Predicting precipitation

The reaction quotient (`Q`) can be used in the context of solubility. It is calculated in exactly the same way as `K_"sp"`, except the solution is not necessarily saturated. We call this value the trial ion product, and we compare it with `K_"sp"` 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 `V_1c_1 = V_2c_2`. 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") = 1.3xx10^-5`.

Since `Q > K_"sp"`, the answer is yes—a precipitate will form.