A 0.050 M solution of the weak acid HA has [H3O+] = 3.77 × 10-4 M. What is the Ka for the acid?

1 Answer

  • HA <–> H^+1 + A^-1

    Ka = [H+][A-]/[HA]

    [H+] = 3.77×10^-4 M

    therefore [A-] = 3.77×10^-4 M

    and [HA] = 0.050 – 3.77×10^-4 = 0.0496 M

    Ka = (3.77×10^-4)^2 / 0.0496 = 2.87 x 10^-6

    Ka = 2.9×10^-6

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