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Answer to Question #31346 in Other Chemistry for Robert

Question #31346
A 0.1645-g sample of an unknown monoprotic acid was dissolved in 16.7 mL of water and titrated with 0.0657 M NaOH solution. The volume of base required to bring the solution to the equivalence point was 18.9 mL.

After 10.1 mL of base had been added during the titration, the pH was determined to be 5.28. What is the Ka of the unknown acid? Molar mass of acid is 133 g/mol.
0.1645 / 133g/mole = 0.0012 moles HA, 0.0012 moles / 0.0167 L = 0.07 M HA
0.0167L H2O
titrant = 0.0657 M NaOH, 0.0189 L = 0.0015 moles
add 0.01 L of 0.0657 M NaOH = 0.000657 moles OH- added
if pH after 0.000657 moles OH- added = 5.28
[H+] = 5.25*10^-6 M so [OH-] = 1.91*10^-9 M
these are the amounts after HA reacted with KOH.
Ka = [H+][A-] = (5.25*10^-6)^2... = 2.76*10^-11

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