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What would the ratio of deprotonated to protonated histidines be if the phph decreased to 6.40?

Sagot :

The ratio of deprotonated to protonated histidines is [tex]10^{6.4 - pK_{a} }[/tex] : 1

We know that,

pH = p[tex]K_{a}[/tex] + ( log [ [tex]A^{-1}[/tex] ] / [ HA ] )

where,

[ [tex]A^{-1}[/tex] ] represents deprotonated form

[ HA ] represents protonated form

Given that,

pH = 6.4

pH = p[tex]K_{a}[/tex] + ( log [ [tex]A^{-1}[/tex] ] / [ HA ] )

6.4 = p[tex]K_{a}[/tex] + ( log [ [tex]A^{-1}[/tex] ] / [ HA ] )

[ [tex]A^{-1}[/tex] ] / [ HA ] = [tex]10^{6.4 - pK_{a} }[/tex]

The relationship between pH and p[tex]K_{a}[/tex] shown above is known as Henderson-Hasselbach equation

Therefore, the ratio of deprotonated to protonated histidines is [tex]10^{6.4 - pK_{a} }[/tex] : 1

To know more about Henderson-Hasselbach equation

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