A 0.035 mol sample of a weak acid, HA, is dissolved in 601 mL of water...

4.4 Ratings (929 Votes)

Letâ€™s write out the acid dissociation:
HA â†” Aâ€“ + H+

Ka= [A-] [H+] / [HA]

The inital moles of acid = 0.035 moles

Moles of base added = moalrity X volume = 0.41 X 0.033 = 0.0135 moles

this much base will neutralize 0.0135 moles of acid

The equation of acid and base will be

Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â HAÂ Â Â Â Â Â Â +Â Â Â Â Â Â Â Â Â Â Â Â OH^{- Â Â} -->Â Â Â Â Â Â Â Â Â H2O + A-

Initial moleÂ Â Â Â Â Â Â Â Â 0.035Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â 0.0135Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â 0

ChangeÂ Â Â Â Â Â Â Â Â -0.0135Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â -0.0135Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â 0.0135

EquilibriumÂ Â Â Â Â Â 0.0215Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â 0Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â 0.0135

pH = pKa + log [salt /acid]

5.281 = pKa + log [0.0135 / 0.0215]

Pka = 5.281 - log(0.627) = 5.281 - (-0.202) = 5.483

pKa = -logKa

taking antilog

Ka = 3.28 X 10^-6

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