The correct answer is 2.5 g.
Let x be the mass of NaOH in grams
The molar mass of NaOH = 40.0 gms/mol
Moles of NaOH = mass / molecular mass of NaOH
= x grams / 40.0 gms /mol = 0.025 x mol
Initial moles of CH₃COOH = volume × concentration of CH₃COOH
= 500 / 1000 × 0.20 = 0.1 mol
CH₃COOH + NaOH ⇒ CH₃COONa + H2O
Moles of CH₃COOH left = initial moles of CH₃COOH - moles of NaOH = 0.1 - 0.025x mol
Moles of CH₃COONa formed = moles of NaOH = 0.025x mol
Henderson-Hasselbalch equation:
pH = pKa + log ([CH₃COONa] / [CH₃COOH])
= pKa + log (moles of CH₃COONa /moles of CH₃COOH)
5.0 = 4.76 + log [0.025a / (0.1 - 0.025a)]
log [0.025a / (0.1 - 0.025a)] = 0.24
0.025a / (0.1 - 0.025a) = 10^0.24 = 1.738
0.068445a = 0.17378
a = 0.17378 / 0.068445
= 2.54 g
Mass of NaOH = a = 2.54 g or 2.5 g
