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Dissolution of Solids and Gases in Liquid

pH of some so...

Question

pH of some solution is given by $p H = \frac{p K_{a_{1}} + p K_{a_{2}}}{2}$ . This formula is valid for the compound $N a H_{2} B O_{3}$ .

A

True

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B

False

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Solution

The correct option is B False
$N a_{2} B O_{3}$ is a strong base - weak acid salt and its $p H$ will be

$p H = \frac{(14 + p K_{2} + l o g C)}{2}$

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Similar questions

Q. Which of the following relation is correct regarding

p H

of salt of amphiprotic ion,

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

K_{a_{1}}

is first acid dissociation constant of

H_{3} P O_{4}

K_{a_{2}}

is second acid dissociation constant of

H_{3} P O_{4}

.

K_{a 1}

K_{a 2}

are the ionization constants of

H_{3} N^{+} C H I C O O H

H_{3} N^{+} C H I C O O^{-},

respectively, the pH of the solution at the isoelectric point is

Q. Calculate approximate

p H

of the resultant solution formed by titration of

25 m L

0.4 M N a_{2} C O_{3}

50 m L

0.025 M H C l

p K_{a_{1}} = 6.4

p K_{a_{2}} = 10.3

H_{2} C O_{3}

50 m L

0.05 M N a_{2} C O_{3}

is titrated against

0.1 M H C l

40 m L

H C l

p H

of the solution will be
[Given for

H_{2} C O_{3}, p K_{a_{1}} = 6.35; p K_{a_{2}} = 10.33; log 3 = 0.477, log 2 = 0.30

]

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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