Acid/Base equilibrium calculations

Question

How many mol of solid LiA would have to be added to 1.0 L of a 3.726 M HA solution in order to obtain a buffer of 2.99? (The Ka for HA is ).

Answer 1

Henderson Hasselbalch equation: pH = pKa + log [salt]/[acid]

pKa = -log Ka = 5.23

2.99 = 5.23 + log [salt]/[3.726]

2.99 = 5.23 + log [salt] - log 3.726

2.99 = 5.23 + log [salt] - 0.5712

log [salt] = -1.669

[salt] = 0.02144 M = 0.021 moles LiA to be added to 1.0 L

Check:

pH = pKa + log [salt]/[acid]

pH = 5.23 + log (0.021/3.726) = 5.23 + log 0.0056

pH = 5.23 + (-2.25)

pH = 3 ~ 2.99 ... agrees with given pH