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Standard XII

Chemistry

Electrolytes

100 mL of 0.0...

Question

100 mL of 0.02 M benzoic acid $(p K_{a} = 4.2)$ is titrated using 0.02 M $N a O H$ . $p H$ after 50 mL of $N a O H$ have been added is:

4.2

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3.7

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4.7

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Solution

The correct option is B 4.2
$p H = p K_{a} + l o g \frac{[s a l t]}{[a c i d]}$

When we add 50 ml of NaOH, $[s a l t] = [a c i d] = 50 \times 0.02$

So, $p H = p K_{a} = 4.2$

Option A is correct.

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Similar questions

100

mL of

0.02 M

benzoic acid

(p K_{a} = 4.2)

is titrated using

0.02 M N a O H

. The pH after

50

mL of

N a O H

has been added is:

Q. 100 mL of 0.02 M benzoic acid

(p K a = 4.2)

is titrated using

0.02 M N a O H

. pH after 50 mL and 100 mL of

N a O H

have been added are

Q. A

20 m L

0.02 M C H_{3} C O O H

is titrated with

0.2 M N a O H

. Find the

p H

of the solution when

80 m L

N a O H

has been added.
Given

p K_{a} (C H_{3} C O O H) = 4.74

Q. A

20 m L

0.02 M C H_{3} C O O H

is titrated with

0.2 M N a O H

. Find the

p H

of the solution when

80 m L

N a O H

has been added.
Given

p K_{a} (C H_{3} C O O H) = 4.74

Q. A

50 m L

solution of

0.2 M C H_{3} C O O H

is titrated with

0.2 M N a O H

. Find the

p H

of solution when:
(a)

0 m L N a O H

is added (

p H_{i n i t i a l}

)
(b)

10 m L

N a O H

has been added (

p H_{10 m L}

)

Given:

K_{a} (C H_{3} C O O H) = 1.8 \times 10^{- 5}

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100 mL of 0.02 M benzoic acid (pKa=4.2) is titrated using 0.02 M NaOH. pH after 50 mL of NaOH have been added is:

The correct option is B 4.2pH=pKa+log[salt][acid]When we add 50 ml of NaOH, [salt]=[acid]=50×0.02So, pH=pKa=4.2Option A is correct.

100 mL of 0.02 M benzoic acid $(p K_{a} = 4.2)$ is titrated using 0.02 M $N a O H$ . $p H$ after 50 mL of $N a O H$ have been added is:

The correct option is B 4.2
$p H = p K_{a} + l o g \frac{[s a l t]}{[a c i d]}$

When we add 50 ml of NaOH, $[s a l t] = [a c i d] = 50 \times 0.02$

So, $p H = p K_{a} = 4.2$

Option A is correct.