The solubility product of silver bromide $(AgBr)$ is $5.0 \times 10^{-13}$. What is the minimum amount of potassium bromide $(KBr)$ that must be added to $1 \ L$ of $0.05 \ M$ silver nitrate $(AgNO_3)$ solution to initiate the precipitation of $AgBr$? (Molar mass of $KBr = 120 \ g \ mol^{-1}$)

(A) The precipitation of $AgBr$ occurs when the ionic product $[Ag^+][Br^-]$ exceeds the solubility product $K_{sp}$.
Given: $K_{sp}(AgBr) = 5.0 \times 10^{-13}$,$[Ag^+] = 0.05 \ M$.
For precipitation to start: $[Ag^+][Br^-] \geq K_{sp}$.
$0.05 \times [Br^-] = 5.0 \times 10^{-13}$.
$[Br^-] = \frac{5.0 \times 10^{-13}}{0.05} = 1.0 \times 10^{-11} \ M$.
Since $1 \ L$ of solution is used,the moles of $KBr$ required = $1.0 \times 10^{-11} \ mol$.
Mass of $KBr$ = $\text{moles} \times \text{molar mass} = 1.0 \times 10^{-11} \ mol \times 120 \ g \ mol^{-1} = 1.2 \times 10^{-9} \ g$.