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27 January, 12:38

What is the ph of a 0.45 m solution of aniline (c6h5nh2) ? (pkb  9.40) ?

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  1. 27 January, 13:05
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    Answer is: pH of aniline is 9.13.

    Chemical reaction: C ₆H₅NH₂ (aq) + H₂O (l) ⇌ C₆H₅NH₃⁺ (aq) + OH⁻ (aq).

    pKb (C₆H₅NH₂) = 9.40.

    Kb (C₆H₅NH₂) = 10∧ (-9.4) = 4·10⁻¹⁰.

    c₀ (C₆H₅NH₂) = 0.45 M.

    c (C₆H₅NH₃⁺) = c (OH⁻) = x.

    c (C₆H₅NH₂) = 0.45 M - x.

    Kb = c (C₆H₅NH₃⁺) · c (OH⁻) / c (C₆H₅NH₂).

    4·10⁻¹⁰ = x² / (0.45 M - x).

    Solve quadratic equation: x = c (OH⁻) = 0.0000134 M.

    pOH = - log (0.0000134 M.) = 4.87.

    pH = 14 - 4.87 = 9.13.
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