-NH3 - CHR - CO2H has its Ka1 - 10-7- The isoelectric point of the -amino acid occur at pH - 8-2 Hence pKa2 of the conjugate acid of -amino acid is equal to-

The correct option is D 9.4 ∗ #160; Isoelectric point is given by, pI= pKa _ 1 + pKa _ 2 2 Ka _ 1 = 10 ^ 7 , pI=8.2 ∵ #160; #160; ...

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Byju's Answer

Standard XII

Chemistry

Nucleophilic Substitution Reaction

+NH3 - CHR - ...

Question

$^{+} N H_{3} - C H (R) - C O_{2} H$ has its $K a_{1} = 10^{- 7}$ . The isoelectric point of the $α$ -amino acid occur at pH = 8.2 Hence $p K a_{2}$ of the conjugate acid of $α$ -amino acid is equal to:

9.4

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Solution

The correct option is D 9.4
$*$ Isoelectric point is given by,
$p I = \frac{{p K a}_{1} + {p K a}_{2}}{2}$

${K a}_{1} = 10^{- 7}, p I = 8.2$

$∵$ ${p K a}_{1} = - l o g {K a}_{1}$

$∴$ ${p K a}_{1} = - l o g 10^{- 7} = 7$

$∴$ $8.2 = \frac{{p K a}_{1} + {p K a}_{2}}{2} = \frac{7 + {p K a}_{2}}{2}$

$∴$ $16.4 = 7 + {p K a}_{2}$

$∴$ ${p K a}_{2} = 9.4$

Option A is correct.

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Similar questions

+ N H_{3} - C H (R) - C O_{2} H

has its

K a_{1} = 10^{- 7}

. The isoelectric point of the

α -

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p H = 8.2

hence

p K a_{2}

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α -

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Q. In an amino acid, the carboxylic group ionises at

p K_{a 1} = 2.34

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+NH3−CH(R)−CO2H has its Ka1=10−7. The isoelectric point of the α-amino acid occur at pH = 8.2 Hence pKa2 of the conjugate acid of α-amino acid is equal to:

The correct option is D 9.4∗ Isoelectric point is given by,pI=pKa1+pKa22Ka1=10−7,pI=8.2∵ pKa1=−logKa1∴ pKa1=−log10−7=7∴ 8.2=pKa1+pKa22=7+pKa22∴ 16.4=7+pKa2∴ pKa2=9.4Option A is correct.

$^{+} N H_{3} - C H (R) - C O_{2} H$ has its $K a_{1} = 10^{- 7}$ . The isoelectric point of the $α$ -amino acid occur at pH = 8.2 Hence $p K a_{2}$ of the conjugate acid of $α$ -amino acid is equal to: