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Standard XII

Chemistry

Salt of Weak Acid and Weak Base

When 10 mL of...

Question

When 10 mL of 0.1 M acetic acid $(p K_{a} = 5.0)$ is titrated against 10 mL of 0.1 M ammonia solution $(p K_{b} = 5.0)$ the equivalence point occurs at pH of:

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Solution

The correct option is C 7
Given, $V_{C H_{3} C O O H} = V_{N H_{3}} = 10 m L, p K_{a} = 5.0, p K_{B} = 5.0$

10 mL of 0.1 M $C H_{3} C O O H$ will completely react with 10 mL of 0.1 M ammonia solution to give $C H_{3} C O O N H_{4}$ (a salt of a weak acid and a weak base).
The $p H$ of the resulting solution can be calculated as follows:
$p H = \frac{1}{2} [p K_{a} + p K_{w} - p K_{b}]$
$p H = \frac{1}{2} [5.0 + 14 - 5.0] ∵ p K_{w} = 14 a t 25^{o} C$
$p H = \frac{1}{2} \times 14 = 7$

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Similar questions

Q. When

10 m L

0.1 M

acetic acid (

p K_{a} = 5.0

) is titrated against

10 m L

0.1 M

ammonia solution

(p K_{b} = 5.0)

, the equivalence point occurs at pH:

Q. A solution of weak acid

H A

was titrated with base

N a O H

. The equivalence point was reached when

36.12 m L

0.1 M

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has been added. Now

18.06 m L

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H C l

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4.92

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of the solution obtained by mixing

10 m L

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and

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0.2 M

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{C H}_{3} C O O H (50 m L, 0.1 M)

is titrated against

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Calculate the

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Q. Calculate

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Salt of Weak Acid and Weak Base

Standard XII Chemistry

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When 10 mL of 0.1 M acetic acid (pKa=5.0) is titrated against 10 mL of 0.1 M ammonia solution (pKb=5.0) the equivalence point occurs at pH of:

When 10 mL of 0.1 M acetic acid $(p K_{a} = 5.0)$ is titrated against 10 mL of 0.1 M ammonia solution $(p K_{b} = 5.0)$ the equivalence point occurs at pH of: