Calculate the ph of a 0.060 m carbonic acid solution, h2co3 (aq), that has the stepwise dissociation constants ka1 = 4.3 * 10-7 and ka2 = 5.6 * 10-11.

Question

Camden Lee

Chemistry

8 March, 10:56

Calculate the ph of a 0.060 m carbonic acid solution, h2co3 (aq), that has the stepwise dissociation constants ka1 = 4.3 * 10-7 and ka2 = 5.6 * 10-11.

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Answers (1)

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Rose Vincent · Answer 1

1) First dissociation

H2 CO3 = H (+) + HCO3 (-) Ka1 = 4.3 * 10^ - 7

0.06 - x x x

Ka1 = x^2 / (0.06 - x) = 4.3 * 10^ - 7

A low Ka = > x < 0.06 - x ≈ 0.06

=> Ka1 ≈ x^2 / 0.06 = > x^2 ≈ 0.06 * Ka1 = 0.06 * 4.3 * 10^-7

=> x ≈ √[ 2.58 * 10 ^ - 8] = 1.606 * 10^ - 4 = 0.0001606

2) Second dissociation

HCO3 (-) = H (+) + CO3 (2-) Ka2 = 5.6 * 10^ - 11

0.0001606 - y y y

Ka2 ≈ y^2 / 0.0001606 = > y = √[0.0001606 * 5.6 * 10^ - 11]

y = 9.48 * 10^ - 8

3) [H+] = x + y = 1.607 * 10^ - 4

4) pH = - log [H+] = 3.79

Answer: 3.79