1. CHM 108 SUROVIEC SPRING 2014 Chapter 4 Part 2

2. I. Solution Stoichiometry According to the following reaction, how many moles of Fe(OH)2 can form from 175.0 mL of 0.227 M LiOH solution? Assume that there is excess FeCl2. FeCl2(aq) + 2 LiOH(aq) → Fe(OH)2(s) + 2 LiCl(aq) Determine the number of grams H2 formed when 250.0 mL of 0.743 M HCl solution reacts with 3.41 x 1023 atoms of Fe according to the following reaction. 2 HCl(aq) + Fe(s) → H2(g) + FeCl2(aq)

3. II. Aqueous Solution and Solubility Consider salt dissolving in water and sugar dissolving in water.

4. A. Electrolyte The way ionic compounds vs. molecular compounds dissolve in water shows the difference between types of solution.

5. A. Electrolyte Electrolytes – ions that act at charge carriers Solutes that completely dissociate into ions are called strong electrolytes

6. B. Solubility of Ionic Compounds Most ionic compounds when dissolved in water the solute breaks into ions. Not true for all ionic compounds

8. Determine the insoluble compounds AgCl NaNO 3 PbCl 2 Ba(OH) 2

9. III. Precipitation Reactions Precipitate: insoluble solid that separates from solution where no solid existed before reaction Hard water contains Ca 2+ and Mg 2+ Laundry detergent contains Na 2 CO 3

10. Examples 1. silver nitrate and potassium chloride 2. lead (II) nitrate and potassium chromate 3. potassium chromate and silver nitrate 4. sodium carbonate and copper (II) chloride 5. nickel (II) chloride and potassium hydroxide

11. IV. Molecular and Ionic Equations A. Molecular Equations Consider the following equation: CaCl 2 (aq) + Na 2 SO 4 (aq)  CaSO 4 (s) + 2NaCl (aq)

12. B. Ionic Equations In these equations, see that some of the ions are present on both sides of the arrow

13. Example Given: 2AgNO 3 (aq) + MgCl 2 (aq)  2AgCl (s) + Mg(NO 3 ) 2 (aq) What is the ionic equation? Net Ionic?

14. Write NET ionic equations 1. AlCl 3 (aq) + Na 3 PO 4 (aq) 2. lead (II) nitrate and potassium chloride

15. IV. Acid and Base Reactions Bronstead definition of acid: proton donor Bronstead definition of base: proton acceptor A. Acids HCl (aq) + H 2 O (l) → H 3 O + (aq) + Cl - (aq) H 2 SO 4 (aq) H + (aq) + HSO 4 - (aq) HSO 4 - (aq) H + (aq) + SO 4 2- (aq)

16. B. Bases Proton acceptors Strong bases ionize completely to OH -  NaOH (s) →  Ca(OH) 2 → Weak bases ionize only partially  NH 3 (aq) + H 2 O ⇌

18. C. Reactions of Acids and Bases 1. Neutralization acid + base salt + water HCl (aq) + NaOH (aq)

19. C. Reactions of Acids and Bases 2. Weak Acid/Base reactions CH 3 COOH + NaOH

20. V. pH Concentration scale for acids and bases Vinegar: [H + ] = 1.6  10 -3 M Pure Water:[H + ] = 1.0  10 -7 M Ammonia: [H + ] = 1.0  10 -11 M pH = -log[H + ] Determine the pH of the above. What is the trend of acids and bases?

21. VI. Acid-Base Titrations Commonly used to determine the concentration of a dissolved species or its percentage in a mixture Titration  Measuring the volume of a standard solution (known concentration) needed to react with a measured quantity of a sample Titrant (in the buret) Analyte (in the Erlenmeyer flask)

22. VI. Acid-Base Titrations  Equivalence point is where the number of moles of acid equals the number of moles of base  The endpoint is indicated by a color change in the acid-base indicator

23. Example 1. What volume (in mL) of a 1.420 M NaOH solution is required to titrate 25.00 mL of a 4.50 M H 2 SO 4 solution? 2. What volume (in mL) of 0.955 M HCl is required to titrate 2.152g of Na 2 CO 3 to the equivalence point?

24. VI. Redox Reactions A. Oxidation Numbers Needed when we are looking at reactions between 2 nonmetals. The oxidation number of an atom in a compound is the “charge” that it would have it all shared electrons were assigned to the atom with higher electronegativity.

25. III. Oxidation-Reduction Reactions Short name: Redox reactions Electron exchange  Oxidation is a loss of electrons  Reduction is a gain of electrons

26. III. Redox Reactions Fe (s)  Fe ? (aq) + 2e - 2H ? (aq)+ 2e -  H 2 (g)

27. Examples A. Fe 3+ (aq) + H 2 (g) Fe 2+ (aq) + H + (aq) B. Au (s) + F 2 (aq) F - (aq) + Au 3+ (aq) 1. Break into 1/2 reactions 2. Mass balance 1/2 reactions 3. Combine and check for neutrality and check again for mass balance