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Mitchell Simoens · Answer 1 · 2023-03-13T02:13:26+0000

Ksp is a constant that gives the following:

For the solution of AgBr:

$AgBr(s)\to Ag^+(aq)+Br^-(aq)$
$K_(sp)=\lbrack Ag^+\rbrack\lbrack Br^-\rbrack$

So, given an initial pure water, let s be the solubility of AgBr. If we add s of AgBr, it will dossiciate into s Ag⁺ and s Br⁻, so, for a saturated solution of AgBr, we have:

$\begin{gathered} \lbrack Ag^+\rbrack=s \\ \lbrack Br^-\rbrack=s \\ K_(sp)=s\cdot s \\ K_(sp)=s^2 \\ s=\sqrt[]{K_(sp)}=\sqrt[]{5.0*10^(-13)}M=\sqrt[]{0.50*10^(-12)}M=\sqrt[]{0.50}*10^(-6)M=0.70710\ldots*10^(-6)M\approx7.1*10^(-7)M \end{gathered}$

So, the solubility of AgBr is approximately 7.1 x 10⁻⁷ M.

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