Predicting precipitation

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

• $\displaystyle Q = K_{\text{sp}} \Rightarrow$ saturated: no more solute can be dissolved
• $\displaystyle Q < K_{\text{sp}} \Rightarrow$ unsaturated: more solute can still be dissolved
• $\displaystyle Q > K_{\text{sp}} \Rightarrow$ 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 $\displaystyle V_{1} c_{1} = V_{2} c_{2}$. 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:

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

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