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27 October, 20:09

A 0.100 m solution of a monoprotic weak acid has a ph of 3.00. what is the pka of this acid?

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  1. 27 October, 22:37
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    Answer is: pKa for the monoprotic acid is 5.

    Chemical reaction: HA (aq) ⇄ A⁻ (aq) + H⁺ (aq).

    c (monoprotic acid) = 0.100 M.

    pH = 3.00.

    [A ⁻] = [H⁺] = 10∧ (-3).

    [A ⁻] = [H⁺] = 0.001 M; equilibrium concentration.

    [HA] = 0.1 M - 0.001 M.

    [HA] = 0.099 M.

    Ka = [A ⁻]·[H⁺] / [HA].

    Ka = (0.001 M) ² / 0.099 M.

    Ka = 0.00001 M = 1.0·10 ⁻⁵ M.

    pKa = - logKa = 4.99.
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