Solution Chemistry solution homogeneous mix of two or more substances solvent = substance present in major amount solute = substance present in minor amount gas air solvent = N2 solute = O2, Ar, CO2, etc. solid steel solvent = Fe solute = C liquid solvent = H2O solute = salts, covalent compounds

H2O O and H share electrons but not equally separation of charge + H2O - O and H share electrons but not equally separation of charge dipole dipole moment polar solvent hydrogen bonds H-bond need donor H-O acceptor O + - + H-N N - + H-F F -

solvent Aqueous solutions solute H2O NaCl H-bond Ion-dipole Ion-ion H+ O- Cl- H+ Na+ Cl- Na+ O- solvation NaCl (s) + H2O (l)  Na+ (aq) + Cl- (aq)

Non-ionic solutions solvent solute H2O glucose C6H12O6 H-bond H-bond H+ O- H+ O- C6H12O6 (s) + H2O (l)  C6H12O6 (aq) “Likes dissolve likes”

Non-ionic solutions solvent solute H2O octane C8H18 non-polar H-bond H+ O- C8H18 (l) + H2O (l)  no reaction

Properties of aqueous solutions ionic covalent conduct electricity do not conduct electricity NaCl C6H12O6 electrolytes produce ions non-electrolytes mobile, charged salts produce other anions and cations

What is the molarity of a solution prepared Solution Composition molarity concentration = amount of solute = mol = M [ ] volume of solution L What is the molarity of a solution prepared by dissolving 23.4 g sodium sulfate in enough water to give 125 mL of solution? 23.4 g Na SO4 1 mol Na2SO4 = 0.165 mol Na2SO4 2 142.0 g Na2SO4 125 mL 1 L = .125 L M = 0.165 mol Na2SO4 = 1.32 M 1000 mL 0.125 L [Na2SO4] = 1.32 M

How many grams of Na2SO4 are required Solution Composition concentration = amount of solute = mol = M [ ] volume of solution L How many grams of Na2SO4 are required to make 350 mL of 0.500 M Na2SO4? 0.500 mol Na2SO4 0.350 L 142.0 g = 24.9 g Na2SO4 1 mol Na2SO4 L

Solution Composition concentration = amount of solute = mol = M [ ] [ ] volume of solution L stock solution HCl = 12.0 M moles solute before dilution = moles solute after dilution How would you prepare 1.5 L of a 0.10 M HCl solution? 0.10 mol HCl 1.5 L = 0.15 mol HCl moles after dilution L 0.15 mol HCl = 12.0 mol HCl (x) L moles before dilution L (x) = 0.0125 L 12.5 mL of 12.0 M HCl + 1.4875 L H2O = 1.50 L 0.10 M HCl

of a 0.10 M HCl solution, using a 12.0 M stock solution? How would you prepare 1.5 L of a 0.10 M HCl solution, using a 12.0 M stock solution? moles of solute before dilution = moles of solute after dilution Mi x Vi = Mf x Vf (mol/L) (L) 12.0 M HCl x Vi = 0.10 M HCl x 1.5 L Vi = 0.0125 L then add H2O to get to Vf =1.37 L H2O

displacement reaction yellow solid + solution solubility of salts Aqueous Reactions Pb(NO3)2 (s) + H2O (l)  Pb2+ (aq) + NO3- (aq) 2 lead (II) nitrate KI (s) + H2O (l)  K+ (aq) + I- (aq) potassium iodide Pb2+ (aq) + 2 NO3- (aq) + K+ (aq) + I- (aq)  displacement reaction yellow solid + solution solubility of salts PbI2 KNO3

Solubility rules 1. Most NO3- salts are soluble 2. Most salts of alkali metals (IA) and NH4+ are soluble 3. Most Cl-, Br- and I- salts are soluble Exceptions: Ag+, Pb2+ and Hg22+ insoluble 4. Most SO42- are soluble Exceptions: Ca2+, Ba2+, Pb2+, Hg22+ insoluble 5. Most OH- are insoluble Exceptions: IA, Ca2+, Ba2+, Sr2+ soluble 6. Most S2-, CO32-, CrO42-, PO43- insoluble Exceptions: IA, NH4+ soluble

Pb2+ (aq) + 2 NO3- (aq) + K+ (aq) + I- (aq) PbI2 (s) + K+ (aq) + NO3- (aq) yellow solid + solution PbI2 KNO3 1. Most NO3- salts are soluble 2. Most IA salts are soluble 3. Most Cl-, Br- and I- salts are soluble Exceptions: Ag+, Pb2+ and Hg22+ insoluble balance the equation start with molecular equation

balance molecular equation Pb(NO3)2 (aq) + KI (aq) 2  PbI2 (s) + KNO3 (aq) 2 change to ionic equation 1. Most NO32- salts are soluble 2. Most Group I salts are soluble 3. Most Cl-, Br- and I- salts are soluble except Ag+, Pb2+ and Hg22+ Pb2+(aq)+2NO3-(aq) +2K+(aq) +2I-(aq) PbI2 (s) +2K+(aq) +2NO3- (aq) spectator ions they can be cancelled out in net ionic equation Pb2+ (aq) + 2 I- (aq)  PbI2 (s)