Question

Quest. 7 (30 pts). Benzoic acid (C6H5COOH), 40.0 mL of 0.150 M, is being titrated with 0.300 M potassium hydroxide (KOH). Calculate the pH a) before any addition of base, b) at the half neutralization (half equivalent) point, and c) at the equivalent point. The Ka of C6H5COOH is 6.5 x 10-5. Use the appropriate tables when necessary.

Answer 1

a) initially only benzoic acid present

for pH of weak acids we have direct formula

pH = 1/2 [pKa - log C]

pKa = - log Ka = - log 6.5 x 10^-5 = 4.19

C = 0.150 M

pH = 1/2 [4.19 - log 0.150]

pH = 2.51

b) half equivalence point means 50% acid convert to salt

at half equivalence point we have pH = pKa

so

pH = 4.19

c) millimoles of acid = 40.0 x 0.150 = 6.0

6.0 millimoles KOH must be added to reach equivalence point

6.0 = V x 0.300

V = 20.0 mL

total volume = 40 + 20 = 60 mL

[salt] = 6.0 / 60 = 0.10 M

at equivalence point

pH = 1/2 [pKw + pKa + log C]

pH = 1/2 [14 + 4.19 + log 0.10]

pH = 8.60