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Question

# Solubility product of silver bromide is 5.0×10−13. The quantity of potassium bromide (molar mass is 120 gmol−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
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B
1.2×109g
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C
6.2×105g
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D
5.0×108g
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Solution

## The correct option is B 1.2×10−9gGiven, Ksp(AgBr)=5.0×10−13, [Ag+]=0.05 M. Molar mass of KBr = 120 gmol−1 Ag+(aq)+Br−(aq)⇌AgBr(s) Precipitation starts when ionic product just exceeds solubility product. Ksp=[Ag+][Br−] putting the values, ⇒[Br−]=Ksp[Ag+]=5×10−130.05=10−11 Precipitation starts when 10−11 moles of KBr is added to 1L of AgNO3 solution. Number of moles of KBr to be added =10−11 Weight of KBr=number of moles×molar mass putting the values, =10−11×120=1.2×10−9g

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