1. Metal + Acid Displacement

2. Activity Series of Metals

3. metals higher in series react with compounds of those below metals become less reactive to water top to bottom metals become less able to displace H 2 from acids top to bottom

4. Potassium + Water

5. Activity Series of Metals Zn (s) + CuSO 4(aq)  ZnSO 4(aq) + Cu (s) Cu (s) + 2AgNO 3(aq)  Cu(NO 3 ) 2(aq) + 2Ag (s) Fe (s) + 2HCl (aq)  FeCl 2(aq) + H 2(g) Zn (s) + 2HBr (aq)  ZnBr 2(aq) + H 2(g)

6. Metal + Metal Salt Displacement

7. Which of the following reactions does NOT happen? 10 0 0 130 1.Cu (s) +H 2 SO 4(aq)  CuSO 4(aq) +H 2(g) 2.2HNO 3(aq) +2K (s)  2KNO 3(aq) +H 2(g) 3.FeCl 2(aq) +Zn (s)  ZnCl 2(s) +Fe (s) 4.Ca (s) +2H 2 O (l)  Ca(OH) 2(aq) +H 2(g) 5.Cu (s) +2AgNO 3(aq)  2Ag (s) +Cu(NO 3 ) 2(aq)

8. Solution A homogeneous mixture of two or more substances comprising the solvent which is the majority of the mixture and one or more solutes which are the smaller fraction. Concentration – how much solute is present in a given amount of solution

9. Cola Drinks Solvent water Solutes carbon dioxide (gas) sweetener (solid) phosphoric acid (liquid) caramel color (solid)

10. Molarity – a measure of concentration The number of moles of solute per liter of solution. molarity  M moles of solute M = liters of solution units  molar = mol/L = M

11. Preparation of 1.00 L of 0.0100 M KMnO 4 solution from solid

12. Which value do you NOT need to determine the molarity of a solution? 10 0 0 130 1.Mass of solute 2.Molar mass of solute 3.Volume of solvent added 4.Total volume of solution

13. Solution Preparation by Dilution

14. Dilution Molarity (mol/L) × Volume (L) = Moles If we take a sample (say 25.0 mL) from a solution (say 0.372 M) and add extra water (say to a total volume of 500. mL) the moles of solute are unchanged Thus M 1 V 1 = moles = M 2 V 2 0.372 M × 25.0 mL = M 2 × 500. mL M 2 = 0.0186 M

15. How many L of conc HNO 3 (16.0 M) are needed to prepare 0.500 L of 0.250 M nitric acid? 10 0 0 130 1.32.0 L 2.16.0 L 3.0.500 L 4.0.250 L 5.0.00781 L

16. Stoichiometric Relationships

17. Titrations at equivalence mol H + = mol OH -

18. EXAMPLE: A sample of lye, sodium hydroxide, is neutralized by sulfuric acid. How many milliliters of 0.200 M H 2 SO 4 are needed to react completely with 25.0 mL of 0.400 M NaOH? 2 NaOH (aq) + H 2 SO 4(aq)  Na 2 SO 4(aq) + 2 H 2 O (25.0 mL NaOH) #mL H 2 SO 4 = (0.400 mol NaOH) (1 L NaOH) (1 L) (1000 mL) (1 mol H 2 SO 4 ) (2 mol NaOH) (1 L H 2 SO 4 ) (0.200 mol H 2 SO 4 ) = 25.0 mL H 2 SO 4 (1000 mL) (1 L)

19. Ion Concentrations 0.100 M NaCl NaCl (aq)  Na + (aq) + Cl - (aq) Is 0.100 M in both Na + and Cl - 0.100 M Al 2 (SO 4 ) 3 Al 2 (SO 4 ) 3(aq)  2 Al 3+ (aq) + 3 SO 4 2- (aq) Is 0.200 M in Al 3+ and 0.300 M in SO 4 2-

20. Consider the NaOH + H 2 SO 4 reaction. What are the final concentrations of Na + and SO 4 2- ? Stoichiometric reaction, can use either reactant to determine moles of product 25.0 mL NaOH × 0.400 mol NaOH × 1 L 1 L 1000 mL 0.0100 mol NaOH × 1 mol Na 2 SO 4 = 0.0050 mol Na 2 SO 4 2 mol NaOH Final Ion Concentrations

21. Total volume = 25.0 mL + 25.0 mL = 50.0 mL = 0.0500 L [Na 2 SO 4 ] = 0.00500 mol = 0.100 M 0.0500 L [Na + ] = 2 × 0.100 M = 0.200 M [SO 4 2- ] = 0.100 M

22. Which solution has the highest concentration of SO 4 2- ? 10 0 0 130 1.0.20 M CuSO 4 2.0.15 M Na 2 SO 4 3.0.070 M Fe 2 (SO 4 ) 3 4.0.10 M Ce(SO 4 ) 2