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

0.1 mol of CH3NH2 (Kb=5×104) is mixed with 0.08 mol of HCl and diluted to 1 L. Which statement(s) is/are correct?
(take log 2=0.3)

A
The concentration of H+ ions is 8×1011 M
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
B
HCl is left over in the solution because the weak base does not completely neutralise it
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
This forms a basic buffer
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
D
The pOH of solution is 5.9
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct options are
A The concentration of H+ ions is 8×1011 M
C This forms a basic buffer
CH3NH2+HClCH3NH+3Cl
Initial 0.1 0.08
moles
Final (0.1-0.08) 0 0.08
All of the HCl gets neutralised since there is a larger quantity of the base present. The low Kb value of the base does not affect this.
Since weak base CH3NH2 is left along with its salt of strong base, it forms a basic buffer solution so using Henderson Hasselbalch equation:
pOH=pKb+log[salt][base]
or
[OH]=Kb[Base][Salt]=5×104×0.020.08=1.25×104 M
and
[H+]=Kw[OH]=10141.25×104=8×1011 M
so, pH=log(8×1011)
=log(23)+11=3×0.3+11
pH=10.1
pOH=1410.1=3.9

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