Help me out with this pH prob please?

Calculate the pH of a buffer that consists of 0.33 M HF and 0.63 M KF (Ka of HF = 6.8 ´ 10–4).

Answer:
pH = -log (Ka) + log (Base(M) / Acid (M))
pH = - log (6.8 * 10^-4) + log ( 0.63/0.33)
pH = 3.45

Base = KF
Acid = HF
Use the equation :
pH = pKa + log [F-] / [HF]
pKa = - log 6.8 x 10^-4 = 3.17
pH = 3.17 + lof 0.63 / 0.33 = 3.45
Base is KF
Acid is HF

pH = -log (Ka) + log (Base/ Acid)
pH = - log (6.8 * 10^-4) + log ( 0.63/0.33)
= 3.45

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