In a buffer solution of N aH 2 P O 4 and N a 2 HP O 4-A- NaH 2 P O 4 is the acid and N a 2 HP O 4 is the saltB- The pH of the solution can be calculated using the relation-pH - K 2-log- HPO 4-2- H 2 PO 4-C- N a 2 HP O 4 is the acid and N aH 2 P O 4 is the saltD- The pH of the solution can be calculated using the relation-pH - K 3-log- HP O 4- H 2 PO 4-

The correct options are A NaH_2PO_4 is the acid and nbsp;Na_2HPO_4 is the salt B The pH of the solution can be calculated using the relation: pH= K_2+log[HPO_ ...

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Standard XII

Chemistry

Concentration Terms

In a buffer s...

Question

In a buffer solution of $N a H_{2} P O_{4}$ and $N a_{2} H P O_{4}$ :

N a H_{2} P O_{4}

is the acid and

N a_{2} H P O_{4}

is the salt

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The pH of the solution can be calculated using the relation:

p H = K_{2} + l o g \frac{[H P O_{4}^{2 -}]}{[H_{2} P O_{4}^{-}]}

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N a_{2} H P O_{4}

is the acid and

N a H_{2} P O_{4}

is the salt

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The pH of the solution can be calculated using the relation:

p H = K_{3} + l o g \frac{[H P O_{4}^{2 -}]}{[H_{2} P O_{4}^{-}]}

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Solution

The correct options are
A $N a H_{2} P O_{4}$ is the acid and $N a_{2} H P O_{4}$ is the salt
B The pH of the solution can be calculated using the relation:
$p H = K_{2} + l o g \frac{[H P O_{4}^{2 -}]}{[H_{2} P O_{4}^{-}]}$
$N a H_{2} P O_{4}$ is more acidic than $N a_{2} H P O_{4}$ , hence $N a H_{2} P O_{4}$ will be considered as the acid and $N a_{2} H P O_{4}$ will be considered as the salt.
In the Henderson-Hasselbalch equation, the dissociation constant of acid is considered, i.e.,
$H_{2} P O_{4}^{-} ⇌ H P O_{4}^{2 -} + H^{+}$
So, $K_{2}$ is appropriate dissociation constant to be used here.

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Similar questions

p H

of a buffer

0.1

N a_{2} H P O_{4}

and

0.2

N a H_{2} P O_{4}

is:

(

p K_{a}

for

H_{2} P O_{4}

are

2.2, 7.2

and

12.0

)

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)

Q. Calculate the pH of

0.1

M solution of (i)

N a H C O_{3}

, (ii)

N a_{2} H P O_{4}

and (iii)

N a H_{2} P O_{4}

Given that:

C O_{2} + H_{2} O ⇌ H^{+} + C O_{3}^{2 -}

;

K_{1} = 4.2 \times 10^{- 7}

H C O_{3}^{-} ⇌ H^{+} + C O_{3}^{2 -}

;

K_{2} = 4.8 \times 10^{- 11}

H_{2} P O_{4} ⇌ H^{+} + H_{2} P O_{4}^{-}

;

K_{1} = 7.5 \times 10^{- 3}

H_{2} P O_{4}^{-} ⇌ H^{+} + H P O_{4}^{2 -}

;

K_{2} = 6.2 \times 10^{- 8}

H P O_{4}^{2 -} ⇌ H^{+} + P O_{4}^{3 -}

;

K_{3} = 1.0 \times 10^{- 12}

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. Two litre solution of a buffer mixture containing 1.0 M

N a H_{2} P O_{4}

and 1.0M

N a_{2} H P O_{4}

is placed in two compartments (one litre in each) of an electrolytic cell. The platinum electrodes are inserted in each compartment and 1.25 ampere current is passed for 212 minute. Assuming electrolysis of water only at each compartment. What will be pH in each compartment after passage of above charge?

p K_{a}

for

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

= 2.15.

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In a buffer solution of NaH2PO4 and Na2HPO4:

In a buffer solution of $N a H_{2} P O_{4}$ and $N a_{2} H P O_{4}$ :