Solubility product of silver bromide is 5.0×10−13. How much of potassium bromide (120g mol−1) has to be added to 1L of 0.05M solution of silver nitrate to start the precipitation of AgBr ?
A
1.2×10−10g
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
5.0×10−8g
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
6.0×10−5g
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
1.2×10−9g
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
Open in App
Solution
The correct option is D1.2×10−9g AgNO30.05M⇌Ag+0.05M+NO−30.05M KBrxM⇌K+xM+Br−xM AgBr(s) is precipitated if [Ag+][Br−]>Ksp (0.05)(x)>5×10−13 ∴x=5×10−130.05=1×10−11mol L−1 ∴KBr to be added, =1×1011mol L−1=1×10−11×120g mol−1 =1.2×10−9g