CHAPTER 15 – SOLUTIONS SOLUTION – A homogeneous mixture SOLVENT – The major component SOLUTE – The minor component Why will 2 substances form a solution? 7B-1 (of 26)

LIKES DISSOLVE LIKES Polar substances dissolve other polar substances Nonpolar substances dissolve other nonpolar substances Polar and nonpolar substances do not mix Lava Lamps 7B-2

Dissolves polar substances that can δ- δ- δ+ δ- δ- δ- δ+ δ+ δ- Water Carbon tetrachloride Polar Nonpolar Dissolves polar substances that can H-Bond to water, and ionic substances Dissolves nonpolar substances NH3, C6H12O6, C2H5OH, HF, NaCl Grease, Oil, CxHy 7B-3

Soluble polar compounds (such as alcohols or sugars) are NONELECTROLYTES because they do not produce ions in solution Exception: Acids (which are polar compounds that do produce ions in solution) can be STRONG or WEAK ELECTROLYTES Soluble ionic compounds are STRONG ELECTROLYTES because they produce many ions in solution 7B-4

Water dissolves ionic substances because negative ends of H2O’s attract cations, and positive ends of H2O’s attract anions HYDRATION – Attachment of H2O molecules to dissolved ions 7B-5

SOLUBILITY SOLUBILITY – The maximum mass of a solute that dissolves in 100 g of solvent at a particular temperature SOLUBLE – A substance in which at least 1 g dissolves per 100 g of solvent INSOLUBLE – A substance in which less than 0.1 g dissolves per 100 g of solvent 7B-6

SATURATED – A solution that contains as much dissolved solute as possible Dissolving Rate Crystallizing Rate SOLUTION EQUILIBRIUM – When the rates of dissolving and crystallizing are equal A saturated solution is in solution equilibrium 7B-7

PROPERTIES OF SOLUTIONS (1) Equilibrium Vapor Pressure A liquid’s equilibrium vapor pressure is lowered by having a solute dissolved in it Temp (ºC) EVP of pure water (torr) EVP of a dilute salt water solution (torr) 10 20 30 9.2 17.5 31.8 9.1 17.4 31.6 100 760.0 757.2 7B-8

(2) Boiling Points A liquid’s boiling point is raised by having a solute dissolved in it Solution Boiling Point (ºC) Pure H2O Dilute Salt Water Conc. Salt Water 100.0 100.1 101.0 7B-9

(3) Freezing Points A liquid’s freezing point is lowered by having a solute dissolved in it Solution Freezing Point (ºC) Pure H2O Dilute Salt Water Conc. Salt Water 0.0 - 0.4 - 3.7 Arctic Ocean Freezing Point: 28ºF 7B-10

CONCENTATION UNITS MASS PERCENT – The mass of solute per mass of solution, times 100 Mass Percent = Mass Solute x 100 __________________ Mass Solution 5.00 g of salt is dissolved in 100.0 g of water. Calculate the mass percent. 5.00 g NaCl x 100 _____________________ 105.0 g Solution = 4.76 % 7B-11

Calculate the mass of sugar in 50 Calculate the mass of sugar in 50.0 g of a sugar-water solution that is 15% sugar by mass. x g sugar x 100 ___________________ 50.0 g solution = 15 % x g sugar = 15 x 50.0 g solution __________________________ 100 x g sugar = 7.5 g 7B-12

MOLARITY (M) – The MOLES of solute per liter of solution Moles Solute __________________ Liter Solution M = n ___ V Calculate the molarity of a solution that has 4.50 g of potassium nitrate dissolved in enough water to make 100. mL of solution. 4.50 g KNO3 x mol KNO3 __________________ 101.11 g KNO3 = 0.04451 mol KNO3 0.04451 mol KNO3 _______________________ 0.100 L solution = 0.445 M KNO3 7B-13

STANDARD SOLUTION – A solution of known concentration Calculate the molarity of a solution that has 10.0 g of lithium sulfate dissolved in enough water to make 250. mL of solution. 10.0 g Li2SO4 x mol Li2SO4 ____________________ 109.95 g Li2SO4 = 0.09095 mol Li2SO4 0.09095 mol Li2SO4 ________________________ 0.250 L solution = 0.364 M Li2SO4 7B-14

Calculate the mass of magnesium fluoride needed to produce 500 Calculate the mass of magnesium fluoride needed to produce 500. mL of a 3.00 M solution. M = n ___ V MV = n 3.00 mol MgF2 ___________________ L solution x 0.500 L solution = 1.500 mol MgF2 1.500 mol MgF2 x 62.31 g MgF2 _________________ mol MgF2 = 93.5 g MgF2 7B-15

Calculate the mass of sodium bromide dihydrate needed to produce 750 Calculate the mass of sodium bromide dihydrate needed to produce 750. mL of a 0.200 M solution. M = n ___ V MV = n 0.200 mol NaBr 2H2O _____________________________ L solution x 0.750 L solution = 0.1500 mol NaBr  2H2O 0.1500 mol NaBr  2H2O x 138.93 g NaBr  2H2O ___________________________ mol NaBr  2H2O = 20.8 g NaBr  2H2O 7B-16

Calculate the concentrations of each ion in a 0 Calculate the concentrations of each ion in a 0.250 M aluminum fluoride solution. AlF3 → (aq) Al3+ + 3F- (aq) (aq) AlF3 completely dissociates,  0.250 M AlF3 x 1 = 0.250 M Al3+ 0.250 M AlF3 x 3 = 0.750 M F- 7B-17

Calculate the concentrations of each ion in a 0 Calculate the concentrations of each ion in a 0.100 M iron (III) sulfate solution. Fe2(SO4)3 → (aq) 2Fe3+ + 3SO42- (aq) (aq) Fe2(SO4)3 completely dissociates,  0.100 M Fe2(SO4)3 x 2 = 0.200 M Fe3+ 0.250 M Fe2(SO4)3 x 3 = 0.300 M SO42- 7B-18

REACTIONS IN SOLUTION Calculate the mass of silver that can be produced when magnesium metal reacts with 50.0 mL of a 0.400 M silver nitrate solution. Mg + AgNO3 → 2 Mg(NO3)2 + Ag 2 50.0 mL 0.400 M x g 2 mol 2 mol 0.400 mol AgNO3 x 0.0500 L sol’n _____________________ L sol’n x 2 mol Ag ________________ 2 mol AgNO3 x 107.9 g Ag _____________ mol Ag = 2.16 g Ag 7B-19

NEUTRALIZATION REACTION – A reaction between an acid and a base Calculate the mass of sodium hydroxide needed to neutralize 100. mL of 0.175 M nitric acid. NaOH + HNO3 → NaNO3 + H(OH) x g 100. mL 0.175 M 1 mol 1 mol 0.175 mol HNO3 x 0.100 L sol’n ___________________ L sol’n x 1 mol NaOH _______________ 1 mol HNO3 x 40.00 g NaOH _________________ mol NaOH = 0.700 g NaOH 7B-20

Calculate the molarity of copper (I) ions in a 75 Calculate the molarity of copper (I) ions in a 75.0 mL sample if the sample reacts with a solution of sulfide ions to produce 1.25 g of copper (I) sulfide. Cu+ + S2- → 2 Cu2S 75.0 mL x M 1.25 g 2 mol 2 mol 1.25 g Cu2S x 1 mol Cu2S _________________ 159.17 g Cu2S x 2 mol Cu+ ______________ 1 mol Cu2S x 1 _________________ 0.0750 L sol’n = 0.209 M Cu2+ M = n / V 7B-21

Calculate the molarity of lead (II) ions in a 25 Calculate the molarity of lead (II) ions in a 25.0 mL sample if it reacts completely with 20.0 mL of a 0.300 M iodide solution. Pb2+ + I- → 2 PbI2 25.0 mL x M 20.0 mL 0.300 M 1 mol 2 mol 0.300 mol I- x 0.0200 L sol’n _______________ L sol’n x 1 mol Pb2+ _____________ 2 mol I- x 1 ___________________ 0.0250 L sol’n = 0.120 M Pb2+ M = n / V 7B-22

Calculate the volume of 0. 200 M sulfuric acid needed to neutralize 25 Calculate the volume of 0.200 M sulfuric acid needed to neutralize 25.0 mL of a 0.150 M potassium hydroxide solution. H2SO4 + KOH → 2 2 H(OH) + K2SO4 x mL 0.200 M 25.0 mL 0.150 M 1 mol 2 mol 0.150 mol KOH x 0.0250 L sol’n ___________________ L sol’n x 1 mol H2SO4 ________________ 2 mol KOH x L sol’n _____________________ 0.200 mol H2SO4 V = n / M = 0.00938 L H2SO4 solution = 9.38 mL H2SO4 solution 7B-23

THE REVIEW Solubility Rules and Strong Acids Dissociation and Ionization Electrolytes: Strong and Weak Precipitate Molecular, Ionic, and Net ionic Equations (with phases) for Precipitation Reactions Acid-Base Reactions 7B-24

THE REVIEW Solution, Solvent, Solute Likes Dissolve Likes Hydration Solubility Solution Equilibrium, Saturated Solution Soluble, Insoluble, Miscible, Immiscible Mass Percent Molarity Molarity of Ions Calculations Involving reactions in Solution 7B-25