The document discusses chemical equilibrium and reversible reactions. It explains that in a reversible reaction, the reactants can transform into products, and the products can then react in reverse to reform the original reactants. At equilibrium, the rates of the forward and reverse reactions are equal and the concentrations of reactants and products remain constant. Mercury(II) oxide decomposing into mercury and oxygen, and then recombining, is provided as an example of a reversible reaction system at equilibrium. Factors that affect the rate and position of equilibrium, such as temperature, concentration, catalysts and pressure, are also described.

1. Chemical Equilibrium

2. Reversible Reactions In a chemical reaction, Reactants are transformed into products . The products formed can react to re-form the original reactants. A B + C D A D C B + A B + C D A D + B C Reactions that can be traversed in both directions are said to reversible . Theoretically at least, all chemical reactions are reversible .

3. Reversible Reactions HgO HgO HgO HgO HgO HgO HgO Hg O 2 Hg Hg HgO Hg HgO O 2 HgO 2 HgO (s) 2 Hg (l) + O 2 (g) 2 Hg (l) + O 2 (g) 2HgO (g)

4. Reversible Reactions 2 HgO (s) 2 Hg (l) + O 2 (g) 1 2 Upon heating, mercury (II) oxide decomposes to mercury (Hg) and oxygen (O 2 ) [ Equation 1]: Under the same conditions, mercury (Hg) and oxygen (O 2 ) recombine form mercury (II) oxide again [Equation 2]: Mercury and oxygen combine to form mercury oxide just as fast as mercury oxide decomposes into mercury and oxygen 2 HgO (s) 2 Hg (l) + O 2 (g) 2 Hg (l) + O 2 (g) 2HgO (s)

5. Reversible Reactions 2 HgO (s) 2 Hg (l) + O 2 (g) Both reactions continue to occur, but there is no net change in the composition of the system. The amounts of mercury (II) oxide, mercury (Hg), and oxygen (O 2 ) remain constant as long as the reaction conditions remain constant There is a state of equilibrium between the two chemical reactions. Chemical equilibrium is a dynamic state of balance in which the rates of opposing reactions are exactly equal .

6. Reversible Reactions Examine the two sets of plots carefully, noting which substances have zero initial concentrations, and are thus " reactants " in each case. Satisfy yourself that these two sets represent the same chemical reaction system , but with the reactions occurring in opposite directions . Most importantly, note how the concentrations of all the components are identical when the system reaches equilibrium . 2 HI H 2 + I 2 Dissociation of Hydrogen Iodide H 2 + I 2 2 HI Synthesis of Hydrogen Iodide

7. Reversible Reactions The equilibrium state is independent of the direction from which it is approached. Whether we start with an equimolar mixture of H 2 and I 2 (left) or a pure sample of hydrogen iodide (shown on the right, using twice the initial concentration of HI to keep the number of atoms the same), the composition after equilibrium is attained (shaded regions on the right) will be the same . H 2 + I 2 2HI For more information, Click Here

8. Factors affecting the rate of chemical reactions The rate of a chemical reaction is the time needed for a chemical reaction to be complete . Factors that affects the rate of the chemical reactions are: 1. The nature of the reactants Double covalent bond (weak) More reactive Single covalent bond (strong) Less reactive The stronger the bond between the elements of a certain molecule, the harder it is to break in a chemical reaction, and the slower the reaction.

9. Factors affecting the rate of chemical reactions 2. The Temperature A mixture of iron and sulfur doesn’t react unless strongly heated. An elevation in the temperature makes the reaction goes faster by increasing the frequency of collisions between reacting particles. In general all reactions, especially endothermic (because they absorb energy) ones occur much more quickly when heated . This is due to the fact that heat gives enough energy to break or form bonds between different atoms. Heat

10. Factors affecting the rate of chemical reactions 3. The presence of a catalyst A catalyst speeds the rate of any reaction without affecting its products. Hydrogen peroxide is an antiseptic that decomposes spontaneously into water and oxygen: 2H 2 O 2 (l) 2H 2 O (l) + O 2 (g) Because the reaction is slow , H 2 O 2 can be conserved for many months. But when a platinum wire is immersed in H 2 O 2 solution, oxygen gas release is observed on the platinum surface.

11. Factors affecting the rate of chemical reactions 4. The surface area As the surface area of the reaction mixture decreases , the rate of the reaction increases . Interpretation: Who can give an interpretation??? As the surface area decreases , the reacting molecules become closer the frequency of collision between molecules increases New molecules are formed

12. Factors affecting the rate of chemical reactions 5. Change in concentration Increasing the concentration of a reactant shifts an equilibrium to the products (or right hand) side because the rate of the forwards reaction is increased. b) Increasing the concentration of a product shifts an equilibrium to the reactant (or left hand) side because the rate of the reverse reaction is speeded up. c) Decreasing the concentration of a reactant (by removal or by compounding it with something else or by precipitation) shifts an equilibrium to the reactants (or left hand) side because the forwards reaction is slowed down. The reverse reaction will 'overtake' the forwards reaction. d) Decreasing the concentration of a product shifts an equilibrium to the products (or right hand) side because the reverse reaction is slowed and the forwards reaction 'overtakes'.

13. The equilibrium constant The adjacent graph shows the changes in the reaction rates of the forward and backward reactions: A + B C + D Initially (t = 0), [A ] and [B] were maximum , while [C] and [D] were zero . The rate of the forward reaction decreases as A and B are used up. The rate of the reverse reaction increase s as C and D are formed. Equilibrium is attained when the two rates become equal [A], [B], [C], and [D] remain constant at equilibrium.

14. The equilibrium constant At equilibrium , the ratio of the product [C] x [D] to the product [A] x [B] has a definite value at a given temperature . It is known as the equilibrium constant of the reaction and is designated by the letter K . Thus, [C] x [D] K = ------------- [A] x [B] K is independent of the initial concentrations. K is dependent on the fixed temperature of the system.

15. The equilibrium constant K shows the extent to which the reactants are converted to the products of the reaction . If K = 1 , the products of the concentrations of the products and the reactants have the same value. If the value of K is very small , the forward reaction occurs only very slightly before equilibrium is established, and the reactants are favored. A very large value of K indicates an equilibrium in which the original reactants are largely converted to products. [C] x [D] K = ------------- [A] x [B] The numerical value of K for a particular equilibrium system is obtained experimentally.

16. The equilibrium constant Consider the following general balanced equation: a A + b B c C + d D [C] c x [D] d K = ------------------ [A] a x [B] b The equilibrium constant K is the ratio of the product of the concentration of the substances formed at equilibrium to the product of the concentrations of the reacting substances, each concentration being raised to the power that is the coefficient of that substance in the chemical equation.

17. The equilibrium constant Example : Give the expression of the equilibrium constant N 2 (g) + 3 H 2 (g) 2 NH 3 (g) Given [N 2 ]=0.1M,[H 2 ]=0.125M, [NH 3 ]=0.11M The equilibrium constant is given by the expression: [NH 3 ] 2 K = ------------------ [N 2 ] [H 2 ] 3

18. The equilibrium constant Important Notes Pure solids don’t appear in the K’s expression. Pure liquids don’t appear in the K’s expression. Water , as a liquid or a reactant, doesn’t appear in the expression. For example: 2 HgO (s) 2 Hg (l) + O 2 (g) K = [O 2 ]

19. The equilibrium constant Exercise 1: An equilibrium mixture of H 2 , I 2 , and HI gases at 425 ºC is determined to consist of 4.5647 x 10 -3 mole/liter of H 2 , 0.7378 x 10 -3 mole/liter of I 2 , and also 13.544 x 10 -3 mole/liter of HI. What is the equilibrium constant for the system at this temperature given that: H 2 (g) + I 2 (g) 2 HI (g) [HI] 2 [13.544 x 10 -3 ] -2 K = ------------ = ---------------------------------------- = 54.47 [H 2 ] [I 2 ] [4.5647 x 10 -3 ] [0.7378 x 10 -3 ]

20. The equilibrium constant Exercise 2: Find the relationship between K 1 and K 2 , the equilibrium constants of these two reactions: 2A + 2B 2C K 1 C A + B K 2 [C] 2 [A] [B] K 1 = ------------- ; K 2 = --------------- [A] 2 [B] 2 [C] By comparing K 1 and K 2 : K 1 = 1 /K 2 2

21. The equilibrium constant Exercise 3: The following reaction takes place at 460ºC, where the equilibrium constant K has a value of 85. SO 2 (g) + NO 2 (g) NO (g) + SO 3 (g) At a certain moment, the concentrations of the reactant and products were measured to be: [SO 2 ] = 0.04, [NO 2 ] = 0.5M, [NO] = 0.3M, [SO 3 ] = 0.02M Is this system at equilibrium? If not, in which direction must the reaction go to reach equilibrium?

22. The equilibrium constant Solution 3: [NO] [SO 3 ] 0.3 x 0.02 K = ------------------ = -------------- = 0.3 [SO 2 ] [NO 2 ] 0.04 x 0.5 K = 0.3 K equilibrium = 85 } K < K equilibrium The reaction is not at equilibrium ** K = 0.3 < 1 this means that the reactants NO 2 and SO 2 are favored. In order for the system to reach equilibrium, it should move forward , towards the products side .

23. Factors that disturb equilibrium What are the factors that affect the rate of the reaction? Any change that alters the rate of either reactions disturbs the original equilibrium . If the original state of equilibrium is disturbed, the system seeks a new equilibrium state . Equilibrium is shifted in the direction that releases stress from the system.

24. Factors that disturb equilibrium Le Chatelier's principle provides a means of predicting the influence of disturbing factors on equilibrium systems. Le Chatalier’s principle states: If a system at equilibrium is subjected to a stress, the equilibrium is shifted in the direction that relieves the stress . If you are stressed, what do you do ? Of course, you will go to a place where you can relax and relieve the stress. The same concept is applied on the equilibrium of a chemical reaction?

25. Factors that disturb equilibrium 1. Effect of temperature Changes in the temperature of the system affect the position of the equilibrium by changing the magnitude of the equilibrium constant for the reaction. Increasing the temperature of a reaction that gives off heat is the same as adding more of one of the products of the reaction. It places a stress on the reaction, which must be alleviated by converting some of the products back to reactants . If the temperature of the system in equilibrium is lowered , the reaction will move in a direction to produce more heat , i.e. the exothermic reaction is favored . In applying Le chatelier’s principle to chemical equilibrium, three stresses will be considered:

26. Factors that disturb equilibrium 2. Effect of pressure This applies to reactions involving gases . If the pressure is increased , the reaction will move to reduce the pressure by reducing the number of particles present . A reaction at equilibrium was subjected to a stress results in an increase in the total pressure on the system. The reaction then shifted in the direction that minimized the effect of this stress . The reaction shifted toward the products because this reduces the number of particles in the gas, thereby decreasing the total pressure on the system.

27. Factors that disturb equilibrium 3. Effect of concentration If the concentration of one substance is increased , the reaction will move in a direction to use up the substance whose concentration was increased. If one substance is removed from the system, the reaction will move in a direction to produce more of the substance being removed .

28. Factors that disturb equilibrium 4. Effect of catalyst Can you predict the effect of Catalyst on the position of equilibrium? Both the forward and backward reactions are speeded up in the same amount; therefore, there is no effect on the position of equilibrium or on the concentrations of the reacting substances.

29. Factors that disturb equilibrium Exercise 4 : Based on the following system at equilibrium: N 2 (g) + 3 H 2 (g) 2NH 3 (g) + heat How is equilibrium restored in following system in each of the following cases? a) A decrease in the concentration of N 2 b) An increase in temperature c) An increase in the total pressure of the system

30. Factors that disturb equilibrium Exercise 5 : Given the following reaction: 2 IBr (g) I 2 (g) + Br 2 (g) If 0.06 moles of IBr are placed in a 0.5 liter container, and the equilibrium constant K is 8.5x10 -3 , find the concentrations of IBr, I 2 , and Br 2 at equilibrium. Initially (t = 0): # of moles of IBr = 0.06 moles # of moles of I 2 = 0 # of moles of Br 2 = 0

31. At equilibrium (t equi ): # of moles of IBr = 0.06 – 2n # of moles of I 2 = n # of moles of Br 2 = n 2 IBr (g) I 2 (g) + Br 2 (g) At t = 0: 0.06 moles 0 mole 0 mole At t eq. : (0.06 – 2n) mole n mole n mole Note: The coefficient of n is always the same as the coefficient of the substance .

32. [I 2 ] [Br 2 ] K = ---------------- [IB r ] # of moles of I 2 n [I 2 ] =----------------------- = ----------- Volume 0.5 # of moles of Br 2 n [Br 2 ] = ----------------------- = ----------- Volume 0.5 # of moles of IBr (0.06 -2n) [IBr] = ----------------------- = ------------- Volume 0.5

33. [I 2 ] [Br 2 ] [n / 0.5] [n / 0.5] K = ---------------- 8.5 x 10 -3 = ------------------------- [IB r ] [(0.06 – 2n) / 0.5] 2 n 2 8.5 x 10 -3 = ---------------- (0.06 – 2n) 2 n = 4.67 x 10-3 moles [I 2 ] = [Br 2 ] = n / 0.5 = (4.67 x 10 -3 ) / 0.5 = 9.34 x 10-3 M [IBr] = (0.06 – 2n)/0.5 = [0.06 – 2(4.67 x 10 -3 )] / 0.5 = 0.101 M At equilibrium:

34. Reactions that run to completion A reaction may be driven in the preferred direction by applying Le Chatelier principle . A reaction reaches a state of equilibrium unless one of the products escapes or is removed . Some reactions appear to go to completion in the forward direction: Burning a paper ( complete reaction). Decomposition of potassium chlorate to oxygen and potassium chloride.

35. Reactions that run to completion 1. Formation of a gas The reaction between sodium hydrogen carbonate (Baking soda) and hydrochloric acid releases carbon dioxide gas as illustrated in the given figures:

36. Reactions that run to completion Illustration NaHCO 3 + HCl NaCl + H 2 CO 3 The ionic reaction Na + + HCO 3 - + H 3 O + + Cl - Na + + Cl - + H 2 CO 3 The net ionic equation: HCO 3 - + H 3 O + H 2 O + H 2 CO 3 CO 2 (g) + H 2 O The net ionic equation: HCO 3 - + H 3 O + 2H 2 O + CO 2 (g) Weak acid Carbonic acid

37. Reactions that run to completion 2. Formation of precipitate When solutions of sodium chloride and silver nitrate are mixed, a white precipitate of silver chloride immediately forms. The reaction effectively runs to completion because an “insoluble” product is formed.

38. Reactions that run to completion 2. Formation of precipitate (illustration) AgNO 3 + NaCl AgCl (s) + NaNO 3 The ionic equation Ag + + NO 3 - + Na + + Cl - AgCl (s) + Na + + NO 3 - The net ionic equation Ag + + Cl - AgCl (s) White precipitate

39. Reactions that run to completion 3. Formation of a slightly ionized product Water is a typical compound that ionizes slightly into H 3 O + and OH - . Water can be formed as a product in the neutralization reaction . The reaction effectively runs to completion because the product (H 2 O) is only slightly ionized .

40. Reactions that run to completion 3. Formation of a slightly ionized product (illustartion) NaOH + HCl NaCl + H 2 O The ionic equation Na + + OH - + H 3 O + + Cl - Na + + Cl - + 2H 2 O (l) The net ionic equation H 3 O + + OH - 2 H 2 O (l)

41. The common ion effect is an application of Le Chatelier's Principle. The Common Ion Effect If we mix a soluble salt containing an ion common to a slightly soluble salt, we will affect the position of the equilibrium of the slightly soluble salt system. Adding the common ion to the salt solution by mixing the soluble salt will add to the concentration of the common ion. According to Le Chatelier's Principle , that will place a stress upon the slightly soluble salt equilibria (added concentration). The equilibrium will respond so as to undo the stress of added common ion.

42. The Common Ion Effect Example Na + Cl - (aq) Na + (aq) + Cl - (aq) Bubble hydrogen chloride gas (HCl) in a saturated solution of sodium chloride (NaCl) . As sodium chloride dissolves, NaCl separates as a precipitate . Interpretation The concentration of the common ion ( Cl - ) increases on the right side of the reaction, while that of sodium ions decreases. The equilibria will shift so that the common ion will be reduced which means a shift to the left, thus REDUCING the solubility of the slightly soluble salt system (NaCl).

43. The Common Ion Effect Example A 0.1 M acetic acid solution (CH 3 COOH) has a pH of 2.9. When sodium acetate (CH 3 COO - Na + ) is dissolved in the given solution, the pH increases to 4.9. Interpretation CH 3 COOH + H 2 O CH 3 COO - + H 3 O + When sodium acetate is dissolved in the acetic acid solution, the concentration of the acetate ion (CH 3 COO - ) on the right side of the equation will increase . The equilibrium will shift to the right (backward) so as to decrease the concentration of the added ion More CH 3 COOH is formed the concentration of H 3 O + in solution decreases pH increases . Common ion

44. Equilibrium Constant of Weak Acids Weak acids ionize to a slight extend , producing small number of the acidic H 3 O + ions, according to the following general reaction: HA + H 2 O H 3 O + + A - The extend to which a weak acid ionizes into ions is referred to as ionization percentage ( α ). For example, acetic acid (CH 3 COOH) has an ionization percentage of 1.4%. This means that if there are 100 moles of acetic acid , then only 1.4 moles will dissociate into the corresponding ions: CH 3 COOH + H 2 O CH 3 COO - + H 3 O + 100 moles 1.4 moles 1.4 moles

45. Equilibrium Constant of Weak Acids HA + H 2 O H 3 O + + A - t = 0 c 0 0 teq c – x x x The portion of HA that dissociates into the ions depends on the acid ionization percentage and on its concentration: x = c α HA + H 2 O H 3 O + + A - t = 0 c 0 0 t eq c – c α c α c α c is the initial concentration

46. [A - ][H 3 O + ] K= ---------------- [HA][H 2 O] [A - ][H 3 O + ] K [H 2 O] = ---------------- [HA] K and [H 2 O] are constants Ka = K [H 2 O], where Ka is the acid-ionization constant . [A - ][H 3 O + ] Ka = ---------------- [HA] The weaker the acid is, the smaller the value of Ka and α, due to the fewer number of ionized species in the numerator. Equilibrium Constant of Weak Acids

47. Equilibrium Constant of Weak Acids [A - ][H 3 O + ] (c α ) (c α ) Ka = ---------------- = ---------------- [HA] c – c α c 2 α 2 c α 2 Ka = -------------- = --------- c (1- α ) 1- α Since α is very small for weak acids α <<< 1 1 - α ~ 1 Ka = c α 2 / 1 Ka = c α 2

48. Equilibrium Constant of Weak Acids Another relation [A - ][H 3 O + ] Ka = ---------------- [HA] But [A - ] = [H 3 O + ] ; [H 3 O + ] 2 Ka = ---------------- [HA] [H 3 O + ] 2 Ka = ---------------- c [H 3 O + ] 2 = Ka . c [H 3 O + ] = √ Ka . c

49. Equilibrium Constant of Weak Acids In summary [H 3 O + ] = √ Ka . c Ka = c α 2 [A - ][H 3 O + ] Ka = ---------------- [HA]

50. Equilibrium Constant of Weak Acids Exercise 6 : a) Find the acid-ionization constant of a solution of 0.1M acetic acid that has an ionization percentage of 1.4%. b) Find the pH of the solution. a) c = 0.1 M ; α = 1.4 x 10 -2 ; Ka = ???? Ka = c α 2 = (0.1) (1.4 x 10 -2 ) 2 Ka = 1.96 x 10 -5 b) pH = -log [H 3 O + ] [H 3 O + ] = √ Ka . c = √ (1.96 x 10-5) (0.1) [H 3 O + ] = 2.744 x 10 -3 M pH = -log [H 3 O + ] = - log (2.744 x 10 -3 ) pH = 2.56

51. Ionization constant of water The equation of the self-ionization of water: H 2 O (l) + H 2 O (l) H 3 O + (aq) + OH - (aq) The equilibrium constant: [H 3 O + ] [OH - ] K = -------------------- [H 2 O] 2 But [H 2 O] is constant K [H 2 O] 2 = [H 3 O + ][OH - ] Kw = [H 3 O + ] [OH - ] = 1 x 10 -14 The equilibrium constant for water is nothing but the ionization constant of water, Kw .

52. Equilibrium Constant of Weak Bases B + H 2 O BH + + OH - t = 0 c 0 0 teq c – x x x The portion of B that dissociates into the ions depends on the base ionization percentage and on its concentration: x = c α B + H 2 O BH + + OH - t = 0 c 0 0 t eq c – c α c α c α c is the initial concentration

53. [OH - ][BH + ] K= ---------------- [B][H 2 O] [OH - ][BH + ] K [H 2 O] = ---------------- [B] K and [H 2 O] are constants K b = K [H 2 O], where K b is the base-ionization constant . [OH - ][BH + ] K b = ---------------- [B] The weaker the base is, the smaller the value of K b and α, due to the fewer number of ionized species in the numerator. Equilibrium Constant of Weak bases

54. Equilibrium Constant of Weak bases Using the same approach as with weak acids, we can derive a similar set of formulas: K b = c α 2 [H 3 O + ] = √ K b . c

55. Equilibrium Constant of Weak bases Exercise 7: Prove that for any acid-conjugate base pair, Ka of the weak acid and Kb of its conjugate base are related through the following formula: Ka. Kb = Kw Consider any weak acid, HA, in water: HA + H 2 O H 3 O + + A - [A - ][H 3 O + ] Ka = ---------------- [HA]

56. Equilibrium Constant of Weak bases Its conjugate base, A - , would undergo the following reaction: A - + H 2 O HA + OH - [HA] [OH - ] Kb = ---------------- [A - ] Ka x Kb = [A - ][H 3 O + ] ---------------- x [HA] [HA] [OH-] ---------------- [A-] Ka x Kb = [H 3 O + ] [OH - ] = Kw This formula is always true, and can be directly applied.

57. Equilibrium Constant of Weak bases Exercise 8: a) Find the ionization percentage of a weak base, B, of concentration 3M, if the base-ionization constant of 7.8 x 10 -4 ? b) Find the pH of the above solution. a) Kb = c . α 2 Kb 7.8 x 10 -4 α 2 = ------------- = --------------- c 3 α = 1.6 %

58. Equilibrium Constant of Weak bases b) pH = ??? pH = -log [H 3 O + ] [OH - ] = √ Kb . c = √ (7.8 x 10-4) (3) [OH - ] = 0.048 M [H 3 O + ][OH - ] = 10 -14 [H 3 O + ] [0.048] = 10 -14 [H 3 O + ] = 2.06 x 10 -13 M pH = -log [H 3 O + ] = - log (2.06 x 10 -13 ) pH = 12.68

59. Buffers Buffers are special type of solutions made up of a weak acid and the salt of its conjugate base , or of a weak base , mixed with the salt of its conjugate acid . Examples of Buffers may include: 1. Acetic acid (HC 2 H 3 O 2 ) mixed with sodium acetate (NaC 2 H 3 O 2 ) Weak acid Conjugate base 2. Ammonia (NH 3 ) mixed with ammonium chloride (NH 4 Cl) Weak base Conjugate base

60. Buffers Buffers resist changes in pH when an acid or a base is added in small amounts . Suppose a diluted acid is added in small amounts to a buffer made of HC 2 H 3 O 2 and NaC 2 H 3 O 2 . HC 2 H 3 O 2 + H 2 O H 3 O + + C 2 H 3 O 2 - When the diluted acid is added to the buffer, the concentration of [H 3 O + ] on the left side of the equation increases According to Le Chatelier principle, the equilibrium will shift to the right , thus reducing the concentration of H 3 O + ions The pH will remain at its initial level unchanged.

61. HC 2 H 3 O 2 + H 2 O H 3 O + + C 2 H 3 O 2 - Now, when a base is added to the solution, it will react with H 3 O + ions causing a decrease in their concentration According to Le Chatelier principle, the equilibrium will shift to the right to increase the concentration of H 3 O + ions pH will return to its original value . Buffers

62. Solubility equilibrium Solubility is defined as the amount of salt (in grams) that can be dissolved in 100 g of water . In general, salts are classified into 3 broad categories: 1. Soluble , when more than 1g of the salt can dissolve in a 100 g of water. 2. Insoluble , when less than 0.1g of the salt can dissolve in a 100 g of water . 3. Slightly soluble when the mass of salt dissolved in a 100 g of water falls between 0.1g and 1g .

63. Solubility equilibrium A saturated solution is defined as a solution that contains the maximum amount of salt dissolved in water. Saturated solutions exhibit the behavior of equilibrium system , since some of the salt is dissolved in water, while the rest is precipitated in the bottom of the beaker.

64. Solubility equilibrium Consider the case of silver chloride (AgCl) having a solubility of 8.9 x 10 -1 g/100 g of water AgCl is considered insoluble in water AgCl (s) Ag + + Cl - The equilibrium constant: [Ag +] [Cl - ] K = ---------------- [AgCl] K [AgCl] = [Ag + ] [Cl - ] K and [AgCl] are considered to be constant (since AgCl is a solid, so its concentration does not affect the equilibrium): Ksp = [Ag + ] [Cl - ] Ksp = solubility-product constant

65. Solubility equilibrium The solubility-product constant is then the product of the molarities of the ions in a saturated solution, each ion being raised to the power of its coefficient. Ksp = [Ag + ] [Cl - ] Exercise 9 : Find the expression of the Ksp of calcium fluoride, CaF 2 CaF 2 Ca 2+ + 2F - Ksp = [Ca 2+ ] [F - ] 2

66. Solubility equilibrium The lower the Ksp, the less soluble the salt is. Least soluble Most soluble

67. AgCl (s) Ag + + Cl - t = 0 M 0 0 teq M – c c c M = initial concentration Ksp = [Ag + ] [Cl - ] = c . c = c 2 c is the molarity in mole/l , while the solubility is in given in mass of AgCl / 100 g of water. To find the concentration, apply this relation: mass of salt C = ---------------------------------------------- (Molecular weight of the salt) x 0.1L Solubility equilibrium

68. Solubility equilibrium Exercise 10 : Find the Ksp of CaF 2 if its solubility is 1.7 x 10 -3 g/100 g of water. Ca = 40 ; F = 19 CaF 2 Ca 2+ + 2F - t = 0 M 0 0 teq M – c c 2c mass of CaF 2 c = ---------------------------------------------- (Molecular weight of CaF 2 ) x 0.1L Molecular weight of CaF 2 = 40 + 2(19) = 78 g/mole ;

69. Solubility equilibrium 1.7 x 10 -3 c = --------------- 78 x 0.1 c = [Ca 2+ ] = 2.18 x 10 -4 mole/l [F - ] = 2 c = 2 x (2.18 x 10 -4 ) = 4.36 x 10 -4 mole/l Ksp = [Ca 2+ ] [F - ] 2 = (2.18 x 1o -4 ) (4.36 x 10 -4 ) 2 Ksp = 4.14 x 10 -11

70. Solubility equilibrium Exercise 11 : Find the solubility of cadmium sulfide, CdS, in g/100g of water, if its Ksp is 8 x 10 -27 . Cd = 112 ; S = 32 CdS Cd 2+ + S 2- t = 0 M 0 0 teq M – c c c Ksp = [Cd 2+ ][S 2- ] = c . c = c 2 c = √Ksp = √8 x 10 -27 c = 8.944 x 10 -14 M

71. mass of CdS c = ---------------------------------------------- (Molecular weight of CdS) x 0.1L Molecular weight of CdS = 112 + 32 = 144 g / mole mass of CdS 8.944 x 10 -14 = ------------------- 144 x 0.1 mass of CdS = 1.287 x 10 -12 g / 100 g of water Solubility equilibrium

72. Solubility equilibrium Exercise 12 : Find the solubility of CdS in mole/l, given its Ksp to be 8 x 10 -27 Ksp = [Cd 2+ ][S 2- ] = c . c = c 2 CdS Cd 2+ + S 2- t = 0 M 0 0 teq M – c c c c = √Ksp = √8 x 10 -27 c = 8.944 x 10 -14 M

73. Solubility equilibrium Precipitation calculation A precipitate is the formation of an insoluble salt in solution. The precipitate may form when mixing two soluble salts. XY X + + Y - For the precipitate XY to form, [X + ][Y - ] > Ksp If [X + ][Y - ] < Ksp no precipitate XY will form .

74. Solubility equilibrium Exercise 13: Will a precipitate form when 20 ml of 0.01M BaCl 2 is mixed with 20 ml of 0.005M Na 2 SO 4 ? Ksp of BaSO 4 = 1.1 x 10 -10 ? BaCl 2 (aq) + Na 2 SO 4 (aq) BaSO 4 (s) + 2 NaCl (aq) The equation of the reaction between the two salts: The dissolution reaction of the precipitate formed is: BaSO 4 (s) Ba 2+ + SO 4 2-

75. Solubility equilibrium [Ba 2+ ] = 5 x 10 -3 M # moles of BaCl 2 = 2 x 10 -4 moles # moles of BaCl 2 = # moles of Ba 2+ = 2 x 10 -4 moles [Ba 2+ ] = (# moles of Ba 2+ ) / (total volume) = (2 x 10 -4 ) / (40 x 10 -3 L) # moles of BaCl 2 [BaCl 2 ] = ----------------------- volume of BaCl 2 # moles of BaCl 2 0.01 = ------------------------- 20 x 10 -3 L

76. Solubility equilibrium # moles of Na 2 SO 4 = # moles of SO 4 2- = 1 x 10 -4 moles [SO 4 2- ] = (# moles of SO 4 2- ) / (total volume) = (1 x 10 -4 ) / (40 x 10 -3 L) [SO 4 2- ] = 2.5 x 10-3 M # moles of Na 2 SO 4 [Na 2 SO 4 ] = ------------------------- volume of Na 2 SO 4 # moles of Na 2 SO 4 0.005 = ------------------------- 20 x 10 -3 L # moles of Na 2 SO 4 = 1 x 10 -4 moles

77. Solubility equilibrium [Ba 2+ ] [SO 4 2- ] = (5 x 10 -3 ) (2.5 x 10 -3 ) = 1.25 x 10 -5 [Ba 2+ ] [SO 4 2- ] = 1.25 x 10 -5 Ksp = 1.1 x 10 -10 } [Ba 2+ ] [SO 4 2- ] > Ksp A precipitate of BaSO 4 will form in this solution