Answer
$[AgBr]$ solubility in a water solution: $ 7.071 \times 10^{-7}M$
Work Step by Step
$Kps (AgBr) = 5 \times 10^{-13}$
1. Write the Kps equation:
$Kps (AgBr) = [Ag^+] * [Br^-]$
$[Ag^+] = [Br^-] = x$
$Kps (AgBr) = x * x$
2. Find x:
$5 \times 10^{-13} = x^2$
$x = \sqrt {5 \times 10^{-13}}$
$x = 7.071 \times 10^{-7}$
3. Since $[Ag^+] = [AgBr] = x$
$[AgBr] = 7.071 \times 10^{-7}$