SELECT * FROM question_mgmt as q WHERE id=3262 AND status=1 SELECT id,question_no,question,chapter FROM question_mgmt as q WHERE courseId=2 AND subId=9 AND chapterId=48 and ex_no='1' AND status=1 ORDER BY CAST(question_no AS UNSIGNED)
It has been found that the pH of a 0.01M solution of an organic acid is 4.15. Calculate the concentration of the anion, the ionization constant of the acid and its pKa.
Let the organic acid be HA.
⇒ HA ↔ H+ + A-
Concentration of HA = 0.01 M
pH = 4.15
-log[H+] = 4.15
[H+] = 7.08x10-5
Now, Ka = [H+] [A-] / [HA]
[H+] = [A-] = 7.08x10-5
[HA] = 0.01
Then,
Ka = 7.08x10-5 x 7.08x10-5 / 0.01
Ka = 5.01x10-7
pKa = -log Ka
= -log(5.01x10-7)
pKa = 6.3001
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