# College Chemistry (DrBob222)

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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?

• College Chemistry (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
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 = ?

• College Chemistry (DrBob222) - ,

Thank you very much!!! Now it makes sense

• College Chemistry (DrBob222) - ,

Excellent job! Thanks a lot