Question

Calculate the solubility product of silver bromide,if its solubility is 1.6×10^-5moldm^-3

Answer 1

Let the solubility of AgBr be S mollitre^-1

$AgBr⇌Ag ^ + +Br ^ −$

$Hence, [Ag ^ + ][Br ^ − ]=5×10 ^{ −13}$

$[Ag ^ + ]=0.05$

$So [Br ^ − ]=10 ^{−11} M$

It means 10^-11 moles of KBr moles in 1 litre of solution.

molecular mass of KBr=120

So 1.2×10^-9 g of KBr to be added to 1 litre of 0.05M solution of silver nitrate to start precipitation of AgBr.