Solutions and Solubility

What is a Solution? Homogeneous mixture (one phase) Consists of a solvent and one/or more solute –Solvent is the “bigger” part – it is the part that does the dissolving –Solute is the “smaller” part – it is dissolved in the solvent A solution can be acidic (turn litmus RED), basic (turn litmus BLUE) or neutral (no change in litmus colour)

Aqueous (aq) Solutions Water is the solvent  “universal solvent” Can be a(n): –Electrolyte: conducts electricity Ionic compounds (NaCl, KBr, etc) Bases (NaOH, KOH, etc) Acids (HCl, acetic acid, etc) –Non-electrolyte: does NOT conduct electricity Molecular compounds (methanol, sugar)

Electrolytes: Dissociation When you dissolve a solid in a solvent (eg. water), the solid dissociates (breaks apart) into charged IONS. –Dissolve salt (NaCl) into water, separates into charged Na + and Cl - ions. These ions can move from one point in a solution to another. –Solid ionic compounds (salt itself) cannot conduct electricity because it is not mobile. Non-electrolytes can dissolve in water, but do not dissociate = no charged particles = no electricity.

States of Solutions Solutions need not have liquid/liquid has solute and solvent. Solutions can exist in all 3 states (g, l, s) SoluteSolventExample Gas Liquid SolidGas Solid Humidity Scuba diver tank Butane lighter Metal alloys (eg. Sterling silver – Ag/Cu)

Solubility The ability of a solute to dissolve in a solvent Formal definition: the quantity of a solute that will dissolve in a specific amount of solvent at a given temperature **It is NOT “the rate” that a solute will dissolve in a solvent (how fast).**

Saturated solution - a stable solution in which the maximum amount of solute has been dissolved. Unsaturated solution - a solution that contains less solute than a saturated solution under existing condition Supersaturated solution - a solution that temporarily contains more than the saturation amount of solute than the solvent can hold (unstable)

Factors that affect Solubility 1.Nature of solute and solvent “Like dissolves like” = rule of thumb to predict solubility Polar/polar & non-polar/non-polar = SOLUTION Depends on: Intermolecular forces (forces within molecules) Type of bonding (I, C, PC) Polarity

Solubility of Polar Covalent Substances in Water The partial charges in a polar substance are attracted to the opposite partial charge in water’s polar molecules. As a result water molecules surround the polar molecules, causing them to dissolve. Methanol in water  intermolecular forces are hydrogen bonds

Solubility of Ionic Compounds in Water When a soluble ionic compound dissolves in water, the attraction between the two ions is broken and the two ions are surrounded by molecules of water. Solubility therefore is the result of ion-dipole attraction. Some ionic compounds are not soluble in water. This is the result of the very strong attractions of their ions (eg. AgCl)

2.Pressure Little effect on solubility of liquids/solids in liquid solvents However, for a gas in a liquid solvent, the solubility ↑ when pressure ↑. Pressure acts to keep gases from escaping the solution

3.Temperature The solubility of a gas in a liquid solvent ↑ when temperature ↓. –What foams more? –Warm pop or cold pop? The solubility of a solid in a liquid solvent usually ↑ with an ↑ in temperature. –What dissolves more? –Tea in cold water or hot water? –Note - If you have a saturated solution, solubility can be ↑ if you raise the temperature.

Explain the solubility of a cold can of pop that is opened and left at room temperature  pressure and temperature! Pop has gaseous CO 2 in liquid solvent  the solubility ↑ when pressure ↑. So pressure is higher in the sealed can. When opened, the pressure ↓, and solubility ↓ too. So CO 2 escapes over time due to this  less fizz (and less taste). The solubility of a gas in a liquid solvent ↑ when temperature ↓. When kept cold, CO 2 tends to stay in solution = more fizz (and more taste). As pop warms up, solubility ↓, so CO 2 escapes over time  less fizz (and less taste).

Solubility and Formation of a Precipitate In single and double displacement reactions, you can predict the formation of a precipitate (a solid formed in a liquid) by finding out the solubility of the products. Low solubility = generally insoluble (precipitate forms) High solubility = generally soluble (no precipitate) ***Note: If BOTH products are soluble in water, then there is NO reaction!***

Example 1 Determine the products (if any) when a solution of sodium sulfate is mixed with a solution of lead(II) nitrate. If a reaction occurs, summarize the reaction as a balanced chemical equation. Na 2 SO4 (aq) + Pb(NO 3 ) 2 (aq)  NaNO 3 + PbSO 4 2 (aq) (s)

Example 2 Determine the products (if any) when a solution of sodium acetate is mixed with a solution of potassium chloride. If a reaction occurs, summarize the reaction as a balanced chemical equation. NaCH 3 OO (aq) + KCl (aq)  NaCl + KCH 3 COO (aq) NO REACTION!

Total Ionic and Net Ionic Equations When you have a chemical reaction, you know the reactants and products. But the chemical equation does not tell you WHICH ions are actually involved in the reaction.

Total Ionic and Net Ionic Equations Total Ionic Eq’n – shows all the soluble compounds written in their dissolved form and all the insoluble compounds written normally as compounds. Net Ionic Eq’n – shows ONLY the ions that are actually involved in the reaction. The ions that are NOT involved are known as spectator ions.

Example 1 Write the total ionic equation and net ionic equation between sodium chloride and lead(II) nitrate. NaCl (aq) + Pb(NO 3 ) 2 (aq)  NaNO 3 + PbCl 2 22 (aq)(s) TOTAL IONIC EQ’N: 2Na + (aq)+ 2Cl - (aq)+ Pb 2+ (aq)+ 2NO 3 - (aq)  2Na + (aq) + 2NO 3 - (aq)+ PbCl 2 (s) NET IONIC EQ’N: Pb 2+ (aq) + 2Cl - (aq)  PbCl 2 (s)

Example 2 Write the total ionic equation and net ionic equation between barium chloride and silver nitrate. BaCl 2 (aq) + AgNO 3 (aq)  Ba(NO 3 ) 2 + AgCl22 (aq)(s) TOTAL IONIC EQ’N: Ba 2+ (aq)+ 2Cl - (aq)+ 2Ag + (aq)+ 2NO 3 - (aq)  Ba 2+ (aq)+ 2NO 3 - (aq)+ 2AgCl (s) NET IONIC EQ’N: 2Ag + (aq) + 2Cl - (aq)  2AgCl (s)