Answer to Question #31346 in Other Chemistry for Robert
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.
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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