In aqueous solution α-amino acid has the following equilibria. Find the iso-electric point of the α-amino acid given.

Question

In aqueous solution α-amino acid has the following equilibria. <img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/chemistry/fcd9dec097eec6fc6c3f7903024ae485-.png" /> Find the iso-electric point of the α-amino acid given. (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

pK a 1=- log 5- log 10-3 ≈ 2.3 pK b =4.3 ⇒ pK a 2=9.7 ∴ p 1=( pK a 1+ pK a 2/2) =(2.3+9.7/2)=6

Q. In aqueous solution $α$ -amino acid has the following equilibria.

Find the iso-electric point of the $α$ -amino acid given.

$p K_{a_{1}} = - lo g 5 - lo g 1 0^{- 3} \approx 2.3$
$p K_{b} = 4.3$
$\Rightarrow p K_{a_{2}} = 9.7$
$∴ p^{1} = \frac{p K _{a_{1}} + p K _{a_{2}}}{2}$
$= \frac{2.3 + 9.7}{2} = 6$

Q. In aqueous solution α-amino acid has the following equilibria. Find the iso-electric point of the α-amino acid given.

pKa1​​=−log5−log10−3≈2.3 pKb​=4.3 ⇒pKa2​​=9.7 ∴p1=2pKa1​​+pKa2​​​ =22.3+9.7​=6

Q. In aqueous solution $α$ -amino acid has the following equilibria.

Find the iso-electric point of the $α$ -amino acid given.

$p K_{a_{1}} = - lo g 5 - lo g 1 0^{- 3} \approx 2.3$
$p K_{b} = 4.3$
$\Rightarrow p K_{a_{2}} = 9.7$
$∴ p^{1} = \frac{p K _{a_{1}} + p K _{a_{2}}}{2}$
$= \frac{2.3 + 9.7}{2} = 6$