1. Solution Chemistry solution of two or more substances homogeneous mix solvent = solute = substance present in major amount substance present in minor amount gas solute = O 2, Ar, CO 2, etc. airsolvent = N 2 solid solute = C steelsolvent = Fe liquid solvent = H 2 O solute = salts, covalent compounds

2. H2OH2O O and H share electrons separation of chargedipole dipole moment polar solvent hydrogen bondsH-bond need donorH-O H-N H-F acceptorO N F but not equally -- ++ ++ ++ ++ ++ -- -- --

3. Aqueous solutions NaClH2OH2O solventsolute H-bond O-O- H+H+ Ion-ion Na + Cl - Ion-dipole Cl - H+H+ Na + O-O- solvation NaCl (s) + H 2 O (l)  Na + (aq)+ Cl - (aq)

4. Non-ionic solutions C 6 H 12 O 6 glucose solventsolute H2OH2O H-bond O-O- H+H+ O-O- H+H+ C 6 H 12 O 6 (s) + H 2 O (l)  C 6 H 12 O 6 (aq) “Likes dissolve likes”

5. Non-ionic solutions C 8 H 18 octane solventsolute H2OH2O H-bond O-O- H+H+ non-polar C 8 H 18 (l) + H 2 O (l)  no reaction

6. Properties of aqueous solutions ioniccovalent conduct electricitydo not conduct electricity NaCl C 6 H 12 O 6 electrolytes non-electrolytes acids produce H + in aqueous solutions basesproduce OH - in aqueous solutions saltsproduce other anions and cations produce ions mobile, charged

7. Electrolytes RuleExceptions 1. Most acids are weak electrolytesHCl HBrHI HNO 3 H 2 SO 4 HClO 4 2. Most bases are weak electrolytes Ca(OH) 2 – Ba(OH) 2 LiOH – CsOH 3. Most salts are strong electrolytesHgCl 2 Hg(CN) 2

8. Strong Electrolytes dissociate completely form hydrated ions HCl (g) strong acids  strong bases NaOH (s) + H 2 O (l)  Na + (aq) + OH - (aq) salts MgSO 4 (s) + H 2 O (l)  Mg 2+ (aq) + SO 4 2- (aq) + H 2 O (l)  H + (aq) + Cl - (aq)

9. Weak Electrolytes do not dissociate completely HF (g) weak acids + H 2 O (l)  + F - (aq)H + (aq) equilibriumall species present  NH 3 (g) weak bases + H 2 O (l)  + OH - (aq)NH 4 + (aq) HgCl 2 (s) weak electrolytic salts + H 2 O (l)  + 2Cl - (aq)Hg 2+ (aq)

10. Non- Electrolytes do not dissociate to form ions CH 3 CH 2 OH (l)  + H 2 O  CH 3 CH 2 OH (aq)

11. Solution Composition concentration= amount of solute volume of solution = mol L = M molarity 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 SO 4 2 1 mol Na 2 SO 4 142.0 g Na 2 SO 4 = 0.165 mol Na 2 SO 4 125 mL 1 L 1000 mL =.125 LM =0.165 mol Na 2 SO 4 0.125 L = 1.32 M [ ] [Na 2 SO 4 ]= 1.32 M

12. How many moles of HNO 3 are present in 2.0 L of 0.200 M HNO 3 solution? 0.200 mol HNO 3 L 2.0 L= 0.40 mol HNO 3 Solution Composition concentration= amount of solute volume of solution = mol L = M[ ]

13. How many grams of Na 2 SO 4 are required to make 350 mL of 0.500 M Na 2 SO 4 ? 0.500 mol Na 2 SO 4 L 0.350 L 1 mol Na 2 SO 4 142.0 g = 24.9 g Na 2 SO 4 Solution Composition concentration= amount of solute volume of solution = mol L = M[ ]

14. stock solutionHCl = 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 L 1.5 L = 0.15 mol HCl 0.15 mol HCl = moles after dilution moles before dilution12.0 mol HCl L = 0.0125 L 12.5 mL of 12.0 M HCl + 1.4875 L H 2 O = 1.50 L 0.10 M HCl L (x)(x) Solution Composition concentration= amount of solute volume of solution = mol L = M[ ] (x)(x)

15. How would you prepare 1.5 L of a 0.10 M HCl solution, using a 12.0 M stock solution? MiMi x V i = M f x V f 12.0 M HCl x V i 0.10 M HCl x 1.5 L V i = 0.0125 L then add H 2 O to get to V f = moles of solute after dilutionmoles of solute before dilution = (mol/L)(L) =1.37 L H 2 O