- Chemical equilibrium is reached when the rates of the forward and reverse reactions of a reversible reaction are equal. At equilibrium, the concentrations of reactants and products no longer change over time. - The equilibrium constant, K, is a measure of the extent to which a reaction favors products or reactants at equilibrium. If K is large, products are favored. If K is small, reactants are favored. - Le Chatelier's principle states that if a system at equilibrium is disturbed, the equilibrium will shift in a direction that counteracts the applied change.

1. Chemical Equilibrium (Ch. 14) • After a period of time, some reactions reach equilibrium , meaning that the concentrations of reactants and products no longer change with time. This does not mean that the reaction has stopped! A Equilibrium is a dynamic process B • Equilibrium is reached when the forward rate of reaction of A  B is equal to the rate of the reverse reaction of B  A. • Both the forward and reverse reactions are elementary reactions For the reaction rate fwd = rate rev

2. Rate fwd = k fwd [A] 1 For A  B For B  A Rate rev = k rev [B] 1 - At equilibrium the rates of forward and reverse reactions are equal so, k fwd [A] = k rev [B] - Rearranging the equation, we get: • Rate laws for elementary reactions are predictable: k fwd k rev = [B] [A] = K The Equilibrium Constant (unitless)

4. • Criteria for chemical equilibrium 1. The reaction must be reversible. A reacts to form B, but B can also react to form A 2. The reaction must occur in a closed system. - If the reaction occurred in the open, the carbon dioxide escapes into the environment. The carbon dioxide is no longer present to undergo the reverse reaction. flame CaCO 3 CaO CO 2 Eg. CaCO 3 (s) CaO(s) + CO 2 (g)

5. • Because the “concentration” of solids and liquids remains the same throughout a reaction, they do not appear in the equilibrium equation. - In a closed system, equilibrium will establish CaCO 3 (s) CaO(s) + CO 2 (g) K c = [CO 2 ] flame CaCO 3 CaO CO 2 reacts

6. The Reaction Quotient: Q • It has the exact same formula as the equilibrium constant and is used to determine reaction direction - If Q = K, then the reaction is at equilibrium - If Q < K, the reaction will proceed forward to produce more products to reach equilibrium - If Q > K, the reaction will proceed in reverse to produce more reactants to reach equilibrium aA + bB cC + dD “ reactants” “ products” (as written) Q = [C] c [D] d [A] a [B] b (reactants) (products)

8. In General: 1. When the concentration is changed, the equilibrium will shift to consume some of the added substance or produce some of the removed substance . 2. An increase in pressure (decrease in volume) shifts the reaction to the side with less moles of gas, and an decrease in pressure (increase in volume) shifts the reaction to the side with more moles of gas. 3. For exothermic reactions , an increase in temperature will decrease the K value (shift to the left), and a decrease in temperature will increase the K value (shift to the right). For endothermic reactions , an increase in temperature will increase the K value (shift to the right), and a decrease in temperature will decrease the K value (shift to the left).

9. • When it comes to equilibrium in solutions , the volume of the solution itself defines the “closed system” allowing for aqueous reversible reactions to reach equilibrium. Eg. (red/orange) - It is possible to “remove” soluble ions from solution by precipitating them out. Eg. If a hydroxide source is added, it will ppt with Fe 3+ (aq) 500 ml Fe 3+ (aq) SCN - (aq) FeSCN 2+ (aq) +

10. Solving Equilbrium Problems: ICE Table • It’s nearly the same as the IRF table for irreversible reactions except that the C stands for “concentration and E stands for “equilibrium”. • ICE tables require the use of variables to fill in unknown information. • Moles can still be entered instead of concentration, but will need to be converted to concentration in the end. 2 HI(g) H 2 (g) + I 2 (g) Problem 1: - At equlibrium, [HI] = 0.078 M. Calculate K. - 0.200 mol of HI gas is added to a 2.00 L flask and the reaction is allowed to proceed at 453 o C

11. CO(g) + H 2 O(g) CO 2 (g) + H 2 (g) - 2.00 M of CO and 2.00 M of H 2 O are reacted in a sealed flask at 900 o K. What will be the composition of the reaction mixture at equilibrium? - At this temperature the K is 1.56 • Sometimes the quadratic equation is required to solve equilibrium concentrations - When the equation is rearranged such that: a x 2 + b x + c = 0 Problem 2: - b ± √ b 2 - 4 ac 2 a x = Then

12. • When K is a small number (less than ~1.0 x 10 -4 ), a simplifying assumption can be made COCl 2 (g) CO(g) + Cl 2 (g) K = 8.3 x 10 -4 - Calculate the concentration of the reaction mixture at equilibrium when starting with 5.00 mol of COCl 2 in a 10.0 L flask The assumption is considered valid if the percent error is less than 5% Problem 3:

13. • The properties of acids and bases have been known for a long time, and many where characteristics back in 1661 by English scientist Robert Boyle. Acid-Base Equilibrium (Ch. 15) Changes the color of litmus dye to blue Changes the color of litmus dye to red Become less basic when combined with acid Become less acidic when combined with base Can have sharp pungent odour Can have sharp pungent odour Have a soapy feel/disposition Are corrosive to metals Have bitter taste Have sour taste Bases Acids

14. • Classical, or Arrhenius definition of acids and bases in water: - An acid is a substance that contains H in its formula and dissociates in water to yield H + ions (or H 3 O + , hydronium ion). - A base is a substance that contains OH in its formula and dissociates in water to yield OH - ions . Eg. HCl, HNO 3 , and HCN Eg. NaOH, KOH, and Ba(OH) 2

15. • The modern and more accepted view of an acid and base is now called the Br Ø nsted-Lowry acid-base definition: - An acid is a proton donor , any species that donates an H + ion. An acid must contain an H in its formula in order for there to be a possibility of H + donation. - Arrhenius acids are also Bronsted-Lowry acids Lone pair electrons bind the H + HCl(g) + H 2 O(l)  H 3 O + (aq) + Cl - (aq) Eg. H Cl O H H Cl - O H H H +

16. • A base is a proton acceptor , any species that accepts an H + ion. A base must contain a lone pair of electrons to bind the H + ion, forming a new covalent bond. This is why bases are sometimes also thought of as an electron donor . - This is a very different concept compared to Arrhenius’. For example, Arrhenius would call NaOH a base, but in the B-L definition, it’s the OH - that acts as the base. The compound NaOH is simply the source of the hydroxide ion. N H H H O H H O H + N H H H H - NH 3 (aq) + H 2 O(l) NH 4 + (aq) + OH - (aq) Eg.

17. Conjugate acid-base pairs NH 3 (g) + H 2 O(l) NH 4 + (aq) + OH - (aq) base acid NH 3 (g) + H 2 O(l) NH 4 + (aq) + OH - (aq) - However, acid base - The conjugate base has one fewer H and one more minus charge than the acid. - The conjugate acid has one more H and one fewer minus charge than the base. • Conjugate pairs differ by 1 H atom:

18. The strength of conjugate acid-base pairs is inversely proportional.

19. • pH is a measure of how much H + (H 3 O + ) is in solution pH = - log [ H + ] 1. What is the pH of a solution that is 4.2 x 10 -3 M HCl(aq)? 2. A solution has a pH of 5.5. What is the concentration of H 3 O + in this solution? 3. What is the pOH of a solution of 0.125 M sodium hydroxide? • pOH is a measure of how much OH - is in solution pOH = - log [O H - ]

20. The Extent of Dissociation of Weak Acids and Bases HA(aq) + H 2 O(l) H 3 O + (aq) + A - (aq) • For any generic weak acid HA in water, we have - However, since the concentration of water is a constant, it is factored out of the equation to leave a new constant K a , the acid-dissociation constant • Strong acids dissociate 100% into ions (irreversible arrow), but weak acids do not, so their reaction in water is reversible. The bigger the K a , the stronger the acid [H 3 O + ] [A - ] [HA] [H 2 O] K = General equilibrium [H 3 O + ] [A - ] [HA] K a =

21. • Similarly, for any weak base B in water, we have the reaction - Again, since the concentration of water is a constant, it is factored out of the equation to leave a new constant K b , the base-dissociation constant The bigger the K b , the stronger the base B(aq) + H 2 O(l) BH + (aq) + OH - (aq) [BH + ] [OH - ] [B] K b =

22. Autoionization of Water • Water does not exist purely as the neutral molecule H 2 O. It actually reacts with itself in an Bronsted-Lowry acid- base manner to form ions. To describe this equilibrium, we use the ion-product constant for water, K w K w = [H 3 O + ][OH - ] = 1.0 x 10 -14 (at 25 o C)

23. • Some derivations: - Taking the –log of this equation [H 3 O + ][OH - ] = 1.0 x 10 -14 pH + pOH = 14 -log [H 3 O + ] + -log[ OH - ] = -log( 1.0 x 10 -14) K a x K b = K w - For a conjugate pair And so p K a + p K b = 14 (-log K a ) (-log K b ) The smaller the pK a , the stronger the acid

24. pH of Soluble Ionic Salts • Review: Salts are solid ionic compounds that consist of ions (cations and anions) in a crystal lattice. Most salt solutions have a pH close to 7.0 Most metal cations do not have any acid or base properties (No H + to donate, and cannot accept an H + either) No H so cannot be an acid (proton donor), but it could be a base (proton acceptor using lone pair electrons). However, it is the conjugate base of the strong acid HCl, so Cl - has negligible base properties. Eg. NaCl(s)  Na + (aq) + Cl - (aq) H 2 O

25. • Some salts contain the ion of a conjugate base or conjugate acid. Eg. 1.) Sodium acetate: CH 3 COONa (sometimes NaCH 3 COO) 2.) Ammonium chloride: NH 4 Cl - When these types of salts dissolve in water, the pH of the solution will be affected. 1. a) The K a of acetic acid is 1.8 x 10 -5 . What is the K b for its conjugate base acetate? b) 0.15 moles of sodium acetate is dissolved in a total of 1.0 L of water. What is the pH of the solution? 2. The K b for ammonia is 1.8 x 10 -5 . What would the pH be of a solution of 0.23 M ammonium chloride?

26. Acid-Base Buffers • An acid-base buffer is a solution that resists changes in pH when a small amount of strong acid or base is added. - a solution can act as a buffer it in contains an acid- base conjugate pair , in similar concentrations • Buffer range: Buffers work best when there are equal concentrations of conjugate pairs. When this occurs, the pH = pK a A buffer is effective at a pH within approximately ± 1 of the pKa • Buffer capacity: The more highly concentrated the buffer solution, the more acid or base it can “consume”. Buffer pH is NOT affected by addition of water

27. H 2 O + CO 2 H 2 CO 3 HCO 3 - + H + Carbonic acid Bicarbonate • A Biological Buffer: Controlling pH in blood plasma Controlled through respiration Controlled by the liver Basic substances are consumed by the carbonic acid, and acidic substances are consumed by the bicarbonate.

28. • Ways of Making Buffers 1. Adding both conjugates into solution . 2. Generating the conjugate pair through neutralization (more common method) - The ionic species typically comes from a salt. Eg. Acetic acid (solution) with sodium acetate . Ammonia (solution) with ammonium chloride. - When a weak acid is partially neutralized by a strong base, a certain quantity of conjugate base will form. (Weak bases require partial neutralization by strong acids). Eg. CH 3 COOH(aq) + NaOH(aq)  CH 3 COO - (aq) + Na + (aq) + H 2 O

29. A buffer is made by dissolving 7.38 grams of solid sodium acetate to a total of 1.0 L solution with 0.100 mol of acetic acid. What is the pH of the buffer solution? Formula weight of sodium acetate = 82.03 K a of acetic acid is 1.8 x 10 -5 1. a) b) Calculate the pH if 500.0 ml of water is added to the buffer. c) Calculate the pH if 0.400 g of solid sodium hydroxide is added to the buffer in a) Formula weight of sodium hydroxide = 40.00 d) Calculate the pH if 0.400 g of solid sodium hydroxide is added to 1.0 L of pure water.

30. 2. You want to make 1.00 L of a 0.100 M solution buffered at pH 9.0 using an ammonia/ammonium buffer system. K b of ammonia = 1.74 x 10 -5 a) How many moles of each conjugate species would you need? b) The source of ammonia is a 1.00 M solution, and the source of ammonium is from solid ammonium nitrate. (FW = 80.04). How much would you add of each to achieve your buffer? 3. You want to make 1.00 L of a 0.100 M solution buffered at pH 9.0 using using an ammonia/ammonium buffer system. You have only solid ammonium nitrate and 6.00 M NaOH solution. How much of each must be used?

31. Titration Curves • During an acid-base titration, the pH of the solution changes as acid or base is continuously added. • The plot of the pH versus the quantity of acid or base added, generates a titration curve . Important points on the titration curve: 1.) Equivalence point: The mid point of large inflections in the curve. (“S” portion of the curve) 2.) Half equivalence point for weak acids and bases : This is when pH = pKa

32. Automated pH Titration Data Acquisition Example: Strong acid HCl titrated with NaOH

34. HAc = acetic acid Ac - = acetate ion

36. Polyprotic Weak Acids • These acids have multiple ionizable protons (acidic hydrogens) Eg. H 3 PO 4 – Phosphoric acid has 3 acidic hydrogens H 2 CO 3 – Carbonic acid has 2 acidic hydrogens H 2 SO 4 – Sulfuric acid has 2 acidic hydrogens • Interestingly, for weak polyprotic acids, the protons ionize one at a time . The first proton has the greatest ionizability, followed by the next proton, etc. H 2 SO 3 HSO 3 - SO 3 2- K a1 K a2 Where K a1 >> K a2 Nearly all polyprotic acids are weak acids except for sulfuric H + H +

38. Molecular Bases are N Containing Compounds

39. Equilibria of Slightly Soluble Ionic Compounds • On solubility table, these are compounds marked as “ss” Eg. CaSO 4 (s) Ca 2+ (aq) + SO 4 2- (aq) • In solution, small quantities of solid are dissolving into ions. Once a certain ion concentration is reached, the ions react to reform the solid in an equilibrium General K [ Ca 2+ ] [ SO 4 2- ] [ CaSO 4 ] = K sp = [ Ca 2+ ] [ SO 4 2- ] Where K sp stands for the solubility product constant

40. • Shifting solubility is a property of Le Chatelier’s Principle - This will increase the concentration of “products” in solution and shift the equilibrium to the left. - The result is that the salt is even less soluble in a solution containing a common ion. 2.) pH differences - excess H 3 O + (low pH) can often react with one of the ions of the salt, removing it from solution. - This will reduce the concentration of “products” in solution, shifting equilibrium to the right which increases the solubility of the salt. - Excess OH - (high pH) often acts as a common ion 1.) A common ion in solution.