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

Calculate the pH of 0.1 M NH4Cl solution in water :
Kb(NH4OH)=1.8×105

A
4.5
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
7.63
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
6.53
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
5.13
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
Open in App
Solution

The correct option is D 5.13
Kb=1.8×105pKb=log(Kb)pKb=log(1.8×105)=(50.25)=4.75
NH4Cl is a salt of strong acid and weak base.
pH=712(pKb+logC)pH=712(4.75+log(0.1))pH=73.742pH=5.125



Theory :
Salt of strong acid (SA) and weak base(SB):
Consider a salt (MX) added to water and produce cations (M+)and anions (X) in water.
For example :- NH+4(aq)+H2O(l)NH4OH(aq)+H+(aq)
C 0 0
CCh Ch Ch
In this reaction the (Kh) hydrolysis constant is:
Kh=[NH4OH][H+][NH+4]=(Ch)2C(1h)
NH4OH(aq)NH+4(aq)+OH(aq)
In this reaction the Kb constant is:
Kb=[NH4][OH][NH4OH]
H2O(l)H+(aq)+OH(aq)
In this reaction the Kw constant is:
Kw=[H+][OH]
Using he value of KbandKw we can derive the value of Kh
KwKb=[H+][OH][NH4][OH][NH4OH]
KwKb=[NH4OH][H+]NH+4
Kh=KwKb
Form earlier equation of Kh
Kh=(Ch)2C(1h) Generally h<<1. So, neglect(1-h)
If h<<1
Ch2=Kh=KwKb
Then h=KhC=KwKbC
[H+]=Ch
[H+]=C×KwKbC=KwCKb
Taking log on both sides, we get
log[H+]=12log Kw12log Kb+12logC
pH=12pKw12pKb12log C
Since pKb=logKb and pKw=logKw
Valid if h<0.1 or 10%
pH=12[pKwpKblog C]
At 25 pH=712pKb12log C
If h obtained from KhC > 0.1, then solve the quadratic equation and get h. Then, [H+]=Ch and now pH calculation can be done.

flag
Suggest Corrections
thumbs-up
0
Join BYJU'S Learning Program
Join BYJU'S Learning Program
CrossIcon