Question
A solution of an unknown weak acid, HA, is titrated with 0.100 M NaOH solution. The
equivalence point is achieved when 36.12 mL of NaOH have been added. After the
equivalence point is reached, 18.06 mL of 0.100 M HCl are added to the solution and the pH
at that point is found to be 5.20. What is the pKa of this unknown acid?
equivalence point is achieved when 36.12 mL of NaOH have been added. After the
equivalence point is reached, 18.06 mL of 0.100 M HCl are added to the solution and the pH
at that point is found to be 5.20. What is the pKa of this unknown acid?
Answers
DrBob222
If I didn't mess up somewhere the pKa = 5.20. Interesting problem.
mmols NaOH added = 36.12 x 0.1 = 3.612
...........HA + NaOH ==> NaA + H2O
So at the equivalence point you have 3.612 mmols NaA in solution.
Now you add 18.06 mL x 0.1M = 1.806 mmols
..........A^- + H^+ ==> HA
I........3.612...0......0
add............1.806.......
C......-1.806.-1.806....+1.806
E........1.806....0.....+1.806
Use the Henderson-Hasselbalch equation and solve for pKa
5.20 = pKa + log (1.806)/(1.806)
pKa = ?
mmols NaOH added = 36.12 x 0.1 = 3.612
...........HA + NaOH ==> NaA + H2O
So at the equivalence point you have 3.612 mmols NaA in solution.
Now you add 18.06 mL x 0.1M = 1.806 mmols
..........A^- + H^+ ==> HA
I........3.612...0......0
add............1.806.......
C......-1.806.-1.806....+1.806
E........1.806....0.....+1.806
Use the Henderson-Hasselbalch equation and solve for pKa
5.20 = pKa + log (1.806)/(1.806)
pKa = ?
Anonymous
Thank you very much!!! Now it makes sense
Cris
Excellent job! Thanks a lot