4.0 x 10 5.1 x 10-9 and Ka2 For the diprotic weak acid H2A, Kal What is the pH of a 0.0800 M solution of H, A? pH What...

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Question

4.0 x 10 5.1 x 10-9 and Ka2 For the diprotic weak acid H2A, Kal What is the pH of a 0.0800 M solution of H, A? pH What are th

Answer 1

Answer #1

The dissociation that occurs is:

H2A = H + + HA-

You have the expression of Ka1:

Ka1 = [H +] * [HA-] / [H2A]

4x10 ^ -6 = X ^ 2 / 0.08

It clears X = 5.7x10 ^ -4 M

The pH is calculated:

pH = - log 5.7x10 ^ -4 = 3.25

You have the second dissociation:

HA- = H + + A-2

You have the expression of Ka2:

Ka2 = [H +] * [A-2] / [HA-]

5.1x10 ^ -9 = 5.7x10 ^ -4 * [A-2] / 5.7x10 ^ -4

It clears [A-2] = 5.1x10 ^ -9 M

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Answer 2

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