Answer:
Heres my attempt at this and hope it helps friend
.1372L * (.83M/L) = .114 mols propanoic acid
.06862L * (1.1M/L) = 0.0755 mols NaOH
using the henderson hasselbach equation
ph = pKa + log([A-]/[HA])
so
ph = 4.89 + log(.0755/.114) = 4.72
Explanation:
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