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.
1
Expert's answer
2013-06-20T12:07:52-0400
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
Comments
Leave a comment