Question

What is the pH if 0.035 mol NaOH is added to 2.0L of a buffer containing 0.10 M acetic acid, CH3COOH, and 0.10 M sodium acetate, CH3COONa? (K, for CH3COOH - 1.8 x 10-5). O pH = 6.83 O pH = 7.91 O pH = 4.89 O pH = 6.23 O pH = 5.08

Answer 1

mol of NaOH added = 0.035 mol

CH3COOH will react with OH- to form CH3COO-

Before Reaction:
mol of CH3COO- = 0.1 M *2.0 L
mol of CH3COO- = 0.2 mol

mol of CH3COOH = 0.1 M *2.0 L
mol of CH3COOH = 0.2 mol

after reaction,
mol of CH3COO- = mol present initially + mol added
mol of CH3COO- = (0.2 + 0.035) mol
mol of CH3COO- = 0.235 mol

mol of CH3COOH = mol present initially - mol added
mol of CH3COOH = (0.2 - 0.035) mol
mol of CH3COOH = 0.165 mol


Ka = 1.8*10^-5

pKa = - log (Ka)
= - log(1.8*10^-5)
= 4.745

since volume is both in numerator and denominator, we can use mol instead of concentration
use:
pH = pKa + log {[conjugate base]/[acid]}
= 4.745+ log {0.235/0.165}
= 4.89

Answer: 4.89