wiz-icon
MyQuestionIcon
MyQuestionIcon
3
You visited us 3 times! Enjoying our articles? Unlock Full Access!
Question

Calculate pH of mixture of (400 mL,1200M Ba(OH)2)+(400 mL,150M HCl)+(200 mL of water)
log 2=0.3010

A
2.4
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
B
7
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
more than 7
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
4.5
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct option is A 2.4
Given, VBa(OH)2=VHCl=400 mL, [Ba(OH)2]=1200M, [HCl]=150M, VH2O=200 mL
Number of H+ ions from acid N1V1=400×150=8
Number of OH ions from base N2V2=400×1200×2=4
here, N1V1>N2V2 so solution will be acidic in nature.
so, [H+]=N1V1N2V2Total volume
[H+]=(84)400+400+200
[H+]=4×103
so, pH=log [H+]=log 4×103
pH=32 log 2=2.4

flag
Suggest Corrections
thumbs-up
15
Join BYJU'S Learning Program
Join BYJU'S Learning Program
CrossIcon