Predicting precipitation

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

• $Q=K_{\text{sp}}⟹$ saturated: no more solute can be dissolved
• $Q<K_{\text{sp}}⟹$ unsaturated: more solute can still be dissolved
• $Q>K_{\text{sp}}⟹$ supersaturated: extra solute will precipitate

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:

AgNO_3(aq) + KCl_(aq) ⇌ KNO_3(aq) + AgCl_(s).

We can calculate the concentrations of the silver ion and chloride ion using $V_1{c}_1=V_2{c}_2\text{.}$ The concentration of silver ions is equivalent to the concentration of AgNO_3(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:

$Q=[{\text{Ag}}^{+}][{\text{Cl}}^{-}]=(0.0050\text{mol/L})(0.0025\text{mol/L})=1.3\times {10}^{-5}\text{.}$

Since $Q>K_{\text{sp}}\text{,}$ the answer is yes—a precipitate will form.