Question

What is the pH when 10.0 mL of 0.20 M potassium hydroxide is added to 30.0 mL of 0.10 M cinnamic acid, HC9H7O2 (Ka = 3.6 × 10–5)

Answer 1

Given:

M(HC9H7O2) = 0.1 M

V(HC9H7O2) = 30 mL

M(KOH) = 0.2 M

V(KOH) = 10 mL

mol(HC9H7O2) = M(HC9H7O2) * V(HC9H7O2)

mol(HC9H7O2) = 0.1 M * 30 mL = 3 mmol

mol(KOH) = M(KOH) * V(KOH)

mol(KOH) = 0.2 M * 10 mL = 2 mmol

We have:

mol(HC9H7O2) = 3 mmol

mol(KOH) = 2 mmol

2 mmol of both will react

excess HC9H7O2 remaining = 1 mmol

Volume of Solution = 30 + 10 = 40 mL

[HC9H7O2] = 1 mmol/40 mL = 0.025M

[C9H7O2-] = 2/40 = 0.05M

They form acidic buffer

acid is HC9H7O2

conjugate base is C9H7O2-

Ka = 3.6*10^-5

pKa = - log (Ka)

= - log(3.6*10^-5)

= 4.444

use:

pH = pKa + log {[conjugate base]/[acid]}

= 4.444+ log {5*10^-2/2.5*10^-2}

= 4.745

Answer: 4.745