The 𝐾a of a monoprotic weak acid is 0.00289. What is the percent ionization of a 0.179 M solution of this acid?

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  • 4 weeks ago

    HA <--> H+ + A-

    Ka = 2.89X10^-3 = [H+][A-]/[HA]

    Let [H+] = [A-] = x, and [HA] = 0.179 - x. As a first approximation, assume that x<<0.179 and so can be ignored in the denominator. Then,

    2.89X10^-3 = x^2 / 0.179

    x = 0.0227

    Ignoring x compared to 0.179 is not appropriate in this situation. So, you will have to go back to:

    2.89X10^-3 = x^2 / (0.179-x)

    Rearrange this into the form ax^2 + bx + c = 0 and use the quadratic formula to solve for x.

    Once you find x, the % ionization is calculated as:

    % ionization = (x / 0.179) X 100

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