1. Acids and Bases (Chapter 19): Properties of acids: Taste sour React with metals to form hydrogen gas React with carbonates to form CO 2 gas Form electrolyte solutions Properties of bases: Feel slippery Taste bitter Will react with some metals Form electrolyte solutions

2. pH scale 0714acidicbasic [H + ] > [OH  ][H + ] < [OH  ][H + ] = [OH  ] Calculating pH: Ex 1: What is the pH of a 0.15 M solution of HCl? pH =  log(0.15 M) = 0.82 pH = -log[H + ] pOH =  log[OH  ] pH + pOH = 14 Ex 2: What is the pH of a 0.2 M solution of NaOH? pOH =  log(0.2 M) = 0.70 pH = 14  0.70 = 13.30 NaOH  Na + + OH  0.2 M x x

3. Arrhenius model of acids and bases: An acid contains acidic protons that will ionize in water to form hydrogen ions. HCl (aq)  H + (aq) + Cl  (aq) A base contains hydroxide ions that will ionize in water and produce aqueous hydroxide ions. NaOH (s)  Na + (aq) + OH  (aq) H 2 SO 4 (aq)  2 H + (aq) + SO 4 2  (aq) pH of 0.1M soln? pH =  log(0.1 M) = 1 pH =  log(0.2 M) = 0.7 pH of 0.1M soln? pH = 14  pOH pH = 14  1 = 13 Mg(OH) 2 (s)  Mg 2+ (aq) + 2 OH  (aq) pH = 14  0.7 = 13.3 pOH =  log(0.1 M) = 1 pOH =  log(0.2 M) = 0.7 0.1 M x x x2x 0.2 M

4. Examples: 1.A solution has a pH of 2.5. What is its pOH? 2.A solution has a [H + ] = 1 x 10  3 M. What is its [OH  ]? 3.A solution has a [OH  ]= 1 x 10  2 M. What is its pH? pH + pOH = 14 pOH = 14  pH = 11.5 [OH-] = 1 x 10  11 M pOH = 14  3 = 11 pH = 14  2 = 12 pH =  log(1 x 10  3 M) = 3 [OH  ] = 10  pOH pOH =  log(1 x 10  2 M) = 2

5. Brønsted-Lowry model of acids and bases: Acids increase the hydrogen ion concentration in a solution by donating a hydrogen ion. Bases increase the hydroxide ion concentration in a solution by accepting a hydrogen-ion. HX (aq) + H 2 O (l) ⇋ H 3 O + (aq) + X - (aq) Acid H + donor Conjugate acid H + donor Base H + acceptor Conjugate base H + acceptor

6. Another example: NH 3 (aq) + H 2 O (l) ⇋ NH 4 + (aq) + OH - (aq) acidbaseConjugate acid Conjugate base Substances that can act as either an acid or a base (e.g. H 2 O) are called amphoteric.

7. Naming Acids Binary acids: contain hydrogen and one other element 1.First part of name is hydro- 2.Second part is the root of the second element with the suffix –ic Ex: HCl is hydrochloric acid Oxyacids: acid form of a polyatomic anion that contains oxygen Format: Root of anion + suffix acid Suffixes: -ate  -ic-ite  -ous HNO 3 Nitric acid HNO 2 Nitrous acid

8. HC 2 H 3 O 2 HNO 3 H 2 CO 3 HCl H 3 PO 4 H 2 SO 4 H 2 S The hydrogens that appear first in the formula are called ACIDIC PROTONS, or just PROTONS for short. Monoprotic acid:an acid with only one acidic proton Polyprotic acid: an acid with more than one acidic proton (diprotic = 2, triprotic = 3) Acetic acid (vinegar) Nitric acid (acid rain) Carbonic acid (in sodas) Hydrochloric acid (stomach acid) Phosphoric acid (in colas) Sulfuric acid (battery acid) Hydrosulfuric acid (rotten egg smelling toxin) Acid nameAcid examples:

9. Ionization of polyprotic acids proceeds stepwise: First ionization: H 3 PO 4 (aq) ⇋ H 2 PO 4 - (aq) + H + (aq) Second ionization: H 2 PO 4 - (aq) ⇋ HPO 4 -2 (aq) + H + (aq) Third ionization: HPO 4 -2 (aq) ⇋ PO 4 -3 (aq) + H + (aq)

10. Self-ionization of water: Water will self-ionize according to the following reaction: H 2 O (l) ⇋ H + (aq) + OH - (aq) The equilibrium constant for this reaction is given a special symbol, K w, which is: In neutral water [H + ] = [OH - ]= 1 x 10 -7 M The pH for any solution can be found by: pH = -log [H + ] The pH of a neutral solution is therefore = 7 2H 2 O (l) ⇋ H 3 O + (aq) + OH - (aq) Or…

11. Strong acids ionize completely in solution. There is no equilibrium between the reactants and the products. Examples: Hydrochloric acidHCl Hydrobromic acidHBr Hydroiodic acidHI Perchloric acidHClO 4 Nitric acidHNO 3 Sulfuric acidH 2 SO 4 HX  H + + X - The pH can be found directly from the concentration of the acid in the solution. Ex: What is the pH of a 1 x 10 -3 M HCl solution? [HCl] = [H+] = 1 x 10 -3 M pH = 3

12. Weak acids only partially ionize in solution. They establish an equilibrium between the acid and its ions. HA ⇋ H + + A - The acid dissociation equilibrium constant is given its own designation: K a K a = [H + ][A - ] [HA] Ex: What is the pH of a weak acid if [HA]= 1 x 10 -4 M at equilibrium and it has a K a = 1 x 10 -5 ? [H+] = 1 x 10 -3 M pH = 3 Note: [H + ] = [A - ]

13. Like strong acids, strong bases ionize completely in solution. NaOH  Na + + OH - Examples: Group 1 and Group 2 hydroxides Weak bases only partially ionize in solution, and will establish an equilibrium with its conjugate acid and the conjugate base of water. CH 3 NH 2 (aq) + H 2 O (l) ⇋ CH 3 NH 3 + (aq) + OH - (aq) K b = [CH 3 NH 3 + ][OH - ] [CH 3 NH 2 ] Base dissociation constant

14. Acid-Base reactions The H + from the acid reacts with the OH -1 from the base to form water. H + + OH - H2OH2O This is called areaction

15. Examples: What would the products be if HCl and NaOH reacted? HCl + NaOHH 2 O +NaCl If HCl and KOH reacted? HCl + KOHH 2 O +KCl If HCl and Mg(OH) 2 reacted? HCl + Mg(OH) 2 H 2 O +MgCl 2 2 2

16. In general... *A salt is any ionic compound that is NOT an acid or a base.

17. Titrations Titrations allow the concentration of an acid or base to be determined using an acid-base reaction and an indicator. 1. Measure out a volume of the acid or base that has the unknown concentration. 2. Add small volumes of the other reactant of a known concentration until the indicator changes color. 3. Use the ‘magic equation’ to calculate the unknown concentration.

18. Indicators: organic dyes whose color depends upon the pH of the solution.

19. M 1 V 1 = M 2 V 2 M 1 = unknown concentration V 1 = volume used for unknown M 2 = known concentration V 2 = total volume added of known concentration

20. Examples: 25 mL of HCl are titrated with 12.5 mL of 1.0 M NaOH. What is the concentration of the HCl? M 1 (25 mL) = (1.0 M)(12.5 mL) M 1 V 1 = M 2 V 2 M 1 = 0.5 M The concentration of the HCl is 0.5 M

21. 10 mL of NaOH are titrated with 10 mL of 1.0 M H 2 SO 4. What is the concentration of the NaOH? M 1 (10 mL) = 2(1.0 M)(10 mL) M 1 V 1 = M 2 V 2 M 1 = 2.0 M The concentration of the NaOH is 2.0 M There are TWO acidic protons in H 2 SO 4. The [H 3 O + ] will be two times more concentrated than the original acid.

22. Why does the ‘Magic Equation’ work? Recall that Molarity = moles/volume of soln. M A V A = M B V B The ‘Magic Equation’ gives the volume at which the number of moles of H + exactly equal the number of moles of OH -. This volume is called the EQUIVALENCE POINT for the neutralization reaction. Moles acid = Moles base

23. 1. What is the pH of 0.100 M HCl? 2. What is the pH of 0.100 M NaOH? pH = 1 pH = 13 Note: The ONLY time the pH = 7 at the equivalence point is during strong acid : strong base titrations. 3. What is V eq ? V eq = 50.0 mL

24. The pH at V eq for a weak acid is greater than 7. pKa = pH at ½ V eq pK a = 4.9 V eq K a = 1.3 x 10 -5

25. The pH at V eq for a weak base is less than 7. pK a = pH at ½ V eq pK a = 9.0 pK b = 5.0 K b = 1.0 x 10 -5

26. Buffer region pH = 4.74 Titration of Acetic Acid with NaOH Volume NaOH added (mL) pH 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 01020304050

27. Buffers: solutions that resist changes in pH Buffers are mixtures of either: A weak acid and the salt of its conjugate base OR A weak base and the salt of its conjugate acid Buffers will maintain a pH that is ± 1 pH unit of their pK a Example: Acetic acid has a K a = 1.8 x 10 -5 pK a = -log(K a ) = -log(1.8 x 10 -5 ) = 4.74 An acetic acid/sodium acetate mixture will buffer a solution at a pH of 4.74.

28. An important buffer example: The pH of blood must be maintained at 7.4 ± 0.2 or death may occur. There are two main buffering equilibria… H 2 CO 3 + H 2 O ⇋ HCO 3 -1 + H 3 O + H 2 PO 4 -1 + H 2 O ⇋ HPO 4 -2 + H 3 O + K a = 4.3 x 10 -7 pK a = 6.4 K a = 6.2 x 10 -7 pK a = 7.2 Example: What is the pH of a solution containing 0.2 M H 2 CO 3 and 0.4 M NaHCO 3 ? K a = [HCO 3 -1 ][H 3 O + ] [H 2 CO 3 ] 4.3 x 10 -7 = [0.4 M][H 3 O + ] [0.2 M] [H 3 O + ] = 2.15 x 10 -7 MpH = 6.67

29. Common ion effect: the addition of one of the ions in an ionic compound will reduce the compound’s molar solubility. PbCl 2 (s)  Pb +2 (aq) + 2 Cl -1 (aq) K sp = [Pb +2 ][Cl -1 ] 2 Ex: What is the molar solubility of PbCl 2 (K sp = 1.7 x 10 -5 )? Let x = [Pb +2 ], 2x = [Cl -1 ] 1.7 x 10 -5 = (x)(2x) 2 x = 0.016 MThe molar solubility of PbCl 2 What is the molar solubility of PbCl 2 in a 0.2 M solution of NaCl? K sp = [Pb +2 ][Cl -1 ] 2 1.7 x 10 -5 = (x)(0.2) 2 x = 4.25 x 10 -4 M The presence of additional Cl -1 reduced the molar solubility of the PbCl 2 significantly. The new molar solubility.

30. Le Chatelier’s Principle: When a stress is applied to an equilibrium, the equilibrium will shift to alleviate the stress. Fe +3 (aq) + SCN -1 (aq) Ý FeSCN +2 (aq) Colorless Ý Dark red Initial color Left shift = lighter color Right shift = darker color

31. Other ways to cause a Le Châtelier Shift: 3 H 2 (g) + N 2 (g) + heat ⇋ 2 NH 3 (g) What kind of shift would you see if: Pressure increased? Volume increased? Heating temperature increased? Right shift   Left shift Right shift 

32. CH 4 (g) + 2 Cl 2 (g)  CCl 4 (g) + 2 H 2 (g) + heat What kind of shift would you see if: Pressure increased? Heating temperature increased? No Change  Left shift

33. Indicators: Indicators are organic dyes that are also weak acids or weak bases. The color of the dye depends upon the pH of the solution. The indicator will change colors at the pH that corresponds to its own equivalence point. Therefore, it is important to pick an indicator that changes color very close to the pH at the equivalence point for the titration. Since the color change does not exactly match the equivalence point, it is called the ENDPOINT of the titration.