# How to calculate ion concentrations for a precipitation from the solubility product constant? [closed]

The equilibrium constant for dissolving the insoluble substance gold (I) iodide, AuI (s), in aqueous solution is 1.6 × 10^-23 at 25ºC. Write the equilibrium constant expression and calculate the concentration of Au + (aq) and I (aq) ions in a solution in which 0.345g of AuI(s) is in equilibrium with the aqueous ions. The answer is 4x10^-12.

So 1.6x10^-23=products/reactants. The products = x^2, but what does the reactant equal to? I think they got the answer by ignoring the reactants, is this possible?

## closed as off-topic by bon, Todd Minehardt, Jon Custer, Klaus-Dieter Warzecha, WildcatJul 21 '16 at 9:48

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## 1 Answer

The concentration (the activity, to be precise) of solids is assumed 1. [Au+][I-]/[AuI]=[Au+][I-] = Ks. Here Ks is the solubility constant. If water originally didn't have any I- or Au+, then the amounts of [Au+]=[I-]=x. Then you have x*x=Ks and x= square root of Ks. The amount of AuI doesn't seem to play any role because they didn't specify the amount of water it is dissolved in.