MyQuestionIcon

1

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

pH the Power of H

50 mL of 0.2 ...

Question

50 mL of 0.2 M ammonia solution is treated with 25 mL of 0.2 M HCI. If $p K_{b}$ , of ammonia solution is 4.75, the pH of the mixture will be :

A

3.75

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

B

9.25

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

C

8.25

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

D

4.75

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

Open in App

Solution

The correct option is C 9.25
Number of millimoles of $N H_{3} = 50 m L \times 0.2 M = 10 m m o l$
Number of millimoles of $H C l = 25 m L \times 0.2 M = 5 m m o l$
5 mmol $N H_{3}$ will react with 5 mmol $H C l$ to form 5 mmol $N H_{4} C l$
$10 - 5 = 5$ mmol $N H_{3}$ will remain.
$p O H = p K_{b} - l o g \frac{[N H_{4} C l]}{[N H_{3}]}$
$p O H = 4.75 - l o g \frac{5 / V}{5 / V}$
$p O H = 4.75$
$p H = 14 - p O H = 14 - 4.75 = 9.25$

Suggest Corrections

thumbs-up

0

Similar questions

40 m l

0.1 M

ammonia solution is mixed with

20 m l

0.1 M H C l

. What is the pH of the mixture?

p K_{b}

of ammonia solution is

4.74

m L

0.1 M

H_{2} S O_{4}

solution is added to

30 m L

0.2 M

N H_{4} O H

p H

of the resultant mixture is:

[P k_{b} o f N H_{4} O H = 4.11]

20 m L

0.1 M H_{2} S O_{4}

solution is added to

30 m L

0.2 M N H_{4} O H

p H

of the resultant mixture is

[p K_{b} o f N H_{4} O H = 4.7, log 2 = 0.3010]

pH of two solutions :
I. 50 ml of 0.2 M~HCl + 50 ml of 0.2 M HA $(K_{a} = 1.0 \times 10^{- 5})$ and
II. 50 ml of 0.2 M HCl + 50 ml of 0.2 M NaA will be respectively

Q. Calculate pH of a solution of a mixture of 10 ml 0.1 M HCI and 40 ml 0.2 M HBr

Join BYJU'S Learning Program

similar_icon

Related Videos

lock

pH and pOH

CHEMISTRY

Watch in App

explore_more_icon

Explore more

pH the Power of H

Standard XII Chemistry

Join BYJU'S Learning Program

CrossIcon