2. 1 Unit 2: Energy Icons are used to prioritize notes in this section. Make some notes: There are SOME important items on this page that should be copied into your notebook or highlighted in your printed notes. Copy as we go: There are sample problems on this page. I EXPECT YOU to copy the solutions into a notebook— Even if you have downloaded or printed the notes!!! Look at This: There are diagrams or charts on this page. Look at them and make sure you understand them. (but you don’t need to copy them) Extra Information: This page contains background information that you should read, but you don’t need to copy it. R Review: There is review material on this page. It is up to you to decide if you want to make notes or highlight it, depending on how well you remember it.

3. 2 Unit 4 (formerly Module 5) Equilibrium

4. 3 Equilibrium Overview: Equilibrium is the concept of a system remaining “in balance”. A system in equilibrium does not change at the macroscopic level (the level that we can detect with our senses). True equilibrium should not be confused with homeostasis or “steady-state”, a process by which living organisms attempt to maintain a consistent internal conditions by absorbing or excreting materials from and to their environment. 11.0

5. 4 Equilibrium vs. Homeostasis Equilibrium usually exists in a closed system, where materials cannot easily enter or leave. Examples: A reversible chemical reaction occurs inside a closed container. The reaction appears to have stopped, but at a molecular level changes are still going on. A liquid in a sealed bottle does not appear to evaporate. Homeostasis usually exists in an open system, where materials can enter and leave. Examples: A dog lives in a kennel. It eats and excretes roughly equal amounts, and therefore maintains a steady weight and internal conditions. A cell in a human body maintains a balance of nutrients.

6. 5 The Temporary Nature of Equilibrium Although we study equilibrium as if it is an unchanging state, in reality dynamic processes as well as static forces may act upon the equilibrium. Eventually something will upset the equilibrium, temporarily or permanently throwing it “out of balance” Often a new equilibrium will be re-established after the original equilibrium is upset.

7. 6 Examples A precariously balanced rock formation can endure in “equilibrium” for centuries. Suddenly the balanced rock falls, temporarily disturbing the equilibrium. A reversible chemical reaction inside a beaker has reached a state of equilibrium and appears to have stopped. A researcher adds more of one of the reactants, upsetting the equilibrium. The reaction temporarily resumes until a new equilibrium is established.

8. 7 The Old Man of the Mountain For two hundred years a precarious rock formation in New Hampshire was said to resemble the face of an old man. It had become a symbol of New Hampshire, appearing on postcards, road signs and coins On May 3, 2003 The rock face collapsed. In terms of equilibrium, we could say that this was a static equilibrium that endured for centuries, until it was disturbed by a spring storm. Afterwards a new equilibrium was established, that unfortunately no longer resembled a face.

9. 8

10. 9 Chapter 11 Qualitative Aspects of Chemical Equilibrium

11. 10 Qualitative Aspects of Equilibrium What is an equilibrium? What properties does it have? How can we distinguish dynamic, and static equilibria and tell them apart from simple steady states? 11.1

12. 11 Static Equilibrium Note: Gravity is not considered to be a dynamic force. It is a static force.

13. 12 Dynamic Equilibrium

14. 13 A dynamic equilibrium is a bit like a hockey game. Barring penalties, there is always the same number of players on the ice, but some players are constantly leaving the bench as others return to it. 19 11

15. 14 Phase Equilibrium (1 st type of dynamic equilibrium)

16. 15 Solubility Equilibrium (2 nd type of dynamic equilibrium)

17. 16 Chemical Equilibrium (3 rd type of dynamic equilibrium)

18. 17 Irreversible and Reversible Reactions Some chemical reactions are easily reversed, like the electrolysis of water. Others, such as the burning of wood are impossible to reverse under laboratory conditions. 11.2

19. 18 Irreversible Chemical Reactions The growth of a tree does allow wood to be produced from materials that might include wood ashes, but growing a tree takes decades, and requires countless changes involving many complex chemical mechanisms and multiple organic catalysts (enzyme systems). Burning is therefore NOT considered reversible!

20. 19 Reactants  Products Reactants can become products, but products cannot turn back into reactants.

21. 20 Reversible Chemical Reactions Some reactions are easily reversed using common laboratory procedures. For example, it is possible to decompose water into hydrogen and oxygen using electrolysis, and equally possible to synthesize water from hydrogen and oxygen by controlled burning: 2H 2 O + electrical energy  2 H 2 + O 2 2H 2 +O 2  H 2 O + heat energy

22. 21 Reactants Products Reactants can become products, and products can also turn back into reactants. Also show as: reactants  products

23. 22 Reversibility and Equilibrium Only reversible reactions can produce a true dynamic chemical EQUILIBRIUM

24. 23 A Chemical System is at Equilibrium if it meets these Criteria: 1.The System is Closed 1.The System is Closed, for example by being sealed inside a container so material cannot enter or leave. 2.The Change is reversible 2.The Change is reversible. The reaction or change can proceed in both direct and reverse directions. 3.There is no Macroscopic Activity. 3.There is no Macroscopic Activity. Nothing seems to be happening – the properties of the system are constant. These unchanging properties can include: colour, amount of undissolved solute, concentration, pressure etc. 4.There is Molecular Activity 4.There is Molecular Activity. Reactions continue at the microscopic or molecular level

25. 24 Definitions of some easily confused terms Macroscopic: Occurring at the level we can detect with our senses, as opposed to microscopic. Observable changes. Microscopic: Occurring at a level below what we can see. Too small to observe without instruments. Dynamic Equilibrium: A balance that involves two opposing active processes that are occurring at the same rate. This contrasts with Static Equilibrium and Steady State (Homeostasis). Static Equilibrium: A balance that does not involve active processes. Steady State (including Homeostasis): An apparent balance that occurs in an open system. An unchanging set of properties is maintained, but materials enter and leave the system.

26. 25 Page 287 Read all the questions, make sure you understand them, be prepared to answer them verbally next class.

27. 26 Le Châtelier’s Principle Henri Louis Le Châtelier (1850-1937) was a French chemist who is most famous for his studies of chemical equilibrium. In addition he studied metal alloys and, with his father, was involved in the development of methods of purifying aluminum 11.4

28. 27 Equilibrium is not eternal An equilibrium can exist for a long time, only to change (be upset) when certain conditions change. After it is upset, there is a period of adjustment, then a new equilibrium is established. Original Equilibrium New Equilibrium Equilibrium Upset Adjusting...

29. 28 Some factors that might upset an equilibrium. Which of these factors do you think might affect the amount of reactant and product at equilibrium? Maybe Temperature? Maybe Pressure? Maybe Concentration of reactants and products? Maybe Catalyst? Most of them do, but one does not.

30. 29 Henri LeChâtelier studied many of these factors to see how they could effect a system at equilibrium. He found some factors could favour the direct reaction, increasing the amount of product. Others could favour the reverse reaction, increasing the amount of reactant. Regardless, eventually an equilibrium was re- established, but with new amounts of product and reactant.

31. 30 When an Equilibrium is Upset... Henri LeChatelier stated the following generalization: “ If the conditions of a system in equilibrium change, the system will react to partially oppose this change until it attains a new state of equilibrium.” In other words: The system will establish a new equilibrium..

32. 31 Will any reversible reaction will eventually reach equilibrium? Answer: yes, as long as it in a closed system. Will the amount of product always equal the amount of reactant at equilibrium? Short answer: NO! they are not always equal. At equilibrium the amount of “reactant” and “product” may vary depending on several factors.

33. 32 The Effect of a Catalyst Adding a catalyst to a system already at equilibrium will have NO EFFECT. Adding a catalyst to a system that has not yet reached equilibrium will cause it to reach equilibrium faster. Why? A catalyst increases both forward and backward rates equally, so the final result will be the same, but the process of reaching equilibrium will be faster.

34. 33 Increasing the temperature will favour the endothermic reaction. Decreasing the temperature will favour the exothermic reaction. Effect of Temperature

35. 34 Effect of Pressure Increasing the pressure may favour the reaction that produces fewer gas particles Decreasing the pressure may favour the reaction that produces more gas particles. Note: only reactants or products that are in the gaseous state are counted towards the effects of pressure.

36. 35 Effects of Concentration The effects of concentration of the reactants and products are most important of all. Before we can see the effects of changing a concentration we must examine the mathematics of an equilibrium reaction.

37. 36 Remember LeChatelier’s Principle: “If the conditions of a system in equilibrium change, the system will react to partially oppose this change until it attains a new state of equilibrium.” In other words, any “stress” or change that you make to the system will cause it to react in a way that tries to (partially) undo the change that you made.

38. 37 Changes in concentration If you increase the concentration of a reactant, the equilibrium will shift to use up some of the reactant you added. If you increase a product, the equilibrium will shift to reduce the product and make more reactant H2H2 +I2I2 2HI If you increase the amount of Hydrogen… …The system will react to reduce the amount of Hydrogen… …by shifting the reaction towards the product

39. 38 H 2 + I 2  2HI  Sudden increase in[H 2 ].  The equilibrium changes: [HI] goes way up, [I 2 ] goes down.  This causes [H 2 ] to adjust towards its original level  The equilibrium changes: [HI] goes way up, [I 2 ] goes down.

40. 39 Changes in Temperature Increasing the temperature causes the equilibrium to shift in the direction that absorbs heat (the endothermic direction). Decreasing the temperature shifts the equilibrium in the exothermic direction. 2SO 2 +O2O2 2SO 3 If you increase the temperature… …The system will react to reduce the temperature… …by shifting the reaction towards endothermic side Heat+

41. 40 2SO 2 + O 2  2SO 3 + heat  Sudden increase in temperature.  The endothermic reaction kicks in, getting rid of some SO 3, and creating more SO 2 and more O 2 and cooling things off  This causes a lowering of the temperature, moving it towards its original level

42. 41 Changes in Pressure (only affects systems where one or more materials are gases) Increasing the pressure causes the equilibrium to shift in the direction that has the fewest gas particles. Decreasing the pressure shifts the equilibrium in the direction that produces more gas particles. If you increase the pressure……the system will create fewer molecules… …to reduce the pressure. 8 molecules4 molecules

43. 42 N 2 + 3 H 2  2 NH 3  When the pressure increases, the equilibrium adjusts, making more NH 3 (since NH 3 takes up less space than H 2 + N 2 )  Sudden increase in pressure.  Pressure partially adjusts towards the original level..

44. 43 Special Note Re. Gases Only gases can be affected by pressure. If only one side of an equation has gases, then... Increasing pressure will favour the side with no gases. Decreasing the pressure will favour the side with gases. I 2(s) I 2(g) Decreased pressure Increased pressure

45. 44 Page 304, Questions 1 to 4

46. 45 Chapter 12 The Quantitative Aspects of Equilibrium

47. 46 Chapter 12 In this section we will explore the mathematical aspects of equilibrium, including: The Equilibrium Constant (K c ) The Equilibrium Law 12.1

48. 47 The Equilibrium Constant and Equilibrium Law Expressions The equilibrium constant (K c ) is: A number derived from an equilibrium law expression. a ratio between the concentration of products and the concentration of reactants of a reversible reaction at equilibrium With each product/reactant raised to the power of its corresponding coefficient

49. 48 Deriving the Equilibrium Law The next two slides show how the equilibrium law was derived from the rate law. If you want to understand the relationship between rates and equilibrium you should follow this. If you just want to use the equilibrium law to find K c, you can skip forward three slides.

50. 49 A Generalized Reversible Reaction with reactants and products aA + bB ↔ cC + dD reactants products Forward rate: r dir = k dir [A] a [B] b Reverse rate: r rev = k rev [C] c [D] d At equilibrium, r dir = r rev so, through the magic of algebra… Products on top Reactants on bottom Forward reaction = dir Reverse reaction = rev

51. 50 Simplifying the Rate Constant The two separate rate constants (k dir and k rev ) are often replaced by a single equilibrium constant, K c : Becomes:

52. 51 The Equilibrium Law The lowercase letters represent coefficients, the uppercase letters are chemical formulas, the square brackets mean concentration. K c is the equilibrium constant For a chemical equation of the type: aA + bB ↔ cC + dD “Products” always go on the top! “Reactants” always go on the bottom! The equilibrium law expression is:

53. 52 Variants of the Equilibrium Constant k eq KcKc KaKa These symbols are all used for Equilibrium Constants in different text books or for different types of reaction. I usually use K c, since your text book and the study guide both use it. The old textbook used k eq. In problems involving acids and bases, K a and K b are often used. K sp is used for problems involving the solubility product. KbKb K sp

54. 53 What does the Equilibrium Constant mean? It is a ratio between the amount of reactant and product that exist at equilibrium. If k c is greater than 1, then the direct reaction is favoured, and there is more product than reactant at equilibrium* If k c is less than 1, then the reverse reaction is favoured, and there is more reactant than product at equilibrium* If k c is 0, the reaction is impossible. if k c is infinite (∞) the reaction is spontaneous and irreversible so as much reactant as possible will change to product *this is a slight over-simplification, since the formula can be complicated, but it is generally true.

55. 54 Effect of Temperature on an Equilibrium Constant An equilibrium constant relates to concentrations, and only remains constant if the other conditions, such as temperature, remain fixed. If the temperature were to change, so would the value of Kc How the value of Kc might change depends on the type of reaction (exothermic or endothermic) and how the temperature changed (increased or decreased)

56. 55 Table Showing Effects of Temperature on Equilibrium Constants Type of reactionTemperature change Favoured ReactionChange in K c Exothermic*IncreaseReverse (  )Decrease Exothermic*DecreaseDirect (  )Increase EndothermicIncreaseDirect (  )Increase EndothermicDecreaseReverse (  )Decrease *Exothermic means either ΔH < 0 or that Reactants  Products + Energy Notice that any change in temperature that favours the direct reaction will cause the value of K c to increase.

57. 56 Fixed Concentrations Some materials have “fixed” concentrations, ie. Their concentrations cannot change in an equilibrium. Examples: An undissolved solid, (it can’t have a concentration unless it dissolves.) A pure liquid. (a pure substance always has its maximum concentration) These cases, which include all substances with the (s) and (l) phase markers, are not included in equilibrium calculations.

58. 57 Example What is the equilibrium expression for the following equation: CN 1- (aq) + H 2 O (l)  HCN (aq) + OH 1- (aq) Answer: Not this: Why? Because H 2 O (liquid water) is a pure substance and therefore has a fixed concentration. Substances with fixed concentration are not included in an equilibrium expression.

59. 58 Calculating Equilibrium Concentrations To calculate the concentrations of reactants and products at equilibrium we sometimes use a table that records the initial concentration, the change in concentration and the final equilibrium concentration of each reactant or product. We call such a table an I.C.E. table 12.1.4

60. 59 The I.C.E. method A technique that can help solve some equilibrium problems Stands for: I nitial concentration C hange in concentration E quilibrium concentration A technique that can help solve some equilibrium problems Stands for: I nitial concentration C hange in concentration E quilibrium concentration

61. 60 Reactant 1 Reactant 2 (if needed) Product 1 Product 2 (if needed) I(Initial) C(Change) E ( Equilibrium ) Make a table with 3 rows labelled I, C, and E, and enough columns for all the reactants and products you are interested in Concentration Before Concentration After Concentration Difference (ΔC) Negative Ratio (reactant) Sum Positive Ratio (product) Negative Ratio (reactant) Molar ratio Reactant + Reactant  Product + Product Write the chemical equation of the reaction. Show all reactants and Products Identify the molar ratios (based on the coefficients of the equation) Eliminate any unused ratios, such as those based on liquids and undissolved solids. Enter the information you already know into the table. If no values are given for the initial product concentrations, it is usually safe to assume they are zero. Look for a column that you can complete! Fill in the missing squares in “C” row. They will be ratios to the one you know The numbers will be negative for reactants, positive for products. Add to find the equilibrium values (adding a negative number is like subtracting) Use the Equilibrium Concentrations to answer any further questions (such as finding K c, K a K b or K sp )

62. 61 ICE method: Step by Step Write the chemical equation of the reaction. Show all reactants and Products Identify the molar ratios (based on the coefficients of the equation) Eliminate any unused ratios, such as those based on liquids and undissolved solids. Add to find the equilibrium values (adding a negative number is like subtracting) Fill in the missing squares in “C” row. They will be ratios to the one you know The numbers will be negative for reactants, positive for products. Make a table with 3 rows labelled I, C, and E, and enough columns for all the reactants and products you are interested in Look for a column that you can complete! Enter the information you already know into the table. If concentrations are not given as mol/L you may have to convert them. If not told otherwise, you may assume that the initial product concentrations are zero. Use the Equilibrium Concentrations to answer any further questions (such as finding Kc, Ka or Kb)

63. 62 Sample Problem (see p. 316 and 317 in Text Book) At a given temperature, 10 moles of nitrogen oxide (NO) and 8 moles of Oxygen (O 2 ) are placed in a 2 litre container. After a given period of time, the following equilibrium is obtained: 2 NO (g) + O 2(g)  2NO 2(g) Once equilibrium is attained, 8 moles of NO 2 have been producedCalculate the equilibrium constant. Once equilibrium is attained, 8 moles of NO 2 have been produced. Calculate the equilibrium constant. Preliminary work: (convert to moles per litre) Initial concentrationsC i [NO] = 10 mol in 2 L = 5 mol/L C i [O 2 ] = 8 mol in 2L = 4 mol/L assume: C i [NO 2 ] = 0 mol/L Final Concentration: C f [NO 2 ] = 8 mol in 2L = 4 mol/L

64. 63 NO 2 O 2(g) NO 2 I (Initial C) C(Change) E ( Equilibrium ) Make a table with 3 rows labelled I, C, and E, and enough columns for all the reactants and products you are interested in 5 5 mol/L (C i [NO 2 ]) 4 4 mol/L (C i [O 2 ]) 0 0 mol/L (C i [NO 2 ]) 4 4 mol/L (C f [NO 2 ]) +4 mol/L (ΔC [NO 2 ]) -4 mol/L (-ratio) 1 mol/L 2 mol/L -2 mol/L (-ratiot) 212 2 NO (g) + O 2(g)  2NO 2(g) Write the chemical equation of the reaction. Show all reactants and Products Identify the molar ratios (based on the coefficients of the equation) Enter the information you already know into the table Look for a column that you can complete! Fill in the missing squares in “C” row. They will be ratios to the one you know The numbers will be negative for reactants, positive for products. Add to find the equilibrium values (adding a negative number is like subtracting)

65. 64 Finishing the Problem Now you need to find the K c value. To do this, use the values from the “E” line of your table (the equilibrium concentrations) and substitute them into the K c formula: The value of the equilibrium constant is 8 (although you do not need to give the units of an equilibrium constant, since our concentrations were in moles/Litre, and our time is assumed to be in seconds it would be safe to call it mol/(L∙s))

66. 65 A Tougher Sample Problem (see p. 316 and 317 in Text Book) At 1100K the equilibrium constant of the following reaction is 25 H 2(g) + I 2(g)  2HI (g) 2 moles of hydrogen and 3 moles of iodine are placed in a 1 L container. What is the concentration of each substance when the reaction attains equilibrium. 2 moles of hydrogen and 3 moles of iodine are placed in a 1 L container. What is the concentration of each substance when the reaction attains equilibrium.. This time we have no numbers for equilibrium concentrations, we will have to use algebraic variable for some of the values. Let x represent the change in Hydrogen Concentration Let: ΔC [H 2 ] = -x (why negative? Because hydrogen is a reactant!) Initial concentrations: [H 2 ] = 2 mol/L, [I 2 ] = 3 mol/L

67. 66 H2H2H2H2 I 2(g) HI I (Initial C) C(Change) E ( Equilibrium ) Make a table with 3 rows labelled I, C, and E, and enough columns for all the reactants and products you are interested in 2 2 mol/L 3 3 mol/L 0 0 mol/L 2x 2x mol/L +2x +2x mol/L -x -x mol/L 2-x 2-x mol/L 3-x 3-x mol/L -x -x mol/L 112 H 2(g) + I 2(g)  2HI (g) Write the chemical equation of the reaction. Show all reactants and ProductsIdentify the molar ratios (based on the coefficients of the equation)Enter the information you already know into the tableLook for a column that you can complete! Fill in the missing squares in “C” row. They will be ratios to the one you know The numbers will be negative for reactants, positive for products. Add to find the equilibrium values (adding a negative number is like subtracting)

68. 67 Continuing the Problem Set up the K c formula, substituting in the expression from line “E” So.. This equation can be rearranged in the following steps:  Carried over to next slide

69. 68 Using the Quadratic formula  Equation to be solved  Quadratic formula, where: a=21, b=125, c=150  substitute  Two solutions to formula  But only one of the solutions will give a real answer!

70. 69 Choosing the Best Answers  Two solutions to formula  Try finding the correct concentrations using the solutions At equilibrium: The concentration of hydrogen will be 0.33 mol/L, The concentration of iodine will be 1.33 mol/L and The concentration of hydrogen iodide will be 3.34 mol/L Concentration cannot be negative! Solve using 4.29 Solve using 1.67

71. 70 Simplifying I.C.E. tables (the 5% rule) Sometimes, when doing an ICE table you may have to subtract a very small value from a relatively large value, for example 2.0 mol/L – 1.0x10 -4 mol/L. In this case, don’t bother doing the subtraction, since by the time you change it to show significant digits, the result will be the same: 2.0 mol/L – 0.0001 mol/L = 1.9999 mol/L ≈ 2.0 mol/L. In fact, as a rule of thumb, if the number you subtract is less than 5% of the original number, you can skip the subtraction.

72. 71 Acids and Bases The concept of acids and bases has been with us for a long time, but there has always been difficulty with the exact definitions. In this section we will look at four theories about acids and bases (Don’t worry, you only need to remember 2 of them) We will find out the difference between “strength”, “concentration” and “acidity” of an acid. We will also explore the concepts of pH and pOH a bit deeper than you did in grade 9 or 10. We will explore the mathematical relationships between concentration and pH 12.2

73. 72 Properties of Acids & Bases Acids are solutions that: redclear Turn litmus red, but leave phenolphthalein clear. React with active metals to give off H 2 gas React with carbonate salts to give off CO 2 gas Taste sour (if safe to taste) Have low pH numbers (below 7) Bases are solutions that: blue purple Turn litmus blue, and turn phenolphthalein purple. Taste bitter (if safe to taste) Seldom react with metals or carbonates Have High pH numbers (above 7)

74. 73 Theories of Acids and Bases Hypothesis#1: Lavoisier’s Mistake (optional item) Lavoisier (1776) dealt mostly with strong oxyacids, like HNO 3 and H 2 SO 4. He claimed that: An acid is a substance that contains oxygen. This hypothesis is now known to be completely wrong, but it did give oxygen its name: “oxy” (meaning acid)+ “gen” (former or creator)

75. 74 Theories of Acids and Bases (Hypothesis#2: Arrhenius’ Theory) Arrhenius (c. 1884) claimed that: An acid is a substance that dissociates in water to produce H+ ions A base is a substance that dissociates in water to produce OH- ions Arrhenius’ theory is still used to this day as a “simplified” way of explaining acids and bases. It explains most of the properties of acids... Why their formulas usually begin with H, why they give off hydrogen when reacting with metals, etc. But this theory does not account for acidic and basic salts– substances that act like acids and bases, but don’t have an H or OH in their formula, or for anhydrous acids, or for reactions that occur outside of water.

76. 75 Theories of Acids and Bases (Hypothesis#3: Brønsted-Lowry) Brønsted and Lowry (1923) proposed: An acid is a substance from which a proton (H + ) can be removed. An acid is a proton donor. A base is a substance that can cause a proton to be removed from an acid. A base is a proton acceptor. Although a bit complicated for explaining simple acid/base reactions, this is the main theory in use today.

77. 76 Theories of Acids and Bases (Hypothesis#4: Lewis Acids) Gilbert Lewis (c.1940) proposed: An acid is a substance that can accept an electron pair to form a covalent bond. A base is a substance that can donate a pair of electrons to form a covalent bond. This theory is not explained in the new textbook, and will probably not be required for examinations. It is given here as optional enrichment material. The Lewis theory is the only one that can explain the properties of acidic & basic salts.

78. 77 Comparing the Hypotheses TheoryAn Acid is...A Base is...Details LavoisierA substance with oxygen Incorrect hypothesis ArrheniusA substance that dissociates in water to produce H+ ions A substance that dissociates in water to produce OH- ions. All acids have H All bases have OH Bronsted- Lowry A substance from which an H+ ion can be removed (a proton donor) A substance which can remove the H + from an acid (a proton acceptor) All acids have H Bases can have any negative ion. LewisA substance that can accept a pair of electrons to form a covalent bond A substance that can donate a pair of electrons to form a covalent bond Acids may have H + or some other + ions, a base may have OH - or some other – ions Incorrect Hypothesis Optional Hypothesis

79. 78 Bronsted-Lowry and conjugates One of the results of the Bronsted-Lowry theory is that in a reaction, each acid has a corresponding base and each base has a corresponding acid (called their conjugates) HCl (aq) + H 2 O (l)  H 3 O + (aq) + Cl - (aq) H+H+ H+H+

80. 79 Dissociation Both theories of acids tell us that acids are compounds that “dissociate” in water to give off H + ions (protons), which immediately attach to water molecules to become H 3 O + ions Eg. HCl added to water breaks up into: H + and Cl - ions, the H + ions join H 2 O as shown by: HCl (aq) + H 2 O (l)  H 3 O + (aq) + Cl - (aq) (monoproteic acid) Different acid dissociate differently: H 2 SO 4(aq) + 2H 2 O (l)  2H 3 O + (aq) + SO 4 - (aq) (diproteic acid) H 3 PO 4 (aq) + 3H 2 O  3H 3 O + (aq) + PO 4 - (aq) (triproteic acid)

81. 80 Dissociation of Water In distilled water, at any given moment about 2 molecules out of every billion (0.00000002%) exist as ions: H + and OH - This means there is an equilibrium reaction occurring. H 2 O  H + + OH - Since so few of the molecules are ionized, the K value of this equilibrium must be tiny! This also means that water can act as either a very weak acid or a very weak base! (it can give off H + or OH - )

82. 81 Equivalence of H + and H 3 O + Some textbooks use H + instead of H 3 O + to represent the hydrogen ion content of acids. Although H 3 O + better represents the actual state of the ions in a water solution, H + is simpler to write, and is used in many texts. [H + ]=[H 3 O + ] In most cases, when water is present

83. 82 K a is called the acidity constant K b is called the basicity constant K w is the ionization constant of water but all three are calculated the same way as K c and are used for similar purposes. We usually use K a or K b in acid-base reactions Tables of K a values help us compare the “natural strength” of various acids. Similarity of K a K b K w and K c

84. 83 The Ionization Constant of Water (K w ) Water is H 2 O, but at any given temperature a tiny fraction (≈ 2 / 10 9 ) of water molecules dissociate into ions: H 2 O (l)  H + (aq) + OH - (aq) K w represents a ratio between the ionized and unionized molecules:

85. 84 Because of the equivalence of H + and H 3 O + in water solutions, you could also write: At room temperature [H 3 O + ] and [OH - ] in pure water are both about 1.0 × 10 -7 mol/L So... The value of K w varies a bit with temperature (see p 328), but has an average value of 1.0 × 10 -14 at 25°C.

86. 85 Effect of Temperature on K w At lower temperatures, K w becomes smaller. Less water molecules dissociate when it is cold At higher temperatures, K w becomes larger. More water molecules dissociate when it is hot. Kw affects the pH of water. We say that the pH of pure water is 7, but in reality it varies by temperature: At the boiling point (100°C), the pH of neutral water will be close to 6! At the freezing point (0°C) it will be about 7.5!

87. 86 The Equilibrium Constant of an Acid The acidity constant of an acid (K a ) represents the fraction of the acid which will dissociate and release H + (ie. H 3 O + ) ions. It is one of two factors that affect what the pH of an acid solution will be. (the other factor is the concentration of the acid) Acids with low K a values are considered naturally “weak” acids. Acids with high K a values are considered naturally “strong” acids.

88. 87 If we consider an acid to have the formula HA then its dissociation can be represented by HA (aq) + H 2 O (l)  H 3 O + (aq) + A - (aq) In this case, we can calculate the K a value this way : NOTE: Because of the equivalence of [H + ] and [H 3 O + ] in some texts, this would be shown as 

89. 88 The Ionization Percentage of an Acid If you ever want to work out the percentage of an acid that is ionized (like if you had too much time on your hands, or were bored or something) you can use this formula:

90. 89 Module 5, Lesson 4 Ph, [H 3 O + ] and K a of Acids Strength, concentration and acidity What are pH and pOH pH (by the chart) pH (by calculation) The I.C.E. method For finding concentrations, K a, K b or K c.

91. 90 Strength, Concentration and Degree of Acidity The strength and concentration of an acid determine its degree of acidity STRENGTH The natural strength of an acid, determined by its K a value Concentration The number of moles per litre of the acidic compound in solution. Degree of Acidity The effective strength of an acid, based on its [H 3 O + ] concentration and related to its pH

92. 91 What are pH and pOH (optional background information) The degree of acidity of a solution is directly related to its [H 3 O + ] concentration. Unfortunately this is often a small number expressed in scientific notation, such as 1.3x10-5 mol/L or 2.3x10-13 mol/L. Not easy numbers to remember, compare or write. In 1909 Søren Sørensen suggested an easier way to record the degree of acidity of solutions. It was a logarithmic scale called pH. Each level of the pH scale represents a 10 fold difference in H 3 O + (or H + ) concentration. ie. A acid of pH 2 has 10 times more [H+] than pH3.

93. 92 pH and pOH pH is a measure of the degree of acidity, given by the following formula: pH = - log [H 3 O + ] Or: (since [H 3 O + ] is equivalent to [H + ] in water solutions) pH = - log [H + ] Although less used, there is also a measure of the degree of alkalinity, called pOH pOH = - log [OH - ]

94. 93 pH and Concentration of H + ions For solutions of an exact pH you may use the chart below:pH [H + ] * [OH - ] * pOHpH [H+] * [OH-] * pOH 0 10 -0 10 -14 148 10 -8 10 -6 6 1 10 -1 10 -13 139 10 -9 10 -5 5 2 10 -2 10 -12 1210 10 -10 10 -4 4 3 10 -3 10 -11 1111 10 -3 3 4 10 -4 10 -10 1012 10 -12 10 -2 2 5 10 -5 10 -9 913 10 -13 10 -1 1 6 10 -6 10 -8 814 10 -14 10 -0 0 7 10 -7 7 AcidsBases *Concentration in mol/L Temperature = 25 ° C For simplicity I use [H + ] instead of [H 3 O + ]

95. 94 What about “in-between” pH values? The formula for pH is: pH = -log [H + ] On your calculator you must find out how to calculate the negative logarithm of a number! The formula for [H+] concentration is: [H + ]=10 –pH or… [H + ]=log -1 (-pH) Sometimes log -1 is called antilog:[H + ] =antilog (-pH) Sometimes log -1 is called inverse log:[H + ] =invlog (-pH) Sometimes log -1 is called 10 x :[H + ] =10 (-pH) On your calculator find out raise 10 to a negative number or how to calculate the inverse logarithm (antilog) of a negative number.

96. 95 (TI 83 instructions) Type: log (1.40 x 10 ^ (-) 8) Enter Question: Find the pH if the H 3 O + concentration is 1.40x10 -8 mol/L (–)(–)log ^ (–)(–)Enter The answer should be: pH=7.85… (I’ve rounded to 3 S.D.) Solution: pH = – log(1.40 x 10 – 8 ) Question: Find the H 3 O + concentration if the pH is 4.30 (TI-83 instructions) 10 ^ (-) 4.3 ^ (–)(–) Solution: [H + ] =10 – 4.30 Enter The answer should be: pH=5.01… x 10 -5 (I’ve rounded to 3 S.D.)

97. 96 Using the Windows Calculator Switch to scientific view To find pH, use: (1.4 Exp 8 ) = To find [H 3 O + ] use: 4.30 = or: 10 4.30 = Finding the pH if H 3 O+ concentration is 1.40x10 -8 mol/L Finding [H 3 O + ] concentration if pH is 4.30 7.85… 5.01…x10 -5

98. 97 Try:(1.4 Exp 8 +/- ) log +/- = Or: (1.4 Exp 8 +/- ) log +/- = Traditional Scientific Calculators Question: Find the pH if the H 3 O + concentration is 1.40x10 -8 mol/L Solution: pH = – log(1.40 x 10 – 8 ) log +/-+/- Exp +/-+/- EE +/-+/- log +/-+/- Question: Find the H + concentration if the pH is 4.30 Try:10 y x 4.30 +/- = Or:4.30 +/- Inv log = Or4.30 +/- 2 nd F 10 x = Or: 10 ^ (-) 4.3 Solution: [H+] = 10 - 4.30

99. 98 A Half-Evil Example Calculate the pH of an aqueous solution of formic acid HCOOH at 0.20 mol/L if its acidity constant is 1.8x10 -4.The equation of this reaction is as follows HCOOH (aq) + H 2 O (l)  H 3 O + (aq) + HCOO - (aq) See the solution on page 333 (I told you, it’s only half-evil)

100. 99 Unit 4: Chapter 12.2.5 Solubility Product Constant Solubility Calculating the solubility constant Examples

101. 100 Solubility A saturated solution contains: Dissolved solute (usually ions) in the solution Some non-dissolved solute (crystal)at the bottom Dissolved ions  Solid solute crystals 

102. 101 In a saturated solution... There is an equilibrium situation: Solute (s)  ion + (aq) + ion - (aq) Some of the solute dissociates (dissolves into ions) and at the same time, some of the ions crystallize back into solid.

103. 102 Solubility The solubility of a substance is the maximum amount of the substance that dissolves in a given volume of the solvent Usually given in g/L or sometimes in g/100mL For calculation purposes you should use molar solubility (mol/L) To convert g/100 mL to g/L, multiply by 10 To convert g/L to mol/L, change grams to moles using the mole formua:  mass (g)  molar mass(g/mol) Moles 

104. 103 The solubility depends on the temperature Standard tables give solubility at 25°C Solids generally have a higher solubility at higher temperatures Gases generally have a lower solubility at higher temperatures. There are a few exceptions to these generalizations!

105. 104 What Is K sp The solubility product constant is a number used to compare the solubilities of different solutes. The higher the K sp, the more of the solute can be dissolved before the solution becomes saturated. K sp is calculated in a similar way to other equilibrium constants, but it’s a bit easier!

106. 105 General Formula for K sp For a solute that dissociates like this: X m Y n(s)  mX + (aq) +nY - (aq) The K sp can be calculated by this way: K sp = [X + ] m [Y - ] n

107. 106 Calculating K sp Example 1: BaSO 4(s)  Ba 2+ (aq) + SO 4 2- (aq) At equilibrium there will be a certain concentration of Ba 2+ ions and SO 4 2- ions K sp = [Ba 2+ ][SO 4 2- ] Example 2: CaCl 2(s)  Ca 2+ (aq) + 2Cl - (aq) At equilibrium there will be a certain concentration of Ca 2+ ions and Cl - ions K sp = [Ca 2+ ][Cl - ] 2 Eliminate solid. 1 : 1 : 1 1 : 1 : 2

108. 107 K sp = [Ba 2+ ][SO 3 2- ] If the solubility of BaSO 3 is 0.0025g/L, what is the K sp of this compound? Since it dissociates equally, the concentration of both ions will be equal to the moles of BaSO 3 that dissolved (0.0025 g/L). We have to convert that to mol/L M BaSO3 = 137.3 + 32.1 + 3(16.0) = 217.4 n BaSO3 = 0.0025 / 217.4 =1.15x10 -5 mol So... [Ba 2+ ] = [SO 3 2- ] = 1.15x10 -5 mol/L K sp = [Ba 2+ ][SO 3 2- ] =(1.15x10 -5 )(1.15x10 -5 ) =1.32x10 -10 1 :1 Dissociates equally (1:1)

109. 108 Example The solubility of silver carbonate (Ag 2 CO 3 ) is 3.6x10 -3 g/100mL at 25C. Calculate the value of the solubility product constant of silver carbonate. Data: Solubility = 3.6×10 -3 g/100mL Molar Solubility =? [ + ions] = ? [ – ions] = ? K sp = ? we want the solubility in g/L, so multiply by 10... 3.6×10 -3 g/100mL = 3.6×10 -2 g/L Now we need it in mol/L, so... n= 3.6×10 -2 g /L 275.8 g/mol n= 1.3×10 -4 mol/L Molar solubility = 1.3x10 - 4 mol/L M = 2(107.8)+12+3(16) Values from periodic table and formula of Ag 2 CO 3. Carry over to the next slide. Use solubility: m = g (per litre)

110. 109 Data: Solubility = 3.6x10 -3 g/100mL Molar Solubility= 1.3x10 -4 mol/L [CO 3 2- ] [Ag + ] K sp =? Equation: Ag 2 CO 3(s)  2 Ag + + CO 3 2- K sp = [Ag + ] 2 [CO 3 2- ] [CO 3 2- ]=1.3×10 - 4 mol/L [Ag + ]= 2.6×10 - 4 mol/L K sp =(2.6×10 - 4 ) 2 (1.3×10 -4 ) K sp =8.8×10 -12 [Ag + ] is double [CO 3 2- ]. Multiply by 2  1 : 2 : 1 = 1.3x10 - 4 mol/L = 2.6x10 - 4 mol/L The molar concentration of CO 3 2- (aq) in solution will equal the molar solubility! Info carried over from previous slide

111. 110 At the annual chemistry Christmas party, a careless chemist spills a whole bottle of calcium fluoride, CaF 2, into a 2 litre punch bowl filled with fruit punch. Most of the calcium fluoride dissolves, but a little powder settles to the bottom of the punch bowl. The K sp of calcium fluoride is 3.4*10 -11 CaF 2 dissociates: CaF 2(s)  Ca 2+ (aq) + 2F - (aq) The lethal dose of fluoride ions is 1g. Will all of the eight chemists at the party die if each drinks a cup of the punch? (1 cup ≈ 250mL or ¼ L.) Would one chemist die if he drank all the punch?

112. 111 K sp = [Ca 2+ ] [F - ] 2 The dissociation formula is: CaF 2(s)  Ca 2+ (aq) + 2F - (aq) So there will be twice as much F - produced as Ca 2+ Let’s choose a variable to represent the concentration of Ca 2+ ions produced at equilibrium. Double that variable to represent the F- ion concentration. Let x = [Ca 2+ ] and 2x = [F - ] Note: The units of x will be mol/L

113. 112 K sp = [Ca 2+ ] [F - ] 2 Simplify x(2x) 2 Reverse the equation and divide by 4 to isolate the x 3 Simplify: Cube root of both sides......Gives us “x”, but we aren’t finished...

114. 113 But x represents [Ca 2+ ], and since the [F - ] is twice as high: So to change this to grams per litre we must multiply by the molar mass of F-, which is 19.0 g/mol (note: this is not F 2 ) A cupful (250mL) is one quarter of this, so...

115. 114 Each cup contains only 0.00194 g of fluoride ions, so nobody would die from drinking one cup. If somebody drank the whole punchbowl (2 litres = 8 cups) he would still only get 0.0155 g of fluoride ions. Still well below the deadly limit. The chemists would survive and have sparkling white teeth! Note: Although this dosage might not be lethal it could have long- term side effects. Never drink contaminated punch! The maximum recommended concentration of fluoride in water is 1 mg/L, about one eighth what is in the punch.

116. 115 Exercises on Solubility Page 340 # 36 to 39