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13 May, 18:54

A weak acid, HA, has a pKapKa of 4.3574.357. If a solution of this acid has a pH of 4.0054.005, what percentage of the acid is not ionized? Assume all H+H + in the solution came from the ionization of HA.

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  1. 13 May, 21:27
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    69.3%

    Explanation:

    The question should read as follows:

    A weak acid, HA, has a pKa of 4.357. If a solution of this acid has a pH of 4.005, what percentage of the acid is not ionized? Assume all H⁺ in the solution came from the ionization of HA.

    The Henderson-Hasselbalch equation relates the pKa and pH of a solution to the ratio of ionized (A⁻) and unionized (HA) forms of a weak acid:

    pH = pKa + log ([A⁻]/[HA])

    Substituting and solving for [A⁻]/[HA]:

    4.005 = 4.3574 = log ([A⁻]/[HA])

    -0.3524 = log ([A⁻]/[HA])

    [A⁻]/[HA] = 0.444/1

    The percentage of acid that is not ionized (i. e. the percentage of acid in the HA form) is calculated:

    [HA] / ([A⁻] + [HA]) x 100% = 1 / (1+0.444) x 100% = 69.3%
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