# Problem: Calculate the molar solubility of AgBr in 0.035 M NaBr (Ksp = 5 × 10–13).

###### FREE Expert Solution

Notice that there is a common ion present, Br-. We can construct an ICE table for the dissociation of AgBr. Remember that solids are ignored in the ICE table. The Ksp expression for AgBr is:

$\overline{){{\mathbf{K}}}_{{\mathbf{sp}}}{\mathbf{=}}\frac{\mathbf{products}}{\overline{)\mathbf{reactants}}}{\mathbf{=}}\mathbf{\left[}{\mathbf{Ag}}^{\mathbf{+}}\mathbf{\right]}\mathbf{\left[}{\mathbf{Br}}^{\mathbf{-}}\mathbf{\right]}}$

Note that each concentration is raised by the stoichiometric coefficient: both [Ag+] and [Br] are raised to 1.

85% (128 ratings) ###### Problem Details

Calculate the molar solubility of AgBr in 0.035 M NaBr (Ksp = 5 × 10–13).

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