1. AP Chapter 11 Notes

2. Reactions and Calculations with Acids and Bases Neutralization Reactions - when stoichiometrically equivalent amounts of acid and base react to form salt and water. (The net ionic equation of a neutralization is H + + OH - → H 2 O) Molarity = mol or mmols or mols LmL1000 mL

3. Examples of Neutralization Reactions: If 100. mL of 0.100 M HCl and 100. mL of 0.100 M NaOH react, then what is the molarity of the salt produced? Rice Table (use mols!)

4. The salt is NaCl, the moles of salt formed is 0.0100 mols, and the volume of the solution is 200. mL (be sure to add your volumes for the results of a reaction). = Molarity = mol = 0.0100 mol = 0.0500 M NaCl L0.200 L Molarity = mmol = 10.0 mmol = 0.0500 M NaCl mL200. mL

5. If 100. mL of 1.00 M HCl and 100. mL of 0.800 M NaOH react, then what is the molarity of the solutes? (remember water is a solvent) Rice Table, then Molarity

6. If 100. mL of 1.00 M HCl and 100. mL of 1.00 M Ca(OH) 2 react, then what is the molarity of the solutes?

7. Your turn: If 100. mL of 1.00 M H 2 SO 4 and 200. mL of 1.00 M KOH react, then what is the salt and what is the concentration of the salt?

8. What Volume of 0.00300 M HCl would neutralize 30.0 mL of 0.00100 M Ca(OH) 2 ? 2HCl + Ca(OH) 2  CaCl 2 + 2HOH 0.00100 mol Ca(OH) 2 30.0 mL2 mol HCl1000 mL = 20.0 mL 1000 mL 1 mol Ca(OH) 2 0.00300 mol HCl

9. Your turn: What volume of 0.0150 M acetic acid would neutralize 18.7 mL of 0.0105 M Ba(OH) 2 ?

10. Titration – adding a known solution (titrant) to an unknown solution, to determine the unknown’s concentration.

11. Standard Solution – titrating one solution with another to get the exact concentration of a titrant (that will later be used for titrating).

12. Often our common acids and bases can decompose or become contaminated while they sit in the storeroom, so even though you might have measured out enough solute to make a particular concentration of solution, it may not end up at the molarity you expected. Once standardized, the exact concentration is known and that is used to calculate your titration results, not the amount you were originally expecting to use. Typical standizing solutions are sodium bicarbonate for acids and potassium hydrogen phthalate (KHP) for bases.

13. Typical standardizing solutions are sodium bicarbonate for acids and potassium hydrogen phthalate (KHP) for bases.

14. Equivalence point – When [H + ] = [OH - ] End Point – point in a titration where the indicate color changes infers you have reached the equivalence point. Pick indicators wisely!

15. Examples of Titration Reactions:

16. What is the molarity of HCl if 36.7 mL of HCl was titrated with 43.2 mL of 0.236 M NaOH? HCl + NaOH → NaCl + H 2 O

17. What is the molarity of HCl if 36.7 mL of HCl was titrated with 43.2 mL of 0.236 M NaOH? HCl + NaOH → NaCl + H 2 O 43.2 mL NaOH 0.236 mols NaOH 1 mol HCl = 0.0102 mols HCl 1000 mL 1 mol NaOH

18. What is the molarity of HCl if 36.7 mL of HCl was titrated with 43.2 mL of 0.236 M NaOH? HCl + NaOH → NaCl + H 2 O 43.2 mL NaOH 0.236 mols NaOH 1 mol HCl = 0.0102 mols HCl 1000 mL 1 mol NaOH For HCl, 0.0102 mol HCl = 0.278 M HCl 0.0367 L

19. Your turn: What is the concentration of H 2 SO 4 if 43.2 mL of 0.236 M NaOH neutralized 36.7 mL of H 2 SO 4 ?

20. Normality – the number of equivalents per liter Good for: qualitative work – like determining the freshness of milk Not good for: quantitative work – like the titrations of complete unknowns Examples: HCl = 1 eq. H 2 SO 4 = 2 eq. NaOH = 1 eq. Ca(OH) 2 = 2 eq.

21. Redox reactions – not only does the mass balance (like before) but the charges (oxidation numbers) must also. To balance redox reactions, the half- reaction method is commonly used.

22. To Balance by Half Reactions: 1.Write unbalanced net ionic equations, as completely as possible 2.Split into two reactions, the oxidation and reduction reactions 3.Balance everything except for O and H 4.Appropriately add O then H, if needed (see list below) 5.Add electrons as needed 6.Balance the number of electrons gained and lost between the two reactions, multiplying as needed 7.Add the two equations together, simplifying as much as possible

23. In acid solution: For O, add H 2 O where O is needed then for H, add H + where needed In basic solution: For O, add 2xOH - and H 2 O on the other side then for H, add H 2 O where needed and OH - to the other side

24. Examples of Redox Reactions:

25. I 2 and S 2 O 3 2- react to form I - and S 4 O 6 2- I 2 + S 2 O 3 2- →I - + S 4 O 6 2-

26. In a basic solution, ClO - will oxidize CrO 2 - to CrO 4 2- and be reduced to Cl -. CrO 2 - + ClO - → CrO 4 2- + Cl -

27. Fe 2+ and MnO 4 - react to form Fe 3+ and Mn 2+ in a sulfuric acid solution Fe 2+ + MnO 4 - → Fe 3+ + Mn 2+

28. Your turn: Balance using HR: Al + NO 3 - + OH - + H 2 O → Al(OH) 4 - + NH 3

29. Examples of Redox Titration Reactions:

30. What volume of 0.0200 M KMnO 4 is required to oxidize 40.0 mL of 0.100 M FeSO 4 ? 5Fe 2+ + MnO 4 - + 8H + → 5Fe 3+ + Mn 2- + 4H 2 O

31. What volume of 0.0200 M KMnO 4 is required to oxidize 40.0 mL of 0.100 M FeSO 4 ? 5Fe 2+ + MnO 4 - + 8H + → 5Fe 3+ + Mn 2- + 4H 2 O 40.0 mL FeSO 4 0.100 mol FeSO 4 1 mol Fe 2+ 1 mol MnO 4 - 1 mol KMnO 4 1000 mL = 40.0 mL KMnO 4 1000 mL FeSO 4 1 mol FeSO 4 5 mols Fe 2+ 1 mol MnO 4 - 0.0200 mol KMnO 4

32. What is the molarity of Na 2 SO 3 if 20.00 mL Na 2 SO 3 are titrated with 36.30 mL of 0.05130 M K 2 Cr 2 O 7 in H 2 SO 4 ? 8H + + Cr 2 O 7 2- + 3SO 3 2- → 2Cr 3+ + 3SO 4 2- + 4H 2 O

33. What is the molarity of Na 2 SO 3 if 20.00 mL Na 2 SO 3 are titrated with 36.30 mL of 0.05130 M K 2 Cr 2 O 7 in H 2 SO 4 ? 8H + + Cr 2 O 7 2- + 3SO 3 2- → 2Cr 3+ + 3SO 4 2- + 4H 2 O 36.30 mL K 2 Cr 2 O 7 0.05130 mols K 2 Cr 2 O 7 1 mol Cr 2 O 7 2- 3 mol SO 3 2- 1 mol Na 2 SO 3 = 0.005586 mol Na 2 SO 3 1000 mL K 2 Cr 2 O 7 1 mol K 2 Cr 2 O 7 1 mol Cr 2 O 7 2- 1 mol SO 3 2-

34. What is the molarity of Na 2 SO 3 if 20.00 mL Na 2 SO 3 are titrated with 36.30 mL of 0.05130 M K 2 Cr 2 O 7 in H 2 SO 4 ? 8H + + Cr 2 O 7 2- + 3SO 3 2- → 2Cr 3+ + 3SO 4 2- + 4H 2 O 36.30 mL K 2 Cr 2 O 7 0.05130 mols K 2 Cr 2 O 7 1 mol Cr 2 O 7 2- 3 mol SO 3 2- 1 mol Na 2 SO 3 = 0.005586 mol Na 2 SO 3 1000 mL K 2 Cr 2 O 7 1 mol K 2 Cr 2 O 7 1 mol Cr 2 O 7 2- 1 mol SO 3 2- For Na 2 SO 3 0.005586 mol Na 2 SO 3 = 0.2793 M Na 2 SO 3 0.02000 L

35. Your turn: 40.00 mL of 0.1442 M Na 2 S 2 O 3 reacts with 26.36 mL of I 2. What is the molarity of the I 2 solution? 2 Na 2 S 2 O 3 + I 2 → Na 2 S 4 O 6 + 2NaI