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Get Started Now- Intro Lesson: a2:53
- Intro Lesson: b5:30
- Intro Lesson: c3:39
- Lesson: 1a4:40
- Lesson: 1b5:43
- Lesson: 1c3:44

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.

- As mentioned in the lesson Solubility product;
__the value of K__. It doesn’t matter how the ion concentrations are made up (whether by one solution or multiple combined);_{sp}for a saturated solution of a given compound is constant__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.

- Using this fact, we can predict if a precipitate forms in the solution or not before making it by using the K
- 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__, then there are less ions in solution than necessary to form a saturated solution. Without a saturated solution,_{sp}__a precipitate will not form__.__If Q is equal to K__, then there are just enough ions in solution necessary for a saturated solution to form. In this scenario,_{sp}__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__, then there are more ions in solution than are necessary to form a saturated solution. In this scenario, an equilibrium will be established to_{sp}__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 MQ = [0.057] x [0.086] ^{2}$\,$ =__4.22*10__^{-4}K _{sp}(CaF_{2}) = 3.45*10^{-11}M__Q > K__therefore a precipitate will form._{sp}

- Introduction
__Predicting if a solution will form a precipitate.__a)Recalling K_{sp}.b)The ionic product Q (with worked example).c)Predicting precipitates using K_{sp}and Q. - 1.
**Use the solubility product expression to predict a precipitate.**^{1}a)A solution was made by combining 75 mL of 0.2M Mg^{2+}_{(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.b)A solution was made by combining 20 mL of 0.01M Fe^{3+}_{(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.c)A solution was made combining 50 mL of 0.003M Ca^{2+}_{(aq)}and 40 mL of 0.008M SO_{4}^{2-}_{(aq)}. Use these values and the solubility product expression to predict whether a precipitate of CaSO_{4}will form in the resultant mixture.^{1}**Source for K**http://www4.ncsu.edu/~franzen/public_html/CH201/data/Solubility_Product_Constants.pdf_{sp}solubility constant data at 25^{0}C:

12.

Solubility Equilibria

12.1

Solubility and ion concentration

12.2

Predicting solubility of salts

12.3

Ionic equations and formulae

12.4

Separating mixtures by precipitation

12.5

The solubility product

12.6

Predicting a precipitate

12.7

Pollution and hard water treatment by precipitation

12.8

The common ion effect