For an amino acid,the ionization constant for | vedclass

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(A) The isoelectric point $(pI)$ of an amino acid is the $pH$ at which the molecule carries no net electric charge.For a neutral amino acid,the $pI$ i

Vedclass · Accepted Answer

$5.97$

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(A) The isoelectric point $(pI)$ of an amino acid is the $pH$ at which the molecule carries no net electric charge.For a neutral amino acid,the $pI$ is calculated as the average of the $pK_a$ values of the two ionizable groups:$pI = \frac{pK_{a_1} + pK_{a_2}}{2}$Given $pK_{a_1} = 2.34$ and $pK_{a_2} = 9.6$:$pI = \frac{2.34 + 9.6}{2} = \frac{11.94}{2} = 5.97$Therefore,the isoelectric point is $5.97$.

For an amino acid,the ionization constant for the carboxyl group is $pK_{a_1} = 2.34$ and for the ammonium ion is $pK_{a_2} = 9.6$. At what $pH$ will the isoelectric point of the amino acid be?

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