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Question

Solubility product of silver bromide is 5.0×1013. How much of potassium bromide (120 g mol1) has to be added to 1 L of 0.05 M solution of silver nitrate to start the precipitation of AgBr ?

A
1.2×1010 g
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B
5.0×108 g
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C
6.0×105 g
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D
1.2×109 g
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Solution

The correct option is D 1.2×109 g
AgNO30.05 MAg+0.05 M+NO30.05 M
KBrx MK+x M+Brx M
AgBr(s) is precipitated if
[Ag+][Br]>Ksp
(0.05)(x)>5× 1013
x=5×10130.05=1×1011 mol L1
KBr to be added, =1×1011 mol L1=1×1011×120 g mol1
=1.2×109 g

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