Answer
Molar solubility $(AgBr)$ = $7.31 \times 10^{-7}M$
Work Step by Step
1. Write the $K_{sp}$ expression:
$ AgBr(s) \lt -- \gt 1Ag^{+}(aq) + 1{Br}^{-}(aq)$
$5.35 \times 10^{-13} = [Ag^{+}]^ 1[{Br}^{-}]^ 1$
2. Considering a pure solution: $[Ag^{+}] = 1x$ and $[{Br}^{-}] = 1x$
$5.35 \times 10^{-13}= ( 1x)^ 1 \times ( 1x)^ 1$
$5.35 \times 10^{-13} = 1x^ 2$
$5.35 \times 10^{-13} = x^ 2$
$ \sqrt [ 2] {5.35 \times 10^{-13}} = x$
$7.31 \times 10^{-7} = x$
- This is the molar solubility value for this salt.
