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12 May, 19:23

The ka for hydrocyanic acid (hcn) is 4.9 * 10-10. what is [oh-] in a 0.075 m solution of sodium cyanide (nacn) ?

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  1. 12 May, 23:14
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    by using ICE table:

    CN - + H2O ↔ HCN + OH-

    initial 0.075 0 0

    change - X + X + X

    Equ (0.075-X) X X

    when Kb = Kw/Ka

    ∴Kb = 1 x 10^-14 / 4.9 x 10^-10

    = 2 x 10^-5

    and when Kb = [HCN][OH]/[CN-]

    by substitution:

    2 x 10^-5 = X^2 / (0.075 - X) by solving for X

    ∴ x = 0.0012 M

    ∴[OH] = 0.0012 M
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