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?

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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