Ask Question
6 June, 12:24

For the diprotic weak acid h2a, ka1 = 2.2 * 10-6 m and ka2 = 8.2 * 10-9 m. what is the ph of a 0.0500 m soluti? for the diprotic weak acid h2a, ka1 = 2.2 * 10-6 m and ka2 = 8.2 * 10-9 m. what is the ph of a 0.0500 m solution of h2a? what are the equilibrium concentrations of h2a and a2 - in this solution?

+2
Answers (1)
  1. 6 June, 12:31
    0
    In the first dissociation of H2A:

    molarity H2A (aq) ↔ (HA) ^ - (aq) + H^ + (aq)

    initial 0.05 m 0 m 0 m

    change - x + x + x

    equilibrium 0.05-x x x

    we can neglect X in [H2A] as it so small compared to the 0.05

    so by substitution in Ka equation:

    Ka1 = [HA][H] / [H2A]

    2.2x10^-6 = X^2/0.05

    X = √ (2.2x10^-6) * (0.05) = 1.1x10^-7

    X = 3.32x10^-4 m

    ∴ [H2A] = 0.05 - 3.32x10^-4 = 0.0497 m

    [HA] = 3.32x10^-4 m

    [H] = 3.32x10^-4 m

    the second dissociation of H2A: when ka2 = 8.2x10^-9

    HA - (aq) ↔ A^2 - (aq) + H + (aq)

    at equilibrium 3.32x10^-4 y 3.32x10^-4

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

    8.2x10^-9 = Y (3.32x10^-4) / (3.32x10^-4)

    ∴y = 8.2x10^-9 m

    ∴[A] = 8.2x10^-9 m

    PH = - ㏒[H+]

    = - ㏒ (3.32x10^-4) = 3.479

    [A]=8.2x10^-9 m

    [H2A] = 0.0497 ≈ 0.05 m
Know the Answer?
Not Sure About the Answer?
Find an answer to your question 👍 “For the diprotic weak acid h2a, ka1 = 2.2 * 10-6 m and ka2 = 8.2 * 10-9 m. what is the ph of a 0.0500 m soluti? for the diprotic weak acid ...” in 📗 Chemistry if the answers seem to be not correct or there’s no answer. Try a smart search to find answers to similar questions.
Search for Other Answers