# Chapter 15 - Additional Aqueous Equilibria - Problem Solving Practice 15.9 - Page 679: a

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.

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