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25 March, 10:17

A 0.100 M solution of chloroacetic acid 1ClCH2COOH2 is 11.0% ionized. Using this information, calculate 3ClCH2COO-4, 3H 4, 3ClCH2COOH4, and Ka for chloroacetic acid.

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  1. 25 March, 11:23
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    [CICH2COOH] = 0.089 M

    [CICH2COO-] = 0.011 M

    [H+] = 0.011 M

    Ka = 1.36 x 10^-3

    Explanation:

    The reaction

    CICH2COOH ⇄ H + (aq) + CICH2COO - (aq)

    The initial concentration of CICH2COOH is 0.10 M, chloroacetic acid is 11.0% ionized.

    let us calculate Ka

    First, find change in concentration

    since, 11% ionized

    change in concentration = 0.10 X 11% = 0.011 M

    Initial Concentration of CICH2COOH = 0.10 M

    change in concentration of CICH2COOH = - 0.011 M

    Equilibrium Concentration of CICH2COOH = 0.10 M - 0.011 M = 0.089 m

    Initial Concentration of CICH2COO - = 0 M

    change in concentration of CICH2COO - = + 0.011 M

    Equilibrium Concentration of CICH2COO - = 0.011 M

    Initial Concentration of H + = 0 M

    change in concentration of H + = + 0.011 M

    Equilibrium Concentration of H + = 0.011 M

    Therefore,

    [CICH2COOH] = 0.089 M

    [CICH2COO-] = 0.011 M

    [H+] = 0.011 M

    Ka = [H+][CICH2COO-] / [CICH2COOH]

    Ka = (0.011 * 0.011) / (0.089)

    Ka = 1.36 x 10^-3
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