Solubility product of silver bromide is 5.0×10−13. How much of pota – InterviewSolution

1.	Solubility product of silver bromide is 5.0×10−13. How much of potassium bromide (120 g mol−1) has to be added to 1 L of 0.05 M solution of silver nitrate to start the precipitation of AgBr ?
Answer» Solubility product of silver bromide is $5.0 \times 10^{- 13}$ . How much of potassium bromide $(120 {g mol}^{- 1})$ has to be added to $1 L$ of $0.05 M$ solution of silver nitrate to start the precipitation of $A g B r$ ?

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