Connect with a global community of knowledgeable individuals on IDNLearn.com. Receive prompt and accurate responses to your questions from our community of knowledgeable professionals ready to assist you at any time.
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
#SPJ4
We appreciate every question and answer you provide. Keep engaging and finding the best solutions. This community is the perfect place to learn and grow together. Your questions deserve precise answers. Thank you for visiting IDNLearn.com, and see you again soon for more helpful information.
