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13 November, 11:10

The salt formed by the reaction of the weak acid hydrocyanic acid, HCN, with the strong base potassium hydroxide is

potassium cyanide, KCN. What is the hydroxide ion concentration of a 0.255 M solution of potassium cyanide at 25 °C

given that the value of for hydrocyanic acid is 4.9 X 10-102

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  1. 13 November, 11:31
    0
    2.28 * 10^-3 mol/L

    Explanation:

    The equation for the equilibrium is

    CN^ - + H2O ⇌ HCN + OH^-

    Ka = 4.9 * 10^-10

    KaKb = Kw

    4.9 * 10^-10 Kb = 1.00 * 10^-14

    Kb = (1.00 * 10^-14) / (4.9 * 10^-10) = 2.05 * 10^-5

    Now, we can set up an ICE table

    CN^ - + H2O ⇌ HCN + OH^-

    I / (mol/L) 0.255 0 0

    C / (mol/L) - x + x + x

    E / (mol/L) 0.255 - x x x

    Ka = x^2 / (0.255 - x) = 2.05 * 10^-5

    Check for negligibility

    0.255 / (2.05 * 10^-5) = 12 000 > 400. ∴ x ≪ 0.255

    x^2 = 0.255 (2.05 * 10^-5) = 5.20 * 10^-6

    x = sqrt (5.20 * 10^-6) = 2.28 * 10^-3

    [OH^-] = x mol/L = 2.28 * 10^-3 mol/L
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