.0 mL of 1.00M NaOH solution? [K_a=1.8*10^-5)

Am I correct if my solution is:

pH = -log(K)+log([conjugate base]/[acid])

M_{acetic acid 1}V₁ = M_{acetic acid 2}V₂
(0.2)(0.5) = M_{acetic acid 2}(0.52)
M_{acetic acid 2} = 0.19

M_{sodium acetate 1}V₁ = M_{sodium acetate2}V₂
(0.3)(0.5) = M_{sodium acetate 2}(0.52)
M_{sodium acetate 2} = 0.29

pH = -log(1.8*10^-5)+log([0.29/[0.19])

Then, pH = 4.93

? Correct or not ?

Thank you

[n4k]

pH=-lgK-lg([conjugate base]/[acid])

[Borek]

M_{acetic acid 1}V₁ = M_{acetic acid 2}V₂

It is not about dilution, it is about neutralization. Assume acid was neutralized stoichiometrically by the added base.

Note, that dilution doesn't matter, as volumes cancel out and only thing that is important is ratio of numbers of moles.

[Borek]

pH=-lgK-lg([conjugate base]/[acid])

No, +log(). See

http://www.chembuddy.com/?left=pH-calculation&right=pH-buffers-henderson-hasselbalch

Also - if it will be -log() increasing concentration of conjugated base you will decrease pH, and increasing concentration of acid you will increase pH. Obviously something is wrong.