1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

# Step-I: 0.5 L of saturated Mg(OH)2 (aq.) is in contact with Mg(OH)2(s). Step-II: 0.5 L of H2O is added to 0.5 L of solution in step (I) and the solution is vigorously stirred, undissolved Mg(OH)2(s) remains. Step-III: 100 mL of the clear solution in step (II) is removed and added to 0.5 L of 0.1 M HCl(aq.). Step-IV: 25 mL of the clear solution in step (II) is removed and added to 225 mL of 0.06 M MgCl2(aq.) (Ksp,Mg(OH)2=3.2×10−11). Which of the following are correct choices about above steps?

A
Concentration of Mg2+ (aq.) in step (II) is 2×104 M
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
B
Concentration of Mg2+ (aq.) in step (III) is 3.33×105 M
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
C
Concentration of Mg2+ (aq.) in step (IV) is nearly equal to 0.054 M
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
D
Precipitation of Mg(OH)2(s) will occurs in step (III)
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

## The correct options are A Concentration of Mg2+ (aq.) in step (II) is 2×10−4 M B Concentration of Mg2+ (aq.) in step (III) is 3.33×10−5 M C Concentration of Mg2+ (aq.) in step (IV) is nearly equal to 0.054 MStep-I: [Mg2+]=3√3.2×10−114=2×10−4M Step-II: [Mg2+]=2×10−4M since it will again gives saturated solution. Step-III: [Mg2+]=2×10−4×100(100+500)=3.33×10−5 M Step-IV: Mg(OH)2 (25 mL of 2×10−4 M or 0.005 m mole) is mixed with MgCl2 (225 mL of 0.06 M or 13.5 m mole) so [Mg2+]mainly from MgCl2≃13.5250M=0.054M

Suggest Corrections
0
Join BYJU'S Learning Program
Related Videos
Gases in Liquids and Henry's Law
CHEMISTRY
Watch in App
Explore more
Join BYJU'S Learning Program