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27 December, 14:57

The acid-dissociation constant of hydrocyanic acid (hcn) at 25.0 °c is 4.9 ⋅ 10-10. what is the ph of an aqueous solution of 0.080 m sodium cyanide (nacn) ?

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  1. 27 December, 16:36
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    According to the reaction equation:

    and by using ICE table:

    CN - + H2O ↔ HCN + OH-

    initial 0.08 0 0

    change - X + X + X

    Equ (0.08-X) X X

    so from the equilibrium equation, we can get Ka expression

    when Ka = [HCN] [OH-]/[CN-]

    when Ka = Kw/Kb

    = (1 x 10^-14) / (4.9 x 10^-10)

    = 2 x 10^-5

    So, by substitution:

    2 x 10^-5 = X^2 / (0.08 - X)

    X = 0.0013

    ∴ [OH] = X = 0.0013

    ∴ POH = - ㏒[OH]

    = - ㏒0.0013

    = 2.886

    ∴ PH = 14 - POH

    = 14 - 2.886 = 11.11
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