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15 June, 07:39

Given that the pka for the imidazole ring of a histidine has a pKa of 6.5 how many histidine side chains will be charged at a pH of 7.2?

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  1. 15 June, 10:21
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    83.4%

    Explanation:

    The imidazole ring is a weak acid, and so, in solution it must dissociate, losing protons (H⁺), and forming the anion, which will be the conjugate base. Calling the acid as HA, the charged histidine side chains will be the A⁻, and, by the Handerson-Halsebach equation:

    pH = pKa + log[A⁻]/[HA], where [X] is the concentration of X. So:

    7.2 = 6.5 + log[A⁻]/[HA]

    log[A⁻]/[HA] = 7.2 - 6.5

    log[A⁻]/[HA] = 0.7

    [A⁻]/[HA] =

    [A⁻]/[HA] = 5.012

    So, the histidines sides that will be charged are 5 times higher then the ones that wouldn't be. If the initial concentrantion of the acid was 1 mol/L, and so, in equilibrium:

    [A⁻] + [HA] = 1

    [HA] = 1 - [A⁻], so:

    [A⁻] / (1 - [A⁻]) = 5.012

    [A⁻] = 5.012 - 5.012[A⁻]

    6.012[A⁻] = 5.012

    [A⁻] = 0.834 mol/L

    So, 83.4% of it will be charged (0.834/1 * 100%).
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