# 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.