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

A 1.458g of Mg reacts with 80.0ml of a HCl solution whose pH is 0.477, The change is pH when all Mg has reacted will be:

Assume constant volume Mg=24.3g/mol(log3=0.47,log2=0.301)

A
0.176
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
+0.477
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
0.2385
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
0.3
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
Open in App
Solution

The correct option is D 0.3
pH=0.477
[H+]=10pH=3M=[HCl]
Moles of HCl=3×0.080
=0.24 moles
Moles of Mg=1.45824.3=0.06 moles
Mg+2HClMgCl2+H2
0.06 2×0.06
0.12 moles of HCl is required for 0.06 moles of Mg
Moles of HCl left =0.240.12
=0.12
New concentration of [H+]=0.120.08=1.5M
New pH=log[H+]=log(1.5)=log(32)
=log3+log2
=0.47+0.301
=0.169
Change in pH=0.169(0.477)
=0.3

flag
Suggest Corrections
thumbs-up
0
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Laws of Mass Action
CHEMISTRY
Watch in App
Join BYJU'S Learning Program
CrossIcon