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

A student prepares a buffer solution which contains 0.05 mol dm3 HCOOH and 0.250 mol dm3 HCOONa+. The Ka of HCOOH is 1.78×104. Calculate the pH of the solution.
log 1.78=0.250, log 5=0.699

A
12.5
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
4.45
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
C
6.2
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
7.5
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct option is B 4.45
Given, [HCOOH]=0.05 mol dm3, [HCOONa+]=0.250 mol dm3, Ka(HCOOH)=1.78×104
Using the Henderson-Hasselbalch equation for weak bases:
pH=pKa+log[[conjugate base][acid]]
pH=pKa+log[[HCOONa+]][HCOOH]]
pH=log Ka+log[[HCOONa+][HCOOH]]
pH=log Ka=+log[[HCOONa+][HCOOH]
pH=log (1.78×104)+log[[0.250][0.05]]
pH=4log (1.78)+log (5)
pH=40.250+0.699
pH=4.45

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