Question: Calculate the molar solubility of \(AgBr\;in\; 0.035MNaBr\left( {{K_{sp}} = 5 \times 1{0^{ - 13}}} \right)\).

The molar solubility is the chemical property of a solute and represents the maximum concentration of a solute in a solution at a certain temperature and pressure. Such a solution is called saturated.

AgBr dissociates according to the equation:

\(AgBr({\rm{s}}) \leftrightarrow {\rm{Ag}} + ({\rm{aq}}) + {\rm{Br}} - ({\rm{aq}})\)

The solubility product \({\rm{\;(Ksp) of\;AgBr}}\)

\({\rm{Ksp}} = ({\rm{Ag}} + )({\rm{Br}} – )\)

As we have the 0.035 M NaBr we can conclude that we have 0.035 M Br ions in the solution.

\(\begin{array}{*{20}{c}}{5.00 \times {{10}^{ - 13}} = ({\rm{Ag}} + ) \cdot 0.035{\rm{M}}}\\{({\rm{Ag}} + ) = \frac{{5.00 \times {{10}^{ - 13}}}}{{0.035{\rm{M}}}}}\end{array}\)

\(({\rm{Ag}} + ) = 1.43 \times {10^{ - 11}}{\rm{M}}\)

The solution for molar solubility is \(1.43 \cdot {10^{ - 11}}{\rm{M}}\)