CameraIcon

SearchIcon

MyQuestionIcon

1

You visited us 1 times! Enjoying our articles? Unlock Full Access!

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:

A

7.2

No worries! We‘ve got your back. Try BYJU‘S free classes today!

B

12.3

No worries! We‘ve got your back. Try BYJU‘S free classes today!

C

8.5

No worries! We‘ve got your back. Try BYJU‘S free classes today!

D

9.75

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

Open in App

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 $N a_{2} H P O_{4}$

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

Suggest Corrections

thumbs-up

0

Similar questions

Q. A one litre solution contains

0.04 M

N a_{2} H P O_{4}

0.02 M

N a H_{2} P O_{4}

. The correct statement(s) is/are:

H_{3} P O_{4} :

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

l o g 2 = 0.3

Q. A buffer solution

0.04 M

N a_{2} H P O_{4}

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

p H

of resulting mixture where

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

p K_{a_{3}}

H_{3} P O_{4}

5, 8, 13

p H

N a_{2} H P O_{4}

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)

Q. A buffer solution

0.04 M

N a_{2} H P O_{4}

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 mL

of the buffer. The reaction is

R N H O H + H_{2} O \to 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}} = 7.20; p K_{a_{3}} = 12, l o g 2 = 0.30 a n d l o g 4 = 0.60

]

View More

Join BYJU'S Learning Program

explore_more_icon

Explore more

Introduction and Degree of Hydrolysis

Standard XIII Chemistry

Join BYJU'S Learning Program

CrossIcon