Predicting a precipitate  Solubility Equilibria
Predicting a precipitate
Lessons
Notes:
In this lesson, we will learn:
 To recall the ionic product Q and how it relates to the solubility product K_{sp}.
 How to calculate the ionic product and use it to predict whether a precipitate will form in solution.
Notes:
 As mentioned in the lesson Solubility product; the value of K_{sp} for a saturated solution of a given compound is constant. It doesn’t matter how the ion concentrations are made up (whether by one solution or multiple combined); if the solution is saturated then the product of the individual ion concentrations will be equal to K_{sp}.
 Using this fact, we can predict if a precipitate forms in the solution or not before making it by using the K_{sp} value. This K_{sp} product value is the product of the ion concentrations required to make a saturated solution (for the equilibrium to be established).
 If this product of ion concentration is smaller than K_{sp}, the solution will not be saturated and the equilibrium will not be established.
 The ‘product of ion concentration’ is known simply as the ionic product with label Q. As a product, Q is calculated by multiplying the concentrations of ions together. For aqueous ions M^{x}+ and X^{m} forming precipitate M_{m}X_{x}:
Q = [M^{+}]^{m} [X^{}]^{x}  Compare this to K_{sp} which is the ion concentration needed to form a saturated solution. Using these two terms, three situations can be described when two solutions are mixed to create a new ionic species.
 If Q is smaller than K_{sp}, then there are less ions in solution than necessary to form a saturated solution. Without a saturated solution, a precipitate will not form.
 If Q is equal to K_{sp}, then there are just enough ions in solution necessary for a saturated solution to form. In this scenario, a solution that has just reached saturation will form. This means that any further addition of aqueous ions will form a precipitate, as the equilibrium shifts to the left to favor the undissolved state.
 If Q is greater to K_{sp}, then there are more ions in solution than are necessary to form a saturated solution. In this scenario, an equilibrium will be established to form a precipitate with the excess ions; a precipitate will form.
 For example: a solution made by combining 100 mL of 0.1 M Ca^{2+}_{(aq)} and 75mL of 0.2 M F^{}(aq) :
Q = [Ca^{2+}][F^{}]^{2}$\qquad$ To Form CaF_{2}
These solutions dilute each other. Find the new concentration by dividing original volume by combined volume.[Ca^{2+}]_{sol} = 0.1 M * $\large \frac{0.1 \, L}{0.175 \, L}$ = 0.057 M [F^{}]_{sol} = 0.2 M * $\large \frac{0.075 \, L}{0.175 \, L}$ = 0.086 M Q = [0.057] x [0.086]^{2}$\,$ = 4.22*10^{4} K_{sp} (CaF_{2}) = 3.45*10^{11} M
Q > K_{sp} therefore a precipitate will form.

Intro Lesson
Predicting if a solution will form a precipitate.

1.
Use the solubility product expression to predict a precipitate.^{1}