15.1 Precipitation and dissolution  (Page 8/17)

Precipitation of silver halides

Solution

If the solution contained about equal concentrations of Cl ^– and I ^– , then the silver salt with the smallest K _sp (AgI) would precipitate first. The concentrations are not equal, however, so we should find the [Ag ⁺ ] at which AgCl begins to precipitate and the [Ag ⁺ ] at which AgI begins to precipitate. The salt that forms at the lower [Ag ⁺ ] precipitates first.

For AgI: AgI precipitates when Q equals K _sp for AgI (1.5 $\times$ 10 ^–16 ). When [I ^– ] = 0.0010 M :

AgI begins to precipitate when [Ag ⁺ ] is 1.5 $\times$ 10 ^–13 M .

For AgCl: AgCl precipitates when Q equals K _sp for AgCl (1.6 $\times$ 10 ^–10 ). When [Cl ^– ] = 0.10 M :

AgCl begins to precipitate when [Ag ⁺ ] is 1.6 $\times$ 10 ^–9 M .

AgI begins to precipitate at a lower [Ag ⁺ ] than AgCl, so AgI begins to precipitate first.

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Common ion effect

Solution

We can solve this equation using the quadratic formula, but we can also make an assumption to make this calculation much simpler. Since the K _sp value is so small compared with the cadmium concentration, we can assume that the change between the initial concentration and the equilibrium concentration is negligible, so that 0.010 + x ~ 0.010. Going back to our K _sp expression, we would now get:

Therefore, the molar solubility of CdS in this solution is 1.0 $\times$ 10 ^–26 M .

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