Predicting Whether Precipitation Will Occur

In the topic on equilibrium we saw how to predict the direction in which a reaction would move to reach equibirium by using Q the reaction quotient.  Exactly the same techniques can be used to decide whether a precipitate is likely to form from the ions in two electrolyte solutions that are mixed.  In this case the equilibrium constant is the solubility product, K_sp, and the quotient is denoted Q_sp.  Precipitation occurs when Q_sp is equal to or greater than K_sp.

 Example 1 

If 45 mL of 2.6e-4 M Mg(NO₃)₂ is mixed with 51 mL of 7.2e-4 M NaOH, Does a precipitate form? The Ksp for Mg(OH)₂ is 1.8 x 10^-11.

Solution
First find the concentrations of Mg2+ and OH- in the mixture. We will do this by using the dilution equation for each. Recall that the dilution equation is C₁V₁ = C₂V₂
V₂ = 45 mL + 51 mL = 96 mL
initial conc. Mg2+ = (45 mL)(2.6e-4 M)/(96 mL) = 1.2e-4 M
initial conc. OH- = (51 mL)(7.2e-4 M)/(96 mL) = 3.8e-4 M
Qsp = {Mg²⁺}{OH^-} = (1.2e-4)(3.8e-4)2 = 1.73e-11
Since Qsp is less than Ksp the answer is no. A precipitate will not form.

  Example 2 

What is the I^- concentration just as AgCl begins to precipitate when 2.0 M AgNO₃ is slowly added to a solution containing 0.010 M Cl^- and 0.010 M I^-?  The K_sp of AgCl is 1.8 x 10^-10 and the K_sp of AgI is 8.3 x 10^-17.

Overview of problem. We are adding Ag⁺ ions to a solution containing both Cl^- and I^- ions.  Both these ions can form a precipitate with silver ions, but AgI will form first since it is the least soluble. As the AgI precipitates the concentration of I^- in the solution will decrease and the concentration of Ag⁺ will increase until it becomes large enough to precipitate the Cl^- as AgCl.  So the first thing we must do is determine the [Ag⁺] when AgCl starts to precipitate, we then use this concentration of Ag⁺ in the K_sp expression for AgI to determine the [I^-]. 

Solution

1.8 x 10^-10 = [Ag⁺][Cl^-] = [Ag⁺](0.010)

[Ag⁺] = 1.8 x 10^-10 / 0.010 = 1.8 x 10^-8 M   (this is the conc of Ag⁺ when AgCl starts to precipitate)

8.3 x 10^-17 = [Ag⁺][I^-] = (1.8 x 10^-8)[I^-]

 [I^-] = 8.3 x 10^-17 / 1.8 x 10^-8 = 4.6 x 10^-9 M   ( the conc of I^- remaining when AgCl starts to precipitate)