You visited us 1 times! Enjoying our articles? Unlock Full Access!

Electrolytes

100 mL of a 0...

Question

$100 m L$ of a $0.5 M C H_{3} C O O N a (a q)$ is titrated against a $0.5 M H C l (a q)$ . Calculate the $p H$ of the solution initially i.e. when no $H C l$ has been added.
Given: $p K_{a} (C H_{3} C O O H) = 4.74$

6.58

No worries! We‘ve got your back. Try BYJU‘S free classes today!

8.24

No worries! We‘ve got your back. Try BYJU‘S free classes today!

7.96

No worries! We‘ve got your back. Try BYJU‘S free classes today!

9.22

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

Open in App

Solution

The correct option is D 9.22
When only $100 m L$ of a $0.5 M C H_{3} C O O N a (a q)$ is present. It will show simple salt hydrolysis.
The formula to calculate the $p H$ of a salt of strong base and weak acid is :
$p H = \frac{1}{2} (p K_{w} + p K_{a} + log C) p H = \frac{1}{2} (14 + 4.74 + log 0.5) p H = \frac{1}{2} (18.74 - 1 + log 5) p H = \frac{1}{2} (18.44) p H = 9.22$

Suggest Corrections

Similar questions

Q. A

30 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 the solution when

10 m L

N a O H

has been added.
Given

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

20 m L

0.2 M C H_{3} C O O N a

solution is titrated against

0.1 M H C l

. Calculate the

p H

of solution when

10 m L H C l

has been added.
Given:

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

50 m L

of a

0.1 M C H_{3} C O O N a (a q)

is titrated against a

0.1 M H C l (a q)

. Calculate the

p H

of solution when

50 m L H C l

was added.
Given:

p K_{a} (C H_{3} C O O H) = 4.74 a t 25^{o} C

50 m L

of a

0.1 M C H_{3} C O O N a (a q)

is titrated against a

0.1 M H C l (a q)

. Calculate the

p H

of solution when

50 m L H C l

was added.
Given:

p K_{a} (C H_{3} C O O H) = 4.74 a t 25^{o} C

20 m L

1 M H C l

is titrated against the

1 M N a O H

solution. Calculate the

p H

of the solution when

30 m L N a O H

is added?

Join BYJU'S Learning Program

Explore more

Electrolytes

Standard XII Chemistry

Join BYJU'S Learning Program

100 mL of a 0.5 M CH3COONa(aq) is titrated against a 0.5 M HCl(aq). Calculate the pH of the solution initially i.e. when no HCl has been added. Given: pKa(CH3COOH)=4.74

The correct option is D 9.22When only 100 mL of a 0.5 M CH3COONa(aq) is present. It will show simple salt hydrolysis. The formula to calculate the pH of a salt of strong base and weak acid is : pH=12(pKw+pKa+log C)pH=12(14+4.74+log0.5)pH=12(18.74−1+log 5)pH=12(18.44)pH=9.22

$100 m L$ of a $0.5 M C H_{3} C O O N a (a q)$ is titrated against a $0.5 M H C l (a q)$ . Calculate the $p H$ of the solution initially i.e. when no $H C l$ has been added.
Given: $p K_{a} (C H_{3} C O O H) = 4.74$