The reaction quotient () can be used in the context of solubility. It is calculated in exactly the same way as , except the solution is not necessarily saturated. We call this value the trial ion product, and we compare it with to predict precipitation:
- saturated: no more solute can be dissolved
- unsaturated: more solute can still be dissolved
- supersaturated: extra solute will precipitate
If 25.0 mL of 0.010 mol/L silver nitrate is mixed with 25.0 mL of 0.0050 mol/L potassium chloride, will a precipitate form? The solubility product constant for silver chloride is 1.8 × 10−10.
First, write the balanced double displacement reaction, consulting the solubility guidelines to figure out which products are aqueous:
AgNO3(aq) + KCl(aq) ⇌ KNO3(aq) + AgCl(s).
We can calculate the concentrations of the silver ion and chloride ion using . The concentration of silver ions is equivalent to the concentration of AgNO3(aq) once it is diluted to 50.0 mL, which we calculate to be 0.0050 mol/L. Using the same method, the concentration of chloride ions is 0.0025 mol/L.
Now we can write the solubility equilibrium equation:
AgCl(s) ⇌ Ag+(aq) + Cl−(aq).
And calculate the trial ion product:
Since , the answer is yes—a precipitate will form.