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

# Solubility product of silver bromide is 5.0×10−13. The quantity of potassium bromide (molar mass taken as 120 g mol−1) to be added to 1 litre of 0.05 M solution of silver nitrate to start the precipitation of AgBr is:

A
1.2×1010g
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
1.2×109g
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
C
6.2×105g
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
5.0×108g
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

## The correct option is B 1.2×10−9gAg++Br−→AgBrPrecipitation starts when ionic product just exceeds solubility productKsp=[Ag+][Br−][Br−]=Ksp[Ag+]=5×10−130.05=10−11i.e., precipitation just starts when 10−11 moles of KBr is added to 1 L of AgNO3 solution. No. of moles of KBr to be added =10−11Weight of KBr to be added =10−11×120 =1.2×10−9g

Suggest Corrections
0
Join BYJU'S Learning Program
Related Videos
Degree of Dissociation
CHEMISTRY
Watch in App
Explore more
Join BYJU'S Learning Program