Question

The Ksp of silver bromide is 5.0 x 10-13. What is solubility of AgBr?

Answer 1

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:

$\begin{gathered} \lbrack Ag^+\rbrack=s \\ \lbrack Br^-\rbrack=s \\ K_{sp}=s\cdot s \\ K_{sp}=s^2 \\ s=\sqrt[]{K_{sp}}=\sqrt[]{5.0\times10^{-13}}M=\sqrt[]{0.50\times10^{-12}}M=\sqrt[]{0.50}\times10^{-6}M=0.70710\ldots\times10^{-6}M\approx7.1\times10^{-7}M \end{gathered}$

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