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3 June, 18:54

Determine the percent ionization of a 0.145 m hcn solution.

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  1. 3 June, 20:14
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    When Ka of HCN = 6.2x10^-10

    HCN H3O + CN-

    initial 0.145 0 0

    change - X + X + X

    final (0.145-X) + X + X

    Ka = [H3O+][CN] - / [HCN]

    by substitution:

    we can assume [HCN] = 0.145

    6.2x10^-10 = X*X / (0.145) by solving this equation,

    X = 9.5 x 10^-6

    ∴ [ H3O+] = 9.5 X 10^-6

    ∴percent ionization = [H3O+]/[HCN] * 100

    = 9.5X10^-6 / 0.145 * 100

    = 0.0066 %
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