1. Acid-Base Chemistry Arrhenius acid: Substance that dissolves in water and provides H + ions Arrhenius base: Substance that dissolves in water and provides OH - ions Examples: HCl  H + and Cl - Acid NaOH  Na + + OH - Base

2. Bronsted Acid: Substance that donates proton to another substance Bronsted base: Substance that accepts proton from another substance Example: HCl + H 2 O  H 3 O + + Cl - HCl acts as acid; H 2 O acts as base In the Reverse Reaction, H 3 O + acts as an acid; Cl - acts as a base Note: (H 3 O + = hydronium ion = H + = proton)

3. A monoprotic acid contains one acidic proton. HCl A diprotic acid contains two acidic protons. H 2 SO 4 A triprotic acid contains three acidic protons. H 3 PO 4 A Brønsted–Lowry acid may be neutral or it may carry a net positive or negative charge. HCl, H 3 O +, HSO 4 −

4. Conjugate acid: Species formed after base accepts a proton Conjugate base: Species remaining after an acid donates its proton Conjugate acid-base pair: an acid and base on opposite sides of the equation that correspond to each other Examples: HNO 3 + H 2 O H 3 O + + NO 3 - acid base acid base Conjugate pairs: HNO 3 and NO 3 - H 2 O and H 3 O +

5. Example: HS - + H 2 O  H 3 O + + S 2- Conjugate pairs: HS - and S 2- H 2 O and H 3 O + Practice: HClO 4 + H 2 O  H 3 O + + ClO 4 - What are the conjugate pairs? HClO 4 and ClO 4 - H 2 O and H 3 O +

6. HBr+ + gain of H + acidbase acid loss of H + H2OH2O Br − H3O+H3O+ HBr and Br − are a conjugate acid–base pair. H 2 O and H 3 O + are a conjugate acid–base pair. Note: The net charge must be the same on both sides of the equation.

7. Water can act as both an acid and a base (amphiprotic)! HClO 4 + H 2 O  H 3 O + + ClO 4 (base) NH 3 + H 2 O  OH - + NH 4 + (acid) Strengths of Acids and Bases Strong acids/bases: dissociate completely when dissolved in solution Weak acids/bases: dissociate only partially when dissolved in solution

8. Examples: Strong Acid: HCl  H + + Cl - (100% dissociation) Strong Base: NaOH  Na + + OH (100% dissociation) Weak Acid: CH 3 COOH  H + + CH 3 COO - (1.3% dissociation) Weak Base: NH 3 + H +  NH 4 +

9. Table 9.1

10. Naming Acids Binary Acids: hydo + root of anion + ic + “acid” ex. HCl hydrochloric acid, HBr hydrobromic acid HI Polyatomic-based Acids: root of polyatomic ion + ic + “acid” ex. H 2 SO 4 sulfuric acid, H 3 PO 4 phosphoric acid H 2 CO 3 HNO 3 Hydroiodic acid carbonic acid nitric acid

12. The Self-Ionization of Water H 2 O + H 2 O  H 3 O + + OH - Pure water: [H 3 O + ]=[OH - = 10 -7 M (at 25 0 C) Neutral Solution: Any solution in which the concentrations of H 3 O + and OH - ions are equal (10 -7 M) Acidic Solution: Solutions having a greater concentration of H 3 O + than OH - ions ([H 3 O + ] greater than 10 -7 M) Example: A solution with [H 3 O + ] = 10 -5 M Basic Solution: solution having a greater concentration of OH - than H 3 O + ions ([H 3 O + ] less than 10 -7 M) Example: A solution with [H 3 O + ] = 10 -12 M

13. The pH Scale  pH is a measure of acidity  Scale ranges from 0-14 pH = 7Neutral pH < 7 Acidic pH > 7 Basic pH represents the concentration of H + ions in solution Pure water: 1 x 10 -7 moles H + per liter and1 x 10 -7 moles OH - per liter

14. Solutions with equal concentrations of and ions are called Neutral Solutions with more than 1 x 10 -7 moles H + per liter are Acidic Solutions with less than 1 x 10 -7 moles H + per liter are Basic Note: [H + ] x [OH - ] = 10 -14 (always!)

15. pH Scale Summary  pH scale refers to amount of H + ions in solution  pH 7 is neutral, less than 7 is acidic, greater than 7 is basic  Lower pH = more acidic = more H + ions  Higher pH = more basic = less H + ions

16. Each pH unit represents a 10-fold change in H + ion concentration! pH 4 has 10 times more H + ions than pH 5 pH 9 has 10 times fewer H + ions than pH 8 Mathematical equation for pH: pH is the negative log of the H 3 O + concentration pH = -log [H 3 O + ]

17. Any number can be expressed as 10 raised to some exponent: y = 10 x Examples: 100 = 10 2 1000 = 10 3 0.10 = 10 -1 The log is that exponent! 100 = 10 2 ; Log of 100 =2 1000 = 10 3 ; Log of 1000 = 3 0.10 = 10 –1 ; Log of 0.10 = -1

18. We can also take the log of non-whole numbers, but we use our calculators for this. Example: Find the log of 2.4 x 10 -3  Enter 2.4 x 10 -3 into calculator  Press the “log” key 0.0024 “log” = -2.62 Therefore, 10 -2.62 = 2.4 x 10 -3

19. Calculating pH from [H 3 O + ] pH = -log [H 3 O + ]  Enter [H 3 O + ] into calculator  Press the “log” key  Change the sign Example: [H 3 O + ] = 1.0 x 10 -7 M pH = -log [H 3 O + ] pH = -log [1 x 10 -7 ] = 7

20. Example: [H 3 O + ] = 1 x 10 -11 M pH = -log [H 3 O + ] pH = -log [1 x 10 -11 ] = 11 Example: [H 3 O + ] = 1 x 10 -3 M pH = -log [H 3 O + ] pH = -log [1 x 10 -3 ] = 3

21. Example: [H 3 O + ] = 4.2 x 10 -5 pH = -log [H 3 O + ]  Enter [H 3 O + ] into calculator (4.2 x 10 -5 )  Press the “log” key (-4.3767507)  Change the sign (4.3767507) pH = 4.3767507 = 4.4

22. Example: [H 3 O + ] = 8.1 x 10 -9 pH = -log [H 3 O + ]  Enter [H 3 O + ] into calculator (8.1 x 10 -9 )  Press the “log” key (-8.091515)  Change the sign (8.091515) pH = 8.091515 = 8.1

23. Reactions Between Acids and Bases Neutralization: reaction between an acid and a base; always produces salt and water Example: Write a balanced equation for the reaction of hydrochloric acid with sodium hydroxide. HCl(aq) + NaOH(aq) H—OH(l) + NaCl(aq)

24. Example: Write a balanced equation for the reaction of hydrochloric acid with magnesium hydroxide. HCl + Mg(OH) 2 H 2 O +MgCl 2 22

25. Titration Titration: a technique used to determine the concentration of an acid or base in a solution If we want to know the concentration of an acid solution, a base of known concentration is added slowly until the acid is neutralized. When the acid is neutralized: # of moles of acid = # of moles of base This is called the end point of the titration.

26. Acid-Base Titration Titration is a laboratory procedure used to determine the molarity of an acid. In a titration, a base such as NaOH is added to a specific volume of an acid. Base (NaOH) Acid solution

27. A few drops of an indicator is added to the acid in the flask. The indicator changes color when the base (NaOH) has neutralized the acid.

28. At the end point, the indicator has a permanent color. The volume of the base used to reach the end point is measured. The molarity of the acid is calculated using the neutralization equation for the reaction.

29. Determining an unknown molarity from titration data requires three operations: Moles of base Moles of base Molarity of acid solution Molarity of acid solution mole–mole conversion factor mole–mole conversion factor M (mol/L) conversion factor M (mol/L) conversion factor Moles of acid Moles of acid Volume of base solution Volume of base solution M (mol/L) conversion factor M (mol/L) conversion factor [1] [2] [3]

30. HOW TO Determine the Molarity of an Acid Solution from Titration Example: What is the molarity of an HCl solution if 22.5 mL of a 0.100 M NaOH solution are needed to titrate a 25.0 mL sample of the acid? volume of base (NaOH) 22.5 mL conc. of base (NaOH) 0.100 M volume of acid (HCl) 25.0 mL conc. of acid (HCl) ?

31. Step [1] Determine the number of moles of base used to neutralize the acid. Volume of base solution Volume of base solution M (mol/L) conversion factor M (mol/L) conversion factor mL–L conversion factor mL–L conversion factor 22.5 mL NaOH x 1 L 1000 mL x 0.100 mol NaOH 1 L = 0.00225 mol NaOH

32. Step [2] Determine the number of moles of acid that react from the balanced chemical equation. HCl(aq) + NaOH(aq) H 2 O(l) + NaCl(aq) 0.00225 mol NaOH x mole–mole conversion factor mole–mole conversion factor 1 mol HCl 1 mol NaOH = 0.00225 mol HCl

33. Step [3] Determine the molarity of the acid from the number of moles and the known volume. M= mol L = 0.00225 mol HCl 25.0 mL solution x 1000 mL 1 L =0.0900 M HCl Answer mL–L conversion factor mL–L conversion factor

34. Buffers Buffer: a solution whose pH changes very little when acid or base is added. Most buffers are solutions composed of roughly equal amounts of a weak acid the salt of its conjugate base The buffer resists change in pH because added base, − OH, reacts with the weak acid added acid, H 3 O +, reacts with the conjugate base

35. Buffers contain 2 compounds:  Compound with the ability to react with H + ions  Compound with the ability to react with OH - ions Example: HCO 3 - + H +  H 2 CO 3 If acids (H + ) are added, react with HCO 3 - H 2 CO 3 + OH -  HCO 3 - + H 2 O If OH - ions are added, react with H 2 CO 3 H 2 CO 3 is unstable: H 2 CO 3  H 2 O + CO 2

36. More Examples of a Buffer CH 3 COOH + H 2 O H 3 O + + CH 3 COO − weak acid conjugate base If an acid is added to the following buffer equilibrium, Adding more product… …drives the reaction to the left. then the excess acid reacts with the conjugate base, so the overall pH does not change much.

37. CH 3 COOH + − OH H 2 O + CH 3 COO − weak acid conjugate base If a base is added to the following buffer equilibrium, Adding more reactant… …drives the reaction to the right. then the excess base reacts with the weak acid, so the overall pH does not change much.