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Standard XI

Chemistry

Introduction and Degree of Hydrolysis

If pKa1, pKa2...

Question

If $p K_{a_{1}}, p K_{a_{2}} a n d p K_{a_{3}}$ of $H_{3} P O_{4}$ are 2.2, 7.2 and 12.3 respectively. The $p H$ of $0.1 M$ $N a_{2} H P O_{4}$ solution will be:

7.2

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12.3

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8.5

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9.75

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Solution

The correct option is D 9.75
$N a_{2} H P O_{4}$ is an amphiprotic salt. To calculate the pH of $N a_{2} H P O_{4}$
Considering the following reaction:
$H P O_{4}^{2 -} (a q) + H^{+} (a q) \to H_{2} P O_{4}^{-} (a q)$
$H P O_{4}^{2 -} (a q) - H^{+} (a q) \to P O_{4}^{3 -} (a q)$
$p H = \frac{1}{2} [p K_{a_{2}} + p K_{a_{3}}]$
Therefore, $p H = \frac{1}{2} [p K_{a_{2}} + p K_{a_{3}}]$ for $p H$ of $N a_{2} H P O_{4}$

$∴ p H = \frac{p K_{a_{2}} + p K_{a_{3}}}{2} = \frac{1}{2} (7.2 + 12.3) = 9.75$

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Similar questions

Q. If

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

H_{3} P O_{4}

are 2.2, 7.2 and 12.3 respectively. The

p H

0.1 M

N a_{2} H P O_{4}

solution will be:

Q. A one litre solution contains

0.04 M

N a_{2} H P O_{4}

and

0.02 M

N a H_{2} P O_{4}

. The correct statement(s) is/are:

For

H_{3} P O_{4} :

p K_{a_{1}} = 2.14; p K_{a_{2}} = 6.84; p K_{a_{3}} = 12.3

and use

l o g 2 = 0.3

Q. A buffer solution

0.04 M

N a_{2} H P O_{4}

and

0.02 M

N a_{3} P O_{4}

is prepared. The electrolytic oxidation of

1.0

milli-mole of the organic compound

R N H O H

is carried out in

100 m L

of the buffer. The reaction is

R N H O H + H_{2} O ⟶ R N O_{2} + 4 H_{4}^{+} e^{-}

The approximate

p H

of solution after the oxidation is complete is:
[Given: for

H_{3} P O_{4}, p K_{a_{1}} = 2.2; p K_{a_{2}} = 7.2; p K_{a_{3}} = 12

]

50 mL

0.1 M N a_{3} P O_{4}

was titrated with

50 mL

0.2 M H C l

. Find the

p H

of resulting mixture where

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

and

p K_{a_{3}}

H_{3} P O_{4}

are

5, 8, 13

p H

of 0.1 M

N a_{2} H P O_{4}

and 0.2 M

N a H_{2} p O_{4}

are respectively:

(p K_{a} f o r H P O_{4} a r e 2.2, 7.2 a n d 12.0)

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Introduction and Degree of Hydrolysis

Standard XI Chemistry

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If pKa1, pKa2and pKa3 of H3PO4 are 2.2, 7.2 and 12.3 respectively. The pH of 0.1 M Na2HPO4 solution will be:

The correct option is D 9.75Na2HPO4 is an amphiprotic salt. To calculate the pH of Na2HPO4 Considering the following reaction: HPO2−4(aq)+H+(aq)→H2PO−4(aq) HPO2−4(aq)−H+(aq)→PO3−4(aq) pH=12[pKa2+pKa3] Therefore, pH=12[pKa2+pKa3] for pH of Na2HPO4 ∴pH=pKa2+pKa32 =12(7.2+12.3)=9.75

If $p K_{a_{1}}, p K_{a_{2}} a n d p K_{a_{3}}$ of $H_{3} P O_{4}$ are 2.2, 7.2 and 12.3 respectively. The $p H$ of $0.1 M$ $N a_{2} H P O_{4}$ solution will be: