Answer
The aqueous solubility of $AgBr$ at $100 ^{\circ} C$ is equal to $2.236 \times 10^{-5}.$
Work Step by Step
1. Write the $K_{sp}$ expression:
$ 1AgBr(s) \lt -- \gt 1Br^-(aq) + 1Ag^+(aq)$
$K_{sp} = [Br^-]^ 1[Ag^+]^ 1$
2. Considering a pure solution: $[Br^-] = 1x$ and $[Ag^+] = 1x$
$5 \times 10^{-10}= ( 1x)^ 1 \times ( 1x)^ 1$
$5 \times 10^{-10} = x^ 2$
$ \sqrt [ 2] {5 \times 10^{-10}} = x$
$2.236 \times 10^{-5} = x$
- This is the molar solubility value for this salt.
