Predicting precipitation

The reaction quotient (Q) can be used in the context of solubility. It is calculated in exactly the same way as Ksp, except the solution is not necessarily saturated. We call this value the trial ion product, and we compare it with Ksp to predict precipitation:

Example

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 V1c1=V2c2. 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:

Q=[Ag+][Cl]=(0.0050 mol/L)(0.0025 mol/L)=1.3×105.

Since Q>Ksp, the answer is yes—a precipitate will form.