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

A solution has 0.05 M Mg2+ and 0.05 M NH3. Calculate the concentration of NH4Cl required to prevent the formation of Mg(OH)2 in this solution. Ksp of Mg(OH)2=9.0×1012 and ionization constant of NH3=1.8×105
(Given, 1.8=1.34)

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

The correct option is A 0.067 M
The maximum concentration of [OH] ions that precipitate Mg(OH)2 is calculated by equation
Ksp=[Mg2+][OH]2
[OH]2=Ksp[Mg2+]=9.0×10120.05=1.8×1010
[OH]=1.34×105 M
NH3 is present in the solution in the form of NH4OH
NH30.05+H2ONH4OH0.05NH+4+OH
The ionization of NH4OH is suppressed by addition of NH4Cl (strong electrolyte)
KNH3=KNH4OH=[NH+4][OH][NH4OH]
Whole of the concentration of NH+4 ions is provided by NH4Cl
[NH+4]=KNH4OH×[NH4OH][OH]
=1.8×105×0.051.34×105=0.067 M
[NH4Cl]=0.067 M

flag
Suggest Corrections
thumbs-up
1
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
pH and pOH
CHEMISTRY
Watch in App
Join BYJU'S Learning Program
CrossIcon