Let volume of NaOH be V mL
mol of NaOH added = 5.90*V mmol
Before adding NaOH
Before Reaction:
mol of CH3COO- = 0.0280 M *650.0 mL
mol of CH3COO- = 18.20 mmol
mol of CH3COOH = 0.0205 M *650.0 mL
mol of CH3COOH = 13.325 mmol
5.9*V NaOH will react with 5.9*V of CH3COOH to form extra 5.9*V of base
After adding NaOH
mol of CH3COOH = 13.325-5.9*V mmol
mol of CH3COO- = 18.2+5.9*V mmol
Ka = 1.8*10^-5
pKa = - log (Ka)
= - log(1.8*10^-5)
= 4.7447
use:
pH = pKa + log {[conjugate base]/[acid]}
5.75 = 4.7447+log {[CH3COO-]/[CH3COOH]}
log {[CH3COO-]/[CH3COOH]} = 1.0053
[CH3COO-]/[CH3COOH] = 10.1221
So,
(18.2+5.9*V)/(13.325-5.9*V) = 10.1221
18.2+5.9*V = 134.8776 - 59.7206*V
(5.9+59.7206)*V = 134.8776-18.2
V = 1.778 mL
Answer: 1.778 mL
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