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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
https://brainly.com/question/13564840
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