Question

Calculate the ph of a solution that is 0.225m benzoic acid and 0.160m sodium benzoate, a salt whose anion is the conjugate base of benzoic acid.

Answer 1

Answer is: pH of benzoic acid is 4.052.
c(C₆H₅COOH) = ck = 0.225 M.
c(C₆H₅COO⁻) = cs = 0.160 M.
pKa(C₆H₅COOH) = 4.2.
Henderson–Hasselbalch equation for buffers: pH = pKa + log(cs/ck).
pH = 4.2 + log(0.160 M / 0.225 M).
pH = 4.2 - 0.148.
pH = 4.052.