Question

In an amino acid the carboxylic acid group has $K_{a}=10^{- 4}$ and amino group has $K_{b}=10^{- 5}.pH$ at isoelectric point of that amino acid is $\lambda $ . Then value of $2\lambda $ will be.

Answer 1

$K _{ b \left( R - NH _2\right)} \times K _{ a }\left( R - NH _3{ }^{+}\right)=10^{-14}$
$K _{ a }\left( R - NH _3{ }^{+}\right)=10^{-9} pI =$
$=\frac{(p K)_{a(R C O O H)}+(p K)_{a\left(R-\left((N H)_3\right)^{+}\right)}}{2}=\frac{4+9}{2}=6.5$
So, the value of $2 \lambda=13$