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A 0.2 M carboxylic acid (RCOOH) has a Ka = 1.66x10-6. What is the pH of this solution? Enter to 2 decimal places.

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  1. Today, 19:02
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    3.24

    Explanation:

    The dissociation equation for the carboxylic acid can be represented as follows:

    RCOOH - -> RCOO - + H+

    We can use an ICE table to get the value of the concentration of the hydrogen ion. ICE stands for initial, change and equilibrium.

    RCOOH RCOO - H+

    Initial 0.2 0.0. 0.0

    Change - x + x. + x

    Equilibrium 0.2-x. x. x

    We can now find the value of x as follows:

    Ka = [RCOO-][H+]/[RCOOH]

    (1.66 * 10^-6) = (x * x) / (0.2-x)

    (1.66 * 10^-6) (0.2-x) = x^2

    x^2 = (3.32 * 10^-7) - (1.66*10^-6) x

    x^2 + (1.66 * 10^-6) x - (3.32 * 10^-7) = 0

    Solving the quadratic equation to get x:

    x = 0.0005753650094369094 or - 0.0005753650094369094

    As concentration cannot be negative, we discard the negative answer

    Hence [H+] = 0.0005753650094369094

    By definition, pH = - log[H+]

    pH = - log (0.0005753650094369094)

    pH = 3.24
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