1. 1 Chemical Equilibrium & Acids & bases Chemistry I

2. 2 Reversible Reactions Some reactions do not go to completion as we have assumed –They may be reversible – a reaction in which the conversion of reactants to products and the conversion of products to reactants occur simultaneously Forward: 2SO 2(g) + O 2(g) → 2SO 3(g) Reverse: 2SO 2(g) + O 2(g) ← 2SO 3(g)

3. 3 Reversible Reactions The two equations can be combined into one, by using a double arrow, which tells us that it is a reversible reaction: 2SO 2(g) + O 2(g) ↔ 2SO 3(g) A chemical equilibrium occurs, and no net change occurs in the actual amounts of the components of the system.

4. 4 Reversible Reactions Even though the rates of the forward and reverse are equal, the concentrations of components on both sides may not be equal –An equlibrium position may be shown: A B or A B 1% 99% 99% 1% Note the emphasis of the arrows direction It depends on which side is favored; almost all reactions are reversible to some extent

5. 5 Le Chatelier’s Principle The French chemist Henri Le Chatelier (1850-1936) studied how the equilibrium position shifts as a result of changing conditions Le Chatelier’s principle: If stress is applied to a system in equilibrium, the system changes in a way that relieves the stress

6. 6 Le Chatelier’s Principle What items did he consider to be stress on the equilibrium? 1)Concentration 2)Temperature 3)Pressure Concentration – adding more reactant produces more product, and removing the product as it forms will produce more product Each of these will now be discussed in detail

7. 7 Le Chatelier’s Principle Temperature – increasing the temperature causes the equilibrium position to shift in the direction that absorbs heat If heat is one of the products (just like a chemical), it is part of the equilibrium so cooling an exothermic reaction will produce more product, and heating it would shift the reaction to the reactant side of the equilibrium: C + O 2 (g) → CO 2(g) + 393.5 kJ

8. 8 Le Chatelier’s Principle Pressure – changes in pressure will only effect gaseous equilibria Increasing the pressure will usually favor the direction that has fewer molecules N 2(g) + 3H 2(g) ↔ 2NH 3(g) For every two molecules of ammonia made, four molecules of reactant are used up – this equilibrium shifts to the right with an increase in pressure

9. 9 Predicting shifts in equilibrium H 2 (g) + I 2 (s) ↔ 2HI(g) (colorless) (purple) (colorless) ∆H = + 52 kJ/mol 1.In which direction will the equilibrium shift if more H 2 is added? 2.In which direction will the equilibrium shift if HI is removed? 3.In which direction will the equilibrium shift if heat is added to the reaction system?

10. 10 Predicting equilibrium shifts C(s) + H 2 O(g) ↔ CO(g) + H 2 (g) Which way will the equilibrium shift if pressure is increased?

11. 11 Equilibrium Constants: K eq Chemists generally express the position of equilibrium in terms of numerical values, not just percent –These values relate to the amounts (Molarity) of reactants and products at equilibrium –This is called the equilibrium constant, and abbreviated K eq

12. 12 Equilibrium Constants consider this reaction (the capital letters are the chemical, and the lower case letters are the balancing coefficient): aA + bB  cC + dD –The equilibrium constant (K eq ) is the ratio of product concentration to the reactant concentration at equilibrium, with each concentration raised to a power (which is the balancing coefficient).

13. 13 Equilibrium Constants consider this reaction: aA + bB  cC + dD –Thus, the “equilibrium constant expression” has this general form: [C] c x [D] d [A] a x [B] b (brackets: [ ] = molarity concentration) K eq = Note that K eq has no units on the answer; it is only a number because it is a ratio

14. 14 Equilibrium Constants the equilibrium constants provide valuable information, such as whether products or reactants are favored: if K eq > 1, products favored at equilibrium if K eq < 1, reactants favored at equilibrium

15. 15 Expressing and Calculating K eq A colorless gas dinitrogen tetroxide (N 2 O 4 ) and the dark brown gas nitrogen dioxide (NO 2 ) exist in equilibrium with each other. N 2 O 4 ↔ 2 NO 2 1.00 Liter of a gas mixture at equilibrium at 10 o C contains 0.0045 mol of N 2 O 4 and 0.030 mol of NO 2. Write the expression for the equilibrium constant and calculate the equilibrium constant (K eq ) for the reaction.

16. 16 Answer K eq = [NO 2 ] 2 [N 2 O 4 ] K eq = (0.030) 2 = 0.20 (0.0045) Does this reaction favor the reactants or the products?

17. 17 The Chemistry of Acids and Bases `

18. 18 Acid and Bases

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21. 21 Acids Have a sour taste. Vinegar is a solution of acetic acid. Citrus fruits contain citric acid. React with certain metals to produce hydrogen gas. React with carbonates and bicarbonates to produce carbon dioxide gas Have a bitter taste. Feel slippery. Many soaps contain bases. Bases

22. 22 Some Properties of Acids þ Produce H + (as H 3 O + ) ions in water (the hydronium ion is a hydrogen ion attached to a water molecule) þ Taste sour þ Corrode metals þ Electrolytes þ React with bases to form a salt and water þ pH is less than 7 þ Turns blue litmus paper to red “Blue to Red A-CID”

23. 23 Acid Nomenclature Review No Oxygen  w/Oxygen An easy way to remember which goes with which… “In the cafeteria, you ATE something ICky”

24. 24 Acid Nomenclature Flowchart

25. 25 HBr (aq)HBr (aq) H 2 CO 3H 2 CO 3 H 2 SO 3H 2 SO 3  hydrobromic acid  carbonic acid  sulfurous acid Acid Nomenclature Review

26. 26 Name ‘Em! HI (aq)HI (aq) HCl (aq)HCl (aq) H 2 SO 3H 2 SO 3 HNO 3HNO 3 HCIO 4HCIO 4

27. 27 Some Properties of Bases  Produce OH - ions in water  Taste bitter, chalky  Are electrolytes  Feel soapy, slippery  React with acids to form salts and water  pH greater than 7  Turns red litmus paper to blue “Basic Blue”

28. 28 Some Common Bases NaOHsodium hydroxidelye KOHpotassium hydroxideliquid soap Ba(OH) 2 barium hydroxidestabilizer for plastics Mg(OH) 2 magnesium hydroxide“MOM” Milk of magnesia Al(OH) 3 aluminum hydroxideMaalox (antacid) Al(OH) 3 aluminum hydroxideMaalox (antacid)

29. 29 Dissociation Strong acids and bases completely dissociate in water. Acids dissociate to form H + ions. Bases dissociate to form OH -1 ions. The more ions in solution, the stronger the acid or base it will be. Weak acids and bases partially dissociate.

30. 30 Dissociation Reactions Write dissociation reactions for the following acids and bases in water. –HCl (l) → H + (aq) + Cl -1 (aq) –NaOH (s) → Na + (aq) + OH - (aq) –Ca(OH) 2 (s)→ Ca +2 (aq) + 2OH -1 (aq) –H 2 SO 4 (l) → 2H + (aq) + SO 4 -2 (aq)

31. 31 Acid/Base definitions Definition #1: Arrhenius (traditional) Acids – produce H + ions (or hydronium ions H 3 O + ) Bases – produce OH - ions (problem: some bases don’t have hydroxide ions!)

32. 32 Arrhenius acid is a substance that produces H + (H 3 O + ) in water Arrhenius base is a substance that produces OH - in water

33. 33 Acid/Base Definitions Definition #2: Brønsted – Lowry Acids – proton donor Bases – proton acceptor A “proton” is really just a hydrogen atom that has lost it’s electron!

34. 34 A Brønsted-Lowry acid is a proton donor A Brønsted-Lowry base is a proton acceptor acid conjugate base base conjugate acid

35. 35 ACID-BASE THEORIES The Brønsted definition means NH 3 is a BASE in water — and water is itself an ACID

36. 36 Conjugate Pairs

37. 37 Learning Check! Label the acid, base, conjugate acid, and conjugate base in each reaction: HCl + OH -  Cl - + H 2 O H 2 O + H 2 SO 4  HSO 4 - + H 3 O +

38. 38 Acids & Base Definitions Lewis acid - a substance that accepts an electron pair Lewis base - a substance that donates an electron pair Definition #3 – Lewis

39. 39 Formation of hydronium ion is also an excellent example. Lewis Acids & Bases Electron pair of the new O-H bond originates on the Lewis base.Electron pair of the new O-H bond originates on the Lewis base.

40. 40 The pH scale is a way of expressing the strength of acids and bases. Instead of using very small numbers, we just use the NEGATIVE power of 10 on the Molarity of the H + (or OH - ) ion. Under 7 = acid 7 = neutral Over 7 = base

41. 41 pH of Common Substances

42. 42 Calculating the pH pH = - log [H+] or [H 3 O + ] (Remember that the [ ] mean Molarity) Example: If [H + ] = 1 X 10 -10 pH = - log 1 X 10 -10 pH = - (- 10) pH = 10 Example: If [H + ] = 1.8 X 10 -5 pH = - log 1.8 X 10 -5 pH = - (- 4.74) pH = 4.74

43. 43 Try These! Find the pH of these: 1) A 0.015 M solution of Hydrochloric acid 2) A 3.00 X 10 -7 M solution of Nitric acid

44. 44 Answers! 1) 1.82 2) 6.51

45. 45 pH calculations – Solving for H+ If the pH of Coke is 3.12, [H + ] = ??? Because pH = - log [H + ] then - pH = log [H + ] - pH = log [H + ] Take antilog (10 x ) of both sides and get 10 -pH = [H + ] [H + ] = 10 -3.12 = 7.6 x 10 -4 M *** to find antilog on your calculator, look for “Shift” or “2 nd function” and then the log button *** to find antilog on your calculator, look for “Shift” or “2 nd function” and then the log button

46. 46 pH calculations – Solving for H+ A solution has a pH of 8.5. What is the Molarity of hydrogen ions in the solution?A solution has a pH of 8.5. What is the Molarity of hydrogen ions in the solution? pH = - log [H + ] 8.5 = - log [H + ] -8.5 = log [H + ] Antilog -8.5 = antilog (log [H + ]) 10 -8.5 = [H + ] 3.16 X 10 -9 = [H + ] pH = - log [H + ] 8.5 = - log [H + ] -8.5 = log [H + ] Antilog -8.5 = antilog (log [H + ]) 10 -8.5 = [H + ] 3.16 X 10 -9 = [H + ]

47. 47 More About Water H 2 O can function as both an ACID and a BASE. For this reason, water is said to be amphoteric. In pure water there can be AUTOIONIZATION In pure water, the concentration of hydronium and hydroxide have been experimentally measured as 1.00 x 10 -7 M

48. 48 More About Water In a neutral solution [H 3 O + ] = [OH - ] = 1.00 x 10 -7 M K w is the ion product constant of water K w = [H 3 O + ] [OH - ] = 1.00 x 10 -14 at 25 o C Autoionization

49. 49 [H + ] & [OH - ] calculations Given a 6.25 x 10 -2 M solution of HCl, find [OH - ] Since HCl is a strong acid, it dissociates completely to form H + and Cl - ions. »Remember 1 x 10 -14 = [H + ] [OH - ] so, 1 x 10 -14 = [6.25 x 10 -2 ][OH - ] [OH - ] = 1.60 x 10 -13 M

50. 50 Try this! The OH -1 concentration of a solution is 8.60 x 10 -8. Find the H 3 O +

51. 51 Answer 1.16 x 10 -7

52. 52 pOH Since acids and bases are opposites, pH and pOH are opposites!Since acids and bases are opposites, pH and pOH are opposites! pOH does not really exist, but it is useful for changing bases to pH.pOH does not really exist, but it is useful for changing bases to pH. pOH looks at the perspective of a basepOH looks at the perspective of a base pOH = - log [OH - ] Since pH and pOH are on opposite ends, pH + pOH = 14

53. 53 pH [H + ] [OH - ] pOH

54. 54 [H 3 O + ], [OH - ] and pH What is the pH of the 0.0010 M NaOH solution? [OH-] = 0.0010 (or 1.0 X 10 -3 M) pOH = - log 0.0010 pOH = - log 0.0010 pOH = 3 pOH = 3 pH = 14 – 3 = 11 OR K w = [H 3 O + ] [OH - ] [HO + ] = 1.0 x 10 -11 M [H 3 O + ] = 1.0 x 10 -11 M pH = - log (1.0 x 10 -11 ) = 11.00

55. 55 The pH of rainwater collected in a certain region of the northeastern United States on a particular day was 4.82. What is the H + ion concentration of the rainwater? The OH - ion concentration of a blood sample is 2.5 x 10 -7 M. What is the pH of the blood? On your calculator, key in: 2 nd then log then ( - ) 4.82 then press ENTER On your calculator, key in: 2 nd then log then ( - ) 4.82 then press ENTER On your calculator, key in: ( - ) log then 2.5 EE - 7 then press ENTER On your calculator, key in: ( - ) log then 2.5 EE - 7 then press ENTER

56. 56 Answers 1) 1.51 x 10 -5 M 2) 6.59

57. 57 [OH - ] [H + ] pOH pH 10 -pOH 10 -pH -Log[H + ] Log[OH - ] -Log[OH - ] 14 - pOH 14 - pH 1.0 x 10 -14 [OH - ] [OH - ] 1.0 x 10 -14 [H + ] [H + ]

58. 58 Calculating [H 3 O + ], pH, [OH - ], and pOH Problem 1: A chemist dilutes concentrated hydrochloric acid to make two solutions: (a) 3.0 M and (b) 0.0024 M. Calculate the [H 3 O + ], pH, [OH - ], and pOH of the two solutions at 25°C. Problem 2: A solution has a pH of 3.67. Calculate IN ORDER: [H 3 O + ], [OH - ], and pOH. Is this an acid, base, or neutral? Problem 3: Problem #2 with pH = 8.05?

59. 59 Answers:

60. 60 HNO 3, HCl, H 2 SO 4 HI, HBr, and HClO 4 are strong acids. All other acids are WEAK. Strong and Weak Acids/Bases The strength of an acid (or base) is determined by the amount of IONIZATION (dissociation).

61. 61 Strong and Weak Acids/Bases Generally divide acids and bases into STRONG or WEAK ones. Strong acids are strong electrolytes meaning they dissociate completely in water.Generally divide acids and bases into STRONG or WEAK ones. Strong acids are strong electrolytes meaning they dissociate completely in water. STRONG ACID: HNO 3 (aq) + H 2 O (l) ---> H 3 O + (aq) + NO 3 - (aq) HNO 3 is about 100% dissociated in water.

62. 62 Weak acids are much less than 100% ionized in water. Most molecules do NOT dissociate, but stay in tact.Weak acids are much less than 100% ionized in water. Most molecules do NOT dissociate, but stay in tact. One of the best known is acetic acid = HC 2 H 3 O 2 Strong and Weak Acids/Bases

63. 63 Strong Base: 100% dissociated in water. Group IA and most of IIA hydroxides are strong bases.Strong Base: 100% dissociated in water. Group IA and most of IIA hydroxides are strong bases. NaOH (aq) ---> Na + (aq) + OH - (aq) NaOH (aq) ---> Na + (aq) + OH - (aq) Strong and Weak Acids/Bases Common strong bases include KOH and Ca(OH) 2. CaO (lime) + H 2 O --> Ca(OH) 2 (slaked lime) Ca(OH) 2 (slaked lime) CaO

64. 64 Weak base: less than 100% ionized in waterWeak base: less than 100% ionized in water One of the best known weak bases is ammonia NH 3 (aq) + H 2 O (l)  NH 4 + (aq) + OH - (aq) Strong and Weak Acids/Bases

65. 65 Weak Bases

66. 66 pH testing There are several ways to test pHThere are several ways to test pH –Blue litmus paper (red = acid) –Red litmus paper (blue = basic) –pH paper (multi-colored) –pH meter (7 is neutral, 7 base) –Universal indicator (multi-colored) –Indicators like phenolphthalein –Natural indicators like red cabbage, radishes

67. 67 Paper testing Paper tests like litmus paper and pH paperPaper tests like litmus paper and pH paper –Put a stirring rod into the solution and stir. –Take the stirring rod out, and place a drop of the solution from the end of the stirring rod onto a piece of the paper –Read and record the color change. Note what the color indicates. –You should only use a small portion of the paper. You can use one piece of paper for several tests.

68. 68 pH indicators Indicators are dyes that can be added that will change color in the presence of an acid or base. Some indicators only work in a specific range of pH Once the drops are added, the sample is ruined Some dyes are natural, like radish skin or red cabbage

69. 69 A neutralization reaction is a reaction between an acid and a base. ACID + BASE → SALT + WATER SALT – is defined as any ionic compound. Ex. HBr + KOH → H 2 O + KBr NEUTRALIZATION REACTIONS

70. 70 ACID-BASE REACTIONS Titrations An acid-base titration is a procedure which is used to determine the concentration of an acid or base. A measured volume of an acid or base of known concentration is reacted with a sample to the equivalence point.concentration acidbasevolumeacidbaseconcentration Acid-base titration reactions are neutralization reactions.

71. 71 Click here to watch titration movie (no sound)Click here to watch titration movie Setup for titrating an acid with a base

72. 72 TitrationTitration 1.Add solution from the buret. 2. Reagent (base) reacts with compound (acid) in solution in the flask. This is called NEUTRALIZATION. 3.Indicator shows when exact stoichiometric reaction has occurred. (moles of Acid = moles of Base) This is called the EQUIVALENCE POINT.

73. 73 35.62 mL of NaOH is neutralized with 25.2 mL of 0.0998 M HCl by titration to an equivalence point. What is the concentration of the NaOH? Titration problem - accurately determine its concentration.

74. 74 Step One: Write the balanced equation NaOH + HCl → H 2 O + NaCl Step Two: Determine the moles of substance using the Molarity and volume of that substance. –Remember M = moles/Liters so moles = M x L –0.0252 L x 0.0998 M = 0.00251 moles HCl Step Three: –Multiply the moles obtained in step 2 by the mole ratio: »0.00251 moles HCl x 1 mole NaOH = 0.00251 mol NaOH Step Four: Divide moles in Step 3 by volume of NaOH to find the concentration of NaOH »0.00251 mol NaOH = 0.0704 M Titration solution. 1 mole HCl 0.03562 L

75. 75 Example 2 - titration Determine the volume of 0.100M HNO 3 needed to neutralize 50.00 mL of 0.842M Ca(OH) 2.

76. 76 Answer 1.19L