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Basic Buffer Action

A buffer solu...

Question

A buffer solution of $0.80 M$ ${N a}_{2} H {P O}_{4}$ and $0.020 M$ ${N a}_{3} {P O}_{4}$ is prepared. The electrolytic oxidation of $1.0$ milli mole $R N H O H$ 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. $p K_{a_{1}}, p K_{a_{2}}$ and $p K_{a_{3}}$ of $H_{3} {P O}_{4}$ are $2.12, 7.20$ and $12.0$ respectively.

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Solution

The electrolytic oxidation of 1 millimole will give 4 mill moles of hydrogen ions. The molar concentration of hydrogen ions is $\frac{4 \times 10^{- 3}}{100 \times 10^{- 3} = 0.04 M}$
0.02 M hydrogen ions will react with 0.02 M $N a_{3} P O_{4}$ to form 0.02 M of $H N a_{2} P O_{4}$ . The total concentration of $H N a_{2} P o_{4}$ will be $0.8 + 0.02 = 0.82 M$ . Again 0.02 M of hydrogen ions will combine with 0.02 M of $H N a_{2} P O_{4}$ to form 0.02 M of $N a H_{2} P O_{4}$ . The total concentration of $H N a_{2} P O_{4}$ will be $0.082 - 0.02 = 0.8 M$ .
The pH of the buffer solution is given by the expression $p H = p K a + l o g (\frac{S a l t}{A c i d})$ .
Substitute values in the above expression.
$p H = 7.20 + l o g 0.8 0.02 = 8.81 \approx 9$

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Similar questions

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 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

]

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. In a buffer solution consisting

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

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

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

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

is salt
(2) The pH of solution can be calculated using the relation:

p H = p K_{2} + {log}_{10} \frac{[H {P O}_{4}^{2 -}]}{[H_{2} {P O}_{4}^{-}]}

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

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

is salt
(4) The pH cannot be calculated.

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

is a tribasic acid and one of its salt is

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

1 M N a O H

solution should be added to

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

to convert it into

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

a t . w t

P = 31

)

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