Predicting precipitation

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


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\displaystyle V_{1} c_{1} = V_{2} c_{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]=(0.0050mol/L)(0.0025mol/L)=1.3×105\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 Q>Ksp\displaystyle Q > K_{\text{sp}}, the answer is yes—a precipitate will form.