Still Confused?

Try reviewing these fundamentals first

- Home
- Chemistry 12
- Solubility Equilibrium

Still Confused?

Try reviewing these fundamentals first

Nope, got it.

That's the last lesson

Start now and get better math marks!

Get Started NowStart now and get better math marks!

Get Started NowStart now and get better math marks!

Get Started NowStart now and get better math marks!

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: