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2 November, 20:17

Determine the percent ionization of a 0.215 m solution of benzoic acid.

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  1. 2 November, 23:58
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    1.69%

    Explanation:

    The percent ionization of a weak acid is the percent of the original acid that ionizes.

    % ionization = [ concentration of the ion at equilibrium / original concentration of acid ] * 100

    You may calculate the percent ionization of an acid of known concentration, from the equilibrium constant.

    So, to determine the percent ionization of a 0.215 M solution of benzoic acid, you must look for the equilibrium (dissociation or ionization) constant.

    Equilibrium constants depend on temperature. So, you must know the temperature.

    For this question, I will assume 25°C, for which you can find that the dissociation constant, Ka, for benzoic acid is 6.25*10⁻⁵.

    With that, you follow these steps:

    1. Write the ionization equation:

    C₆H₅CO₂H ⇄ C₆H₅CO₂⁻ + H⁺ (simplified version)

    2. Calculate the concentration of the ion C₆H₅CO₂⁻ at equilibrium

    ICE (initial, change, equilibrium) table:

    C₆H₅CO₂H ⇄ C₆H₅CO₂⁻ + H⁺

    I 0.215 0 0

    C - x + x + x

    E 0.215 - x x x

    Equilibrium expression:

    Ka = x² / 0.215 - x = 6.25*10⁻⁵

    Solve for x (assume 0.215 >> x)

    x² = 0.215 * 6.25*10⁻⁵ = 0.0000078125

    x = 0.00367 M

    If you do not make the assumption but solve the quadratic equation you will get x = 0.00363 M

    3. Calculate the percent ionization:

    With x = 0.00363 M (exact calculation)

    % = [0.00363 / 0.215] * 100 = 1.69%

    With x = 0.00367 M

    % = [0.00367 / 0.215] * 100 = 1.71%
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