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The pH of a 0.1 molar solution of the acid HQ is 3. The value of the ionization constant, Ka of this acid is ?

The pH of a 0.1 molar solution of the acid HQ is 3. The value of the ionization constant, Ka of this acid is ?

Grade:12

1 Answers

Avinash
askIITians Faculty 1241 Points
8 years ago
We start with the chemical reaction:
HA <=> H++ A-
for which we can write the usual equilibrium equation:

eq. 1 Ka= [H+][A-]/[HA]

The amount of undissociated acid is equal to the total amount of acid that we began with ("C") minus the amount that has dissociated:

eq. 2 [HA] = C - [A-]
We can substitute eq. 2 into eq. 1 to eliminate the [HA] term:

eq. 3 Ka= [H+][A-]/(C - [A-])
For such a high pH (3.8), it is a good assumption that virtually all of the H+comes from the added HA, so:

eq. 4 [H+] = [A-]
We can now substitute eq. 4 into eq. 3 to eliminate the [A-] term:

Ka= [H+][H+]/(C - [H+]) = [H+]2/(C - [H+])[H+] = 10-pH= 10-3 and C = 0.1 M, so

Ka= (10-3)2/(0.1 - 10-3) = 1.0 x 10-7

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