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17 January, 05:58

A solution of 2.9 m weak acid is 0.52 ionized. what is the ka value of this acid

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  1. 17 January, 06:13
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    We can get the Ka value from this formula:

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

    So we need to know [H+] & [A-] & [Ha] to get the Ka value

    [H+] = 2.9 * 0.52/100 = 0.01508 M

    and when [H+] = [A-]

    ∴[A-] = 0.01508 M

    ∴ [HA] = 2.9 - 0.01508 = 2.88 M

    So by substitution:

    Ka = [0.01508][0.01508] / 2.88

    = 7.9 x 10^-5
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