What is the mass of naoh that would have to be added to 500 ml of a solution of 0.20 m acetic acid in order to achieve a ph of 5.0?

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