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##### Intros
###### Lessons
1. Predicting if a solution will form a precipitate.
2. Recalling Ksp.
3. The ionic product Q (with worked example).
4. Predicting precipitates using Ksp and Q.
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##### Examples
###### Lessons
1. Use the solubility product expression to predict a precipitate.1
1. A solution was made by combining 75 mL of 0.2M Mg2+(aq) and 110 mL of 0.08M OH-(aq). Use these values and the solubility product expression to predict whether a precipitate of Mg(OH)2 will form in the resultant mixture.
2. A solution was made by combining 20 mL of 0.01M Fe3+(aq) and 25 mL of 0.01M OH-(aq). Use these values and the solubility product expression to predict whether a precipitate of Fe(OH)3 will form in the resultant mixture.
3. A solution was made combining 50 mL of 0.003M Ca2+(aq) and 40 mL of 0.008M SO42-(aq). Use these values and the solubility product expression to predict whether a precipitate of CaSO4 will form in the resultant mixture.

1 Source for Ksp solubility constant data at 250C: http://www4.ncsu.edu/~franzen/public_html/CH201/data/Solubility_Product_Constants.pdf
###### Topic Notes

In this lesson, we will learn:

• To recall the ionic product Q and how it relates to the solubility product Ksp.
• 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 Ksp 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 Ksp.
• Using this fact, we can predict if a precipitate forms in the solution or not before making it by using the Ksp value. This Ksp 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 Ksp, 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 Mx+ and Xm- forming precipitate MmXx:

Q = [M+]m [X-]x

• Compare this to Ksp 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 Ksp, 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 Ksp, 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 Ksp, 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 Ca2+(aq) and 75mL of 0.2 M F-(aq) :

Q = [Ca2+][F-]2$\qquad$ To Form CaF2

These solutions dilute each other. Find the new concentration by dividing original volume by combined volume.

[Ca2+]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

Ksp (CaF2) = 3.45*10-11 M

Q > Ksp therefore a precipitate will form.