Solubility Unit 3 Lesson 1 Unit Intro Our

Solubility Unit 3 Lesson 1

Unit Intro • Our focus is on solutions of aqueous ions • As you know; acids, bases and salts form ionic solutions. • This unit is only concerned with salts.

Review Electrolytes: substances that dissolve to give electrically conducting solutions that contain ions. Ex: Mg. Cl 2(s) ® Mg+(aq) + 2 Cl-(aq) Na. NO 3 Cu. SO 4

Salts differ in their ability to dissolve in water. Some salts dissolve well and produce a solution with a high ion concentration. These are strong electrolytes. When salt don’t dissolve as well, they produce solutions with lower ion concentrations and are weaker electrolytes.

Review Non-electrolytes: a substance that dissolves to give non-conducting solutions containing only neutral molecules. Ex: C 2 H 2 OH (ethanol) C 12 H 22 O 11 sucrose

Molecular vs Ionic solutions metals and non-metals compounds that contain polyatomic ions Both are referred to as salts.

Molecular compounds They are covalent compounds Non-metal and non-metal For example. . all organic compounds. (high carbon content. Carbon bonds covalently to other elements).

Why do salts dissolve so well in water? Because of the dipole in the water molecule. This allows for the water molecule to compete with the electrostatic attraction between the salt ions. This process is called dissociation.

Useful hint When it comes to our chem 12 course, you are only going to deal with ionic compounds that have one type of positive ion and one type of negative ion only. - no KNa. FCl ® K+ + Na+ + F- + Cl-

Ionic Solutions Na. Cl(aq) Ca(OH)2(aq) (NH 4)3 PO 4(aq) Ca(CH 3 COO)2(aq) H 2 SO 4(aq) Conduct electricity Molecular/Covalent Solutions C 6 H 12 O 6(aq) metal or polyatomic ion C 12 H 22 O 11(aq) nonmetal or carbon CH 3 OH(aq). Molecular - covalently bonded O 2(aq) N 2 H 4(aq) Do not conduct electricity

Write equations to show the dissolving of the following substances in water Na. Cl(s) ® Na+ C 6 H 12 O 6(s) ® C 6 H 12 O 6(aq) Ca(OH)2(s) ® Ca 2+ C 12 H 22 O 11(s) ® C 12 H 22 O 11(aq) (NH 4)3 PO 4(s) ® 3 NH 4+ CH 3 OH(l) ® CH 3 OH(aq) + + Cl- 2 OH- + PO 43 -

Back to solubility In chem 11, you learnt that… The solubility of a substance is the maximum amount of the substance which can dissolve in a given amount of solvent at a given temperature. Units for this are….

Equilibrium Solubility The solubility of a substance is the equilibrium concentration of the substance in solution at a given temperature. when expressed in moles/L it’s called Molar Solubility.

Solubility at equilibirum Solid Mg. Cl 2 dissolves and enters solution dissolving reaction Mg+2 + Cl-1 ions come together to form Mg. Cl 2 crystallization reaction

Solubility at equilibrium When the rate of dissolving reaction equals the rate of crystallization reaction, we have equilibrium. – A solution at equilibrium is called a saturated solution.

Saturation exists when… • Equilibrium exists between the dissolved (ions) and the undissolved material (solid) • Some undissolved material is till present (crystal solids)

How to saturate a solution & determine solubility • To saturate a solution, add weighed portions of your solid to a volume of (solvent) water and stir until full. – A bit of excess solid will always be present at equilibrium saturation though. • In order to determine the solubility, you must completely fill or saturate the solution!

Determining The Solubility of Mg. Cl 2 Add measured portions of Mg. Cl 2 to 100. 0 m. L and stir to dissolve Rate of dissolving > Rate of crystallization Mg. Cl 22 Amount Mg. Cl 2 Dissolved unsaturated 100. 0 m. L Mg 2+ Cl- Mg. Cl 2(s) saturated 10. 0 g 3. 0 g 0. 0 g 33. 0 g slow very slow equilibrium Rate of dissolving = Rate of crystallization

Calculate the solubility in units of g/L and mole/L Solubility = Molar Solubility = 33. 0 g 0. 100 L Moles/L 33. 0 g = = x 1 mole 95. 3 g 0. 100 L = 3. 46 M 330. g/L

do not use the solid Equilibrium Equation Expression: Mg. Cl 2(s) ⇌ Mg 2+ + 2 Cl- Keq = [Mg 2+][Cl-]2 Ksp = [Mg 2+][Cl-]2 The Ksp or (solubility product) is used for saturated solutions at equilibrium

Un-Saturated, Saturated and Super-satured Review

Unsaturated Solutions Not full – ( more solid can dissolve if you add it) How does it look? Clear solution! The rate of dissolving > the rate of crystallizing Not at equilibrium

Saturated Solutions Full- ( adding more solid will not dissolve ) How does it look? it always has crystals/solids in the solution. The rate of dissolving = the rate of crystallizing At equilibrium

Supersaturated Solutions Over full – ( adding more solid causes precipitation) How does it look? Clear solution! The rate of dissolving < the rate of crystallizing Not at equilibrium Supersaturated video

Predicting the solubility of salts Page 332 in your textbook has the table “ Solubility of common compounds in water” Will be provided for you Use it to predict solubility ( high or low ) Use it to predict if a precipitate will form.

Low Solubility means £. 1 M High Solubility means >. 1 M Na 3 PO 4 High Cu. Cl 2 High Ca(NO 3)2 High Cu. Cl Low K 2 CO 3 High Cu. SO 4 High Ca. SO 4 Low Ag 2 SO 4 Low Fe. SO 4 High Ba. S High

Precipitate Questions Will a precipitate form when 0. 2 M solutions of Ca. S and Na 2 SO 4 are mixed?

Write the equation for equilibrium present in a saturated solution of Al 2(SO 4)3(s) solution. Al 2(SO 4)3(s) ⇌ 2 Al 3+ + Equilibrium Expression Ksp = [Al 3+]2[SO 42 -]3 3 SO 42 -

Write the equation for equilibrium present in a saturated solution of Ca 3(PO 4)2(s) solution. ⇌ Ca 3(PO 4)2(s) 3 Ca 2+ + Equilibrium Expression Ksp = [Ca 2+]3[PO 43 -]2 2 PO 43 -

Practice Time pg 74 #1 -2, pg 76 #3 -7, pg 77 #8 -11 Pg 83 #21, 22 Pg 84 #24.