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What would happen to the predominant protonation state and charge of the his and asp side chains if the ph were to change from 7.40 to 5.00?

Sagot :

The predominant protonation state and charge of the His and Asp side chain will be predominantly protonated and positively charged if both its p[tex]K_{a}[/tex] values is more than 5.

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

Consider pH = p[tex]K_{a}[/tex]

In this case,

[ [tex]A^{-1}[/tex] ] / [ HA ] = 1

This means that protonated and deprotonated forms are equal.

This also means that as pH decreases, the acid's state will be predominantly protonated and its charge will be positive.

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

Therefore, the predominant protonation state and charge of the His and Asp side chain will be predominantly protonated and positively charged if both its p[tex]K_{a}[/tex] values is more than 5.

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