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
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