Answer
$AgBr$ solubility $\approx 1.667 \times 10^{-11}M$
Work Step by Step
$Kps (AgBr) = 5 \times 10^{-13}$
1. Find the inital concentration of $Ag^+$ in the solution:
$[Ag^+]_{(initial)} = [AgNO_3] = 3 \times 10^{-2}M $
2. Write the Kps equation:
$Kps = [Ag^+][Br^-]$
$Ag^+ solubility = Br^- solubility = x$
3. Find x:
$5 \times 10^{-13} = [x + 3 \times 10^{-2}][x]$
*Since x has a very small value compared to $3 \times 10^{-2}$:
$5 \times 10^{-13} = [3 \times 10^{-2}][x]$
$\frac{5 \times 10^{-13}}{3 \times 10^{-2}} = x$
$x \approx 1.667 \times 10^{-11}M$
