INTRODUCTION TO SOLUBILITY There are three classes of compounds which can form ionic solutions: Acids: HCl, H 2 SO 4, HNO 3 Bases: NaOH, KOH, NH 3 Chemistry 12 UNIT 3A SOLUBILITY THEORY Salts: NaCl, KMnO 4, NaClO We will only deal with the solubility principals of salts. RULES FOR CLASSIFYING COMPOUNDS AS IONIC OR MOLECULAR Ionic: A compound made up of a metal and a nonmetal will be ionic in solution (form ions in solution). FeCl 3(s) Fe 3+ + 3 Cl A compound which contains polyatomic ions will be ionic in solution. (NH 4 ) 2 Cr 2 O 7(s) 2 NH + 4 + Cr 2 O 2 7 Al(NO 3 ) 3(s) Al 3+ + 3 NO 3 CaBr 2(s) Ca 2+ + 2 Br 1
Molecular: Compounds made up of a nonmetal and a nonmetal will be molecular in solution (do not form ions, stay as a molecule in solution). CH 3 OH (l) CH 3 OH Usually organic compounds. Beware of acetate (CH 3 COO ) and other organic ions. They will form ionic solutions. DISSOCIATION OF IONS Electrolyte: A substance which dissolves to give an electrically conducting solution containing ions. Produced from an ionic salt. (NH 4 ) 2 SO 4(s) 2 NH 4 + + SO 4 2 Dissociation equation: salt ions. SATURATED SOLUTIONS NonElectrolyte: A substance which dissolves to give an electrically nonconducting solution. Produced from molecules. C 12 H 22 O 11(s) C 12 H 22 O 11 No ions = NO CONDUCTIVITY!!! A solution in which a dissolved substance is in equilibrium with an undissolved substance. In order to have a saturated solution we must meet two conditions: 1. Some undissolved material must be present. 2. Equilibrium must exist between the dissolved and undissolved material. An unsaturated solution contains less than the maximum amount of a substance which can dissolve. There is no undissolved solid present therefore no equilibrium exists. 2
Solubility: The equilibrium concentration of a solute in a particular solvent at a particular temperature. Solubility is expressed as: amount volume Common units include: mol/l (M), g/100 ml of H 2 O, g/l IONIC EQUATIONS We show that a solution is saturated by writing an equilibrium equation in net ionic form: AgBrO 3(s) Ag + + BrO 3 THE EQUILIBRIUM EQUATION CAN BE INTERPRETED THE FOLLOWING WAYS: When solid AgBrO 3(s) dissolves and enters the solution we have: Dissolving Reaction AgBrO 3(s) Ag + + BrO 3 When Ag + and BrO 3 ions come together and form AgBrO 3 we have: Crystalization Reaction Ag + + BrO 3 AgBrO 3(s) IONIC SALT IONS IONS IONIC SALT 3
HOMEWORK: The rate of the dissolving reaction will equal the rate of the crystallization reaction when we have equilibrium: Read: Page 77 Do: #1 7 AgBrO 3(s) Ag + + BrO 3 CALCULATING SOLUBILITY It is experimentally found that 1L of saturated AgBrO 3 contains 1.96g of AgBrO 3. What is the solubility of AgBrO 3 in moles per litre? The solubility of Pbl 2 is 1.37 x 10 3 M. Express this value in grams per litre. It is found that 250mL of saturated CaCl 2 contains 18.6g of CaCl 2 at 20 o C. What is the solubility of CaCl 2 in moles per litre? CONCENTRATION OF INDIVIDUAL IONS When a substance such as Na 2 SO 4(s) dissolves in water, each formula unit dissociates into two Na + ions and one SO 4 2 ion. Na 2 SO 4(s) 2 Na + + SO 4 2 Why is the concentration of Na + twice that of SO 4 2? 4
FOR EXAMPLE: What is the concentration of all the ions present in a saturated solution of Ag 2 CO 3 having a concentration of 1.2 x 10 4 M? Ag 2 CO 3(s) 2 Ag + + CO 3 2 REMEMBER THE DILUTION EFFECT? If we mix two solutions together, they will mix and dilute each other. The use of a DILLUTION FACTOR will allow you to calculate the new concentrations. [...] new =[...] old V initial V final FOR EXAMPLE: Calculate the concentration of all the ions present when 23.0 ml of 0.135 M MgCl 2 is mixed with 15.0 ml of 0.250 M of LiCl. HOMEWORK: Read: Page 81 83 Do: #8 20 Study for quiz!!! 5
SOLUBILITY AND PRECIPITATION Soluble: More than 0.1 mol of a substance dissolves per litre of solvent. Low Solubility: If a saturated solution of a substance is less than 0.1 M. The table Solubility of Common Compounds in Water (page 4 of your Data Booklet) helps us to determine whether a compound will be soluble or insoluble in water. LETS TRY AN EXAMPLE A COUPLE MORE: Determine whether CuCl (s) is soluble or has low solubility. Find the compounds negative ion in the NEGATIVE ION column of the table. Cl in this case. Find the compounds positive ion in the POSITIVE ION column of the table. Cu + in this case. If the positive ion is not specifically listed, it must be included on the ALL OTHERS category. Finally, look in the SOLUBILITY OF COMPOUNDS column to determine the compounds solubility. In this case, CuCl (s) has low solubility. 6
HOMEWORK: Read: Page 84 87 Do: #21 24 EXPRESSING CHEMICAL EQUATIONS REMEMBER We must be able to represent a precipitation reaction in three different ways: Formula Equation (just like Chemistry 11) Total or Complete Ionic Equation (shows all ions) Net Ionic Equation (only shows ions involved in the the reaction) Include all phases, charges, and coefficients!!! 7
FOR EXAMPLE: Write a formula equation, a total ionic equation, and a net ionic equation for the reaction which occurs when 0.2 M solutions of Al(NO 3 ) 3 and MgS are mixed. FORMULA EQUATION A formula equation is just a balanced chemical equation. Make sure to include all the phases and coefficients. for soluble species, (s) for precipitates. We get: 2Al(NO 3 ) 3 + 3MgS Al 2 S 3(s) + 3Mg(NO 3 ) 2 TOTAL OR COMPLETE IONIC EQUATION Shows all the soluble ionic species broken up into their respective ions. We get: 2Al 3+ +6NO 3 +3Mg 2+ +3S 2 Al 2 S 3(s) +3Mg 2+ +6NO 3 This method is not frequently used because it is cumbersome to write out. It is used to emphasize the situation which exists before and after the reaction. NET IONIC EQUATION Shows only the reacting species in the equation. The net ionic equation is formed by omitting the spectator ions from the total ionic equation. Ions that do not react. They maintain the same identity on each side of the chemical equation. They get cancelled out. 8
HOMEWORK: Cancel out the spectators: 2Al 3+ +6NO 3 +3Mg 2+ +3S 2 Al 2 S 3(s) +3Mg 2+ +6NO 3 We get: 2Al 3+ + 3S 2 Al 2 S 3(s) Read: Page 88 89 Do: #25 SEPARATING MIXTURES OF IONS BY PRECIPITATION METHODS Qualitative analysis involves the use of experimental procedures to determine what elements or ions are present in a substance. For many ionic substances the decision can be made using the solubility table. To solve these problems, you need to develop a quantitative analysis scheme. FOR EXAMPLE: A solution contains both SO 2 4 and OH. Outline an experimental procedure to remove each ion individually from the solution and identify the reagents (compounds compounds) used in this procedure. Include netionic equations for any precipitates formed. 9
HOMEWORK: Do: #26 39 Assignment due next class Study for quiz!!! 10