Answer
$AgBr$ $solubility$ $\approx 5 \times 10^{-12}$
Work Step by Step
$Kps (AgBr) = 5 \times 10^{-13}$
1. Find the inital concentration of $Br^-$ in the solution:
$[Br^-]_{(initial)} = [NaBr] = 0.1M $
2. Write the Kps equation:
$Kps = [Ag^+][Br^-]$
$Ag^+ solubility = Br^- solubility = x$
3. Find x:
$5 \times 10^{-13} = [x][x + 0.1]$
*Since x has a very small value compared to $0.1$:
$5 \times 10^{-13} \approx [x][0.1]$
$\frac{5 \times 10^{-13}}{0.1} \approx x$
$x \approx 5 \times 10^{-12}M$
