Question

The structure of alanine is shown at the right in its fully protonated form. The pK_a for the ionizable groups are 2.3 (alpha-carboxyl group), and 9.7 (alpha-amino group). Calculate the pH at which alanine will have a net charge of 0. pH = Select the true statements about alanine when the pH is equal to its pl. If an electric field is applied, alanine will not move toward either electrode. Since alanine does not have an ionizable R group, the zwitterion species will be >50% at any pH. The alpha-amino group will be predominantly neutral. The alpha-carboxyl group will be predominantly negatively charged. For both the alpha-amino group and the alpha-carboxyl group, the protonated form is necessary for the formation of the zwitterion.

Answer 1

At pH=pI, alanine exists as zwitter ion i.e the net charge on the species is 0. So, alanine will not move towards either electrode at this pH.

For alanine, the pI Cph at which net charge is o) is the average of the pka of two ionisable groups py = pl = pka, + pka2 = 2.3 +9.7 - 14:=61 When pH = pi , following form of alanure există - من - پر هم

At any other pH, the conc of zwitter ion will be less. In acidic pH, the carboxylate ion will be protonated, so, positively charged species will exist, while in basic pH, amine and carboxylic group both will be deprotonated, so, negatively charged species will exist. Hence, 2nd statement is incorrect.

At pH=pI, Amino group will be protonated,so it will be positively charged. The 3rd statement is incorrect.

At pH=pI, the acid group is deprotonated so, it will be negatively charged. So, the 4th statement is correct.

For the formation of zwitter, only the Amino group should be protonated while the acid group should be deprotonated. So, the 5th statement is incorrect.