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Reaction with Metals

A buffer solu...

Question

A buffer solution $0.04 M$ in $N a_{2} H P O_{4}$ and $0.02 M$ in $N a_{3} P O_{4}$ 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 $H_{3} P O_{4}, p K_{a_{1}} = 2.2; p K_{a_{2}} = 7.2; p K_{a_{3}} = 12$ ]

A

6.90

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B

7.20

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C

7.5

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D

None of these

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Solution

The correct option is C $7.5$
As we know,
$p H = p K_{a} + l o g [s a l t] / [a c i d]$
$p H = 7.2 + l o g (4 / 2) = 7.2 + 0.3 = 7.5$ .

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Similar questions

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

]

Q. A buffer solution of

0.80 M

{N a}_{2} H {P O}_{4}

0.020 M

{N a}_{3} {P O}_{4}

is prepared. The electrolytic oxidation of

1.0

R N H O H

is carried out in

100 m L

buffer to give.

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

Calculating approximate

p H

(nearest integer) of the solution after oxidation is complete.

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

p K_{a_{3}}

H_{3} {P O}_{4}

2.12, 7.20

12.0

Q. A buffer solution,

0.080 M

{N a}_{2} {P O}_{4}

0.020 M

{N a}_{3} {P O}_{4}

is prepared. The electrolytic oxidation of

1.0

mmol of the organic compound

R N H O H

is carried out in

100 m L

of the buffer. The reaction is as follows:

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

Calculate the approximate

p H

of the solution after the oxidation is complete.

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

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:

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