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7 January, 04:50

Calculate the ph of a 0.021 m nacn solution. [ka (hcn) = 4.9  10-10]

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  1. 7 January, 05:19
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    Answer is: pH of solution of sodium cyanide is 11.3.

    Chemical reaction 1: NaCN (aq) → CN ⁻ (aq) + Na⁺ (aq).

    Chemical reaction 2: CN ⁻ + H₂O (l) ⇄ HCN (aq) + OH⁻ (aq).

    c (NaCN) = c (CN ⁻) = 0.021 M.

    Ka (HCN) = 4.9·10 ⁻¹⁰.

    Kb (CN ⁻) = 10⁻¹⁴ : 4.9·10⁻¹⁰ = 2.04·10⁻⁵.

    Kb = [HCN] · [OH ⁻] / [CN⁻ ].

    [HCN] · [OH ⁻ ] = x.

    [CN ⁻ ] = 0.021 M - x ...

    2.04·10 ⁻⁵ = x² / (0.021 M - x).

    Solve quadratic equation: x = [OH ⁻ ] = 0.00198 M.

    pOH = - log (0.00198 M) = 2.70.

    pH = 14 - 2.70 = 11.3.
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