In an amino acid the carboxylic acid group has Ka=10- 4 and amino group has Kb=10- 5.pH at isoelectric point of that amino acid is λ . Then value of 2λ will be.

Question

In an amino acid the carboxylic acid group has Ka=10- 4 and amino group has Kb=10- 5.pH at isoelectric point of that amino acid is λ . Then value of 2λ will be. <img class="img-fluid question-image" alt="Question" src="https://cdn.tardigrade.in/q/nta/c-rpl8lsbursoq1uvo.jpg" /> (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

K b ( R - NH 2) × K a ( R - NH 3 +)=10-14 K a ( R - NH 3 +)=10-9 pI = =((p K)a(R C O O H)+(p K)a(R-((N H)3)+)/2)=(4+9/2)=6.5 So, the value of 2 λ=13

Q. In an amino acid the carboxylic acid group has $K_{a} = 1 0^{- 4}$ and amino group has $K_{b} = 1 0^{- 5} . p H$ at isoelectric point of that amino acid is $λ$ . Then value of $2 λ$ will be.

$K_{b (R - N H_{2})} \times K_{a} (R - N H_{3}^{+}) = 1 0^{- 14}$
$K_{a} (R - N H_{3}^{+}) = 1 0^{- 9} p I =$
$= \frac{( p K ) _{a (RCOO H)} + ( p K ) _{a (R - ((N H)_{3})^{+})}}{2} = \frac{4 + 9}{2} = 6.5$
So, the value of $2 λ = 13$

Q. In an amino acid the carboxylic acid group has Ka​=10−4 and amino group has Kb​=10−5.pH at isoelectric point of that amino acid is λ . Then value of 2λ will be.

Kb(R−NH2​)​×Ka​(R−NH3​+)=10−14 Ka​(R−NH3​+)=10−9pI= =2(pK)a(RCOOH)​+(pK)a(R−((NH)3​)+)​​=24+9​=6.5 So, the value of 2λ=13

Q. In an amino acid the carboxylic acid group has $K_{a} = 1 0^{- 4}$ and amino group has $K_{b} = 1 0^{- 5} . p H$ at isoelectric point of that amino acid is $λ$ . Then value of $2 λ$ will be.

$K_{b (R - N H_{2})} \times K_{a} (R - N H_{3}^{+}) = 1 0^{- 14}$
$K_{a} (R - N H_{3}^{+}) = 1 0^{- 9} p I =$
$= \frac{( p K ) _{a (RCOO H)} + ( p K ) _{a (R - ((N H)_{3})^{+})}}{2} = \frac{4 + 9}{2} = 6.5$
So, the value of $2 λ = 13$