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28 February, 10:17

Calculate the percent ionization of 0.140 m lactic acid in a solution containing 8.5*10-3 m sodium lactate.

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  1. 28 February, 13:13
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    Chemical reaction 1: C₃H₆O₃ (aq) ⇄ C₃H₅O₃⁻ (aq) + H⁺ (aq).

    Chemical reaction 2: C₃H₅O₃Na (aq) → C₃H₅O₃⁻ (aq) + Na⁺ (aq).

    c (lactic acid - C₃H₆O₃) = 0,140 M.

    c (sodim lactate - C₃H₅O₃Na) = 8,5·10⁻³ M.

    Ka (C₃H₆O₃) = 1,4·10⁻⁴.

    [H⁺] = x.

    [C₃H₅O₃⁻] = 0,0085 M + x.

    [C₃H₆O₃] = 0,140 M - x.

    Ka = [C₃H₅O₃⁻] · [H⁺] / [C₃H₆O₃].

    1,4·10⁻⁴ = (0,0085 M + x) · x / (0,140 M - x).

    1,4·10⁻⁴ = 0,0085x + x² / (0,140 - x).

    x = 0,00188 M.

    percent ionization of lactic acid:

    α = 0,00188 M : 0,140 M · 100% = 1,35%.
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