Question

(4 pts) What is the pH of the a buffer solution containing 0.10 moles of acetic acid and 0.12 moles of sodium acetate in 45 ml after adding 0.030 moles of KOH? 3. 0.049L Acenc acid + KOH CH3CO0K t H20 HOE

Answer 1

mol of KOH added = 0.03 mol

CH3COOH will react with OH- to form CH3COO-

Before Reaction:

mol of CH3COO- = 0.12 mol

mol of CH3COOH = 0.1 mol

after reaction,

mol of CH3COO- = mol present initially + mol added

mol of CH3COO- = (0.12 + 0.03) mol

mol of CH3COO- = 0.15 mol

mol of CH3COOH = mol present initially - mol added

mol of CH3COOH = (0.1 - 0.03) mol

mol of CH3COOH = 0.07 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 {0.15/7*10^-2}

= 5.076

Answer: 5.08