The pH of a 0.0666 M solution of a weak acid is 2.529. What is the Ka of the acid?
Answer:
First, let use assume that the acid has a formula HA. Hence, HA dissociates partially in water to form H+ and A- ions.
HA <---> H+ + A-
Therefore, we can write the Ka expression as follows:
Ka = [H+]x[A-]/[HA]
Now, a little bit of thinking: if 1 mole of HA dissociates, 1 mole of each H+ and A- will be formed. By using this reasoning, [H+]=[A-]
So, the expression can be rewritten as Ka= [H+]^2 / [HA]
Now, let us find [H+]
The pH of the weak acid is 2.529. So, since pH = -lg[H+]
-lg[H+] = 2.529
[H+] = 10^-2.529
[H+] = 2.958 x10^-3 M
Now, just substitute in this value of [H+] and the value of [HA] given in the question (0.0666 M).
Ka= (2.958x10^-3)^2 / 0.0666
Ka= 1.31x10^-4 M (Ans)
Hope this helps =)
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