a. Write the equilibrium reaction to show the amphoteric behaviour of alanine in `H_(2)O`.
b. Calculate the isoelctric `pH` of alanine, given `pK_(a1`, of cation of `"alanine"=2.3 "and' pK_(a_2)` of anion of alanine=`9.7`.
c. Write equilibrium reaction for the dissociation of lysine, a basic amino acid and calculate its isoelectric point.
`pK_(a_1)` of lysine with net `(+2)` charge=`2.18`
`pK_(a_2)` of lysin with net `(+1)` charge=`8.95`
`pK_(a_3)` of lysine with net `(-1)` charge=`10.53`
d. Write the equilibrium reaction for the dissociation of aspartic acid, an acidic amino acid and calculate its isoelectric point.
`pK_(a_1)` of aspartic acid with net `(+1)` charge=`1.88`
`pK_(a_2)` of aspartic acid with net `(-1)` cahrge =`3.65`
`pK_(a_3)` of aspartic acid with net `(-2)` charge=`9.60`

Explanation

a.
b. `pH` at isoelectric point `(pI)=((pK_(a_1)+pK_(a_2)))/(2)=(2.3+9.7)/(2)=6`
The isoelectric point `(pI)` is the `pH` at which the amino acid exists only as a dipolar ion (armpholyte or zwitterion) with zero net charge.
c.
`:. pH` at isolectric point `(pI)=((pK_(a_2)+pK_(a_3)))/(2)=((8.95+10.53))/(2)=9.74`
`:. pI=9.74`
d.
`:. pH` at isoelctric point `(pI)=((pK_(a_1)+pK_(a_2)))/(2)=((1.88+3.65))/(2)=2.77`
`:. pI=2.77`