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Nucleophilic Substitution in Aryl Halides

HOOC-⊕NH3|CH-...

Question

$⎛ ⎜ ⎜ ⎝ H O O C - C | H^{\oplus} N H_{3} - C H_{2} C O O H ⎞ ⎟ ⎟ ⎠$ (A)
The structure of aspartic acid is given.The $p K_{a_{1}}, p K_{a_{2}}$ and $p K_{a_{3}}$ of $(A)$ respectively are: $1.88, 3.65$ and $9.60$ . $p K_{a_{1}}$ corresponds to the ionisation of the $C O O H$ group, of $⎛ ⎜ ⎜ ⎜ ⎜ ⎝ R \oplus N H_{3} | C H - C O O H ⎞ ⎟ ⎟ ⎟ ⎟ ⎠$
$p K_{a_{2}}$ corresponds to the ionisation of $(\oplus N H_{3})$ (ammonium ion). What is the $p H$ at isoelectric points $(p I)$ ?

\frac{p K_{a_{1}} + p K_{a_{2}}}{2}

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\frac{p K_{a_{1}} + p K_{a_{3}}}{2}

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\frac{p K_{a_{2}} + p K_{a_{3}}}{2}

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\frac{p K_{a_{1}} + p K_{a_{2}} + p K_{a_{3}}}{3}

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Solution

The correct option is A $\frac{p K_{a_{1}} + p K_{a_{2}}}{2}$
$p I$ = $\frac{p K_{a_{1}} + p K_{a_{2}}}{2}$
If there are more than one carboxylic group then we will take the average of $p K_{a}$ of carboxyl groups.

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Similar questions

p K_{a_{1}}, p K_{a_{2}}

p K_{a_{3}}

(A)

2.18, 8.95

10.53

p H

(p I)

Lysine has p K_{a_{1}} = 2.18, p K_{a_{2}} = 8.95, p K_{a_{3}} = 10.53

p H = 9.7

p K_{a 1} = 2.34

p K_{a 2} = 9.60

p K_{a_{1}}, p K_{a_{2}} a n d p K_{a_{3}}

H_{3} P O_{4}

p H

0.1 M

N a_{2} H P O_{4}

p H

H_{2} P O_{4}^{-}

K_{a_{1}}

H_{3} P O_{4}

K_{a_{2}}

H_{3} P O_{4}

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