wiz-icon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

The number of grams/weight of NH4Cl required to be added to 3 litres of 0.01 M NH3 to prepare the buffer of pH=9.45 at temperature 298 K (Kb for NH3 is 1.85×105).

A
0.354 gm
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
B
4.55 gm
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
0.455 gm
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
3.55 gm
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct option is A 0.354 gm
pOH=pKb+log[NH4Cl][NH3]
pOH=logKb+log[NH4Cl][NH3]
(pKb=logKb)
Also, pOH=14pH=149.45=4.55
and log[NH4Cl][NH3].Kb=log30.470
Thus, log1014(log109+log3)log[NH4Cl][NH3].Kb
log1014109×3=log[NH4Cl][NH3].Kb
1014109×3[NH4Cl][NH3].Kb
or 1014×[NH3].Kb109×3
=[NH4Cl]1014109×3×0.01×1.85×105
=[NH4Cl]0.354 gm.

flag
Suggest Corrections
thumbs-up
0
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Buffer Solutions
CHEMISTRY
Watch in App
Join BYJU'S Learning Program
CrossIcon