1 General Concepts of Chemical Equilibrium

2 In this chapter you will be introduced to basic equilibrium concepts and related calculations. The type of calculations you will learn to perform in this chapter will be essential for solving equilibrium problems in later chapters. The ease of solving equilibrium problems will depend on the skills you will master in this chapter.

3 The Equilibrium Constant For chemical reactions that do not proceed to completion, an equilibrium constant can be written as the quotient of multiplication of molar concentrations of products divided by that of reactants, each raised to a power equal to its number of moles. aA + bB  cC + dD Where, the small letters represent the number of moles of substances A, B, C, and D. The equilibrium constant is written as: K eq = ([C] c [D] d )/([A] a [B] b )

4 In a chemical reaction, as reactants start to react products start to form. Therefore, reactants continuously decrease and products continuously increase till a point is reached where eventually no change in concentrations can be detected. This is the point of equilibrium which is a point where the rate of the forward reaction (product formation) equals the rate of backward reaction (reactants formation). In fact equilibrium implies continuous transformation between infinitesimally small amounts of products and reactants.

5 Progress of a chemical reaction The rate of the forward reaction diminishes with time, while that of the backward reaction increases, until they are equal. A large K means the reaction lies far to the right at equilibrium.

6

7 Calculation of Equilibrium Constants We know from thermodynamics that a reaction occurs spontaneously if it has a negative  G (Gibbs free energy) where:  G =  H - T  S Where,  H is the change in enthalpy of the reaction and  S is the change in entropy. At standard conditions of temperature and pressure (standard state) we have the standard free energy  G o where:  G o =  H o - T  S o

8 The standard free energy is related to equilibrium constant by the relation:  G o = - RT ln K K = e -  Go/RT R is the gas constant (8.314 deg K -1 mol -1 ) It should be clear that  G o gives us good information about the spontaneity of the reaction but it offers no clue on the rate at which the reaction may occur.

Lecture 14 Equilibrium Reactions, cont… Equilibrium Calculations

10 Le Chatelier's Principle The equilibrium concentrations of reactants and products can be changed by applying an external stress to the system, e.g. increasing or decreasing the concentration of a reactant or product, changing temperature or pressure. The change occurs in a direction which tends to counterpart the applied stress. For example:

11 a. Increasing the temperature of an exothermic reaction will shift the reaction to left (more reactants and less products). The opposite is observed if the reaction is exothermic or if heat is removed from an exothermic reaction. b. In a reaction where the number of gaseous molecules produced is more than the number of reacting gaseous molecules, increasing the pressure of the system will shift the reaction toward the reactants in an attempt to decrease the number of moles and vice versa. Reactions in solutions are usually insensitive to changes in pressure.

12 c. Increasing the concentration of a reactant or removing a product will result in a shift of reaction towards more products and vice versa. d. Some reactions can be facilitated by addition of a catalyst (a substance that is not a part of reactants or products but its presence makes the reaction faster). The catalyst does not change the position of equilibrium but makes the time required to reach this equilibrium point shorter. Le Chatelier's principle can be advantageously used to force reactions that are close to completion to proceed to completion. This is usually done by addition of more reagent in a gravimetric procedure, allowing a gas to escape if one of the products is a gas, etc..

13 Stepwise Equilibrium Constants Sometimes a chemical equilibrium can be written as a multi step equilibrium. The overall equilibrium constant for the reaction is the multiple of stepwise equilibrium constants. Look at the stepwise equilibrium below: A 2 B  A + AB K 1 = [A][AB]/[A 2 B] AB  A + B K 2 = [A][B]/[AB] K 1 K 2 = [A] 2 [B]/[A 2 B]

14 Now look at the equilibrium A 2 B  2 A + B K = [A] 2 [B]/[A 2 B] From K and K 1 K 2 we see that they are equal. This is valid for any multi step equilibrium reaction as will be seen later.

15 Calculations Using Equilibrium Constants Here, we are faced with three situations which should be considered separately: 1. When the equilibrium constant is very small. Most reactants do not undergo a reaction and very little products are produced. Usually the concentration of products can be neglected as compared to reactants concentrations.

16 2. When the equilibrium constant is very high. In this case, most reactants disappear and the reaction container contains mainly the products. In calculations, one can neglect the concentration of remaining reactants as compared to concentrations of products. 3. A situation where the equilibrium constant is moderate. Appreciable amounts of reactants are left and appreciable amounts of products are also formed in the reaction mixture. One can not neglect the reactants or the products concentration. Following are examples on each case.

17 Example Calculate the equilibrium concentration of A and B in a 0.10 M solution of weak electrolyte AB if the equilibrium constant of the reaction is 3.0x Solution AB  A + B First we should look at the value of the equilibrium constant to have an appreciation of what is going on. It is clear that we have a small equilibrium constant which suggests that very little products may have been formed and thus we build our solution on the assumption that AB will mainly remain unreacted except for a very little concentration x.

18 Therefore, AB concentration will decrease by x and A and B will be formed in a concentration equals x for each. The following table represents what is happening.

19 K = [A][B]/[AB] Substitution of equilibrium concentration gives K = (x)(x)/(0.1-x) = 3.0 x In order to solve this equation, we should make a justified assumption which we have discussed above; that is x is very small as compared to 0.10 M and later we will check whether our assumption is valid or not.

20 assume 0.10 >>x. We have 3.0 x = x 2 /0.10, thus we have x = 5.5 x Now we should check the validity of our assumption that 0.10 >>x by calculating the relative error Relative error = (5.5x10 -4 /0.10) x 100 = 0.55%

21 In this course, we will consider our assumption valid if the relative error is below 5%. Therefore, our assumption is valid and we have: [A] =5.5x10 -4 M, [B] = 5.5x10 -4 M, and [AB] = (0.10 – 5.5x10 -4 ) ~ 0.10 M

22 In the reaction A + B  C + D If 0.2 mol of A is mixed with 0.5 mol of B in 1.0 L, find the equilibrium concentrations of A, B, C, and D. The equilibrium constant of the reaction is 4.0x It is clear that we have a small equilibrium constant which suggests that very little products may have been formed and thus we build our solution on the assumption that A and B will mainly remain unreacted except for a very little concentration x.

23 Therefore, A and B concentration will decrease by x and C and D will be formed in a concentration equals x for each. The following table represents what is happening.

24 K = [C][D]/[A][B] Substitution using equilibrium concentrations results in K = (x)(x)/(0.20 – x)(0.50 – x) Now from the value of the equilibrium constant we can assume that 0.20 >> x. Then we have 4.0x10 -9 = x 2 /(0.20x0.50) x = 2.0x10 -5

25 Now check the validity of the assumption by calculating the relative error Relative Error = (2.0x10 -5 /0.20) x 100 = 0.01% The assumption is valid and the concentrations of the different species are [A] = (0.20 – 2.0x10 -5 ) ~ 0.20 M [B] = 0.50 M [C] = [D] = 2.0x10 -5 M

26 In the reaction A + B  C + D If 0.2 mol of A is mixed with 0.5 mol of B in 1.0 L, find the equilibrium concentrations of A, B, C, and D. The equilibrium constant of the reaction is 4.0x10 8. It is clear that we have a high equilibrium constant which suggests that very little reactants may have been left and thus we build our solution on the assumption that A will mainly be of a very little concentration x as a result of the chemical equilibrium. B is already present in excess and an equivalent amount to A reacts while the rest of B remains in unreacted.

27 K = [C][D]/[A][B] Substitution using equilibrium concentrations results in K = (0.20 – x) (0.20 – x)/(x( x)) Now from the value of the equilibrium constant we can assume that 0.20 >> x. Then we have 4.0x10 8 = 0.040/0.3x x = 3.3x10 -10

28 Now check the validity of the assumption by calculating the relative error Relative Error = (3.3x /0.20) x 100 = 0.005% The assumption is valid and the concentrations of the different species are [A] = 3.3x M [B] = ( x ) = 0.30 M [C] = [D] = ( x ) ~ 0.20 M

29 A and B react as follows A + 2 B  2 C Assume 0.10 mol of A reacts with 0.20 mol of B in 1 L. If K = 1.0x10 10, find the equilibrium concentrations of A, B, and C. Solution We have a high equilibrium constant which suggests that most reactants are changed into products except for a tiny equilibrium concentration. Therefore, at equilibrium we mainly have C.

30 K = [C] 2 /[A][B] 2 K = (0.20 – 2x) 2 /x(2x) 2 Again, we are justified to assume that 0.20 >> 2x since the equilibrium constant is high. We then have: 1.0x10 10 = (0.2) 2 /4x 3 x = 1.0x10 -4

31 Now let us check for our assumption by determining the relative error Relative error = (2 x 1.0x10 -4 /0.20) x 100 = 0.10% The assumption is valid and the equilibrium concentrations are [A] = x =1.0x10 -4 M [B] = 2 x = 2.0x10 -4 M [C] = (0.20 – 2.0x10 -4 ) ~ 0.20 M

32 In the reaction A + B  C + D If 0.2 mol of A is mixed with 0.5 mol of B in 1.0 L, find the concentration of A, B, C, and D. The equilibrium constant of the reaction is It is clear that we do not have a clearly small or high equilibrium constant which suggests that both products and reactants will coexist appreciably in the reaction mixture at equilibrium. Therefore, A and B concentration will decrease by x and C and D will be formed in a concentration equals x for each.

33 K = [C][D]/[A][B] K = (x)(x)/(0.20 – x)(0.50 – x) Now from the value of the equilibrium constant we can not assume that 0.20 >> x. However, let us do that and find out what we get 0.30 = x 2 /(0.20x0.50) x = 0.17

34 Now check the validity of the assumption by calculating the relative error Relative Error = ( /0.20) x 100 = 87% The assumption is invalid and we must not neglect x in comparison with 0.20, therefore the equilibrium should be as follows K = (x)(x)/(0.20 – x)(0.50 – x) 0.30 = x 2 /(0.10 – 0.70x +x 2 ) Rearrangement of the relation above results in 0.70x x = 0

35 The solution of the quadratic equation ax 2 + bx + c = 0 is: x = {(- b + (b 2 – 4 ac) 1/2 )/2a} Substitution gives x = {( ( – 4 *0.7*(-0.03)) 1/2 )/2*0.7} x = 0.11 M Therefore, we have [A] = 0.20 – 0.11 = 0.09 M [B] = 0.50 – 0.11 = 0.39 M [C] = [D] = 0.11 M

Lecture 15 Equilibrium Calculations, Cont…. Systematic Approach

37 The Common Ion Effect We have discussed earlier that changing the concentration of a product or reactant has a significant effect on equilibrium. Addition of a reactant forces the reaction to proceed forward. Addition of a product forces the reaction in the backward direction, a consequence of Le Chatelier's principle. A common ion is usually a product ion which means that reactants concentration will increase on the expense of products. Let us look at the following example:

38 Calculate the equilibrium concentrations of A, B, and AB in a 0.10 M AB (weak electrolyte) in presence of 0.20 M B. The equilibrium constant is 3.0x AB  A + B Solution Be aware that we have previously solved this problem but without addition of extra amount of B. Therefore, compare the answers.

39 The equilibrium constant is very small therefore only tiny amounts of AB will dissociate to yield A and B. The initial concentration of B is 0.20, therefore the equilibrium concentration of B will be x. The following table summarizes the situation before and after equilibrium:

40 Writing the equilibrium constant expression we get: K = x ( x)/(0.10 – x) Since K is very small we can assume that 0.1 >> x. This gives: 3.0x10 -6 = 0.20x/0.1 x = 1.5x10 -6 Let us calculate the relative error to check for the validity of our assumption

41 Relative error = (1.5x10 -6 /0.10) x 100 = 1.5x10 -3 % The error is less than 5% therefore our assumption is OK. The concentrations of the different species in present of 0.20 M common ion will be: [A] = 1.5x10 -6 M [B] = x10 -6 ~ 0.20 M [AB] = 0.10 – 1.5x10 -6 ~ 0.10

42 Systematic Approach to Equilibrium Calculations This Type of approach to equilibrium calculations uses what is called mass balance and charge balance concepts. These will be discussed below shortly. By mass balance we mean that when we dissolve an amount of dissociating substance, the analytical concentration of that substance will be equal to the sum of the concentrations of the different species derived from it ( mass conservation). Also, solutions are neutral and thus the concentration of the negative charges is equal to the concentration of the positive ones (charge balance, from electroneutrality).

43 Write mass and charge balance equations for M HOAc in water. Solution First, we write the equilibrium: HOAc  H + + OAc - H 2 O  H + + OH - Now we deal with the problem and find the charge balance by placing the positive ions on one side and the negative ions on the other side. [H + ] = [OH - ] + [OAc - ]

44 The mass balance equation represents the concentration of added HOAc where the analytical concentration added to water was divided into two parts one represents undissociated HOAc and the other represents dissociated HOAc (OAc - ). Therefore: = C HOAc, the mass balance equation is: = [HOAc] + [OAc - ]

45 Write a mass and charge balance equations for M [Ag(NH 3 ) 2 ]Cl. Solution First, write all possible equilibria which might take place in solution. These would be: [Ag(NH 3 ) 2 ]Cl  Ag(NH 3 ) Cl -

46 Ag(NH 3 ) 2 +  Ag(NH 3 ) + +NH 3 Ag(NH 3 ) +  Ag + + NH 3 NH 3 + H 2 O  NH OH - H 2 O  H + + OH - Now we can write the charge balance equation: [H + ] + [Ag + ] + [NH 4 + ] + [Ag(NH 3 ) 2 + ] + [Ag(NH 3 ) + ] = [OH - ] + [Cl - ] The mass balance equation can then be written where all species containing ammonia came from the original compound [Ag(NH 3 ) 2 + ]Cl.

47 Also, the concentration of ammonia in Ag(NH 3 ) 2 + is twice its concentration since each mol of Ag(NH 3 ) 2 + contains two moles of ammonia. Therefore, the mass balance equation for ammonia will be: M = 2[Ag(NH 3 ) 2 + ] + [Ag(NH 3 ) + ] + [NH 3 ] + [NH 4 + ] Another mass balance equation for silver can be written as [Ag + ] + [Ag(NH 3 ) 2 + ] + [Ag(NH 3 ) + ] = [Cl - ] = 0.1 M

48 Write mass and charge balance equations for a solution of H 2 S. Solution First let us set up the equations representing equilibrium of H 2 S in water H 2 S  HS - + H + HS -  H + + S 2- H 2 O  H + + OH -

49 The charge balance equation can be written as before but considering that the charge concentration of S 2- is twice as the concentration of S 2-. This means that each mole of S 2- contains two moles of charges. The charge balance equation is: [H + ] = [OH - ] + [HS - ] + 2[S 2- ] Mass balance equation will be: C H2S = [H 2 S] + [HS - ] + [S 2- ]

50 Write a mass and charge balance equations for CdS in solution. CdS  Cd 2+ + S 2- S 2- + H 2 O  HS - + OH - HS - + H 2 O  H 2 S + OH - H 2 O  H + + OH - The charge balance equation can be simply derived from the equilibria above but taking in consideration that charge concentration on cadmium is twice the concentration of cadmium.

51 The charge concentration on sulfide is also twice the sulfide concentration. The equation will be: [H + ] + 2[Cd 2+ ] = [OH - ] + [HS - ] + 2[S 2- ] Mass balance equation: [Cd 2+ ] = [H 2 S] + [HS - ] + [S 2- ]

52 Write a charge balance equation for a solution containing KI and AlI 3. KI  K + + I - AlI 3 = Al I - H 2 O  H + + OH - The equation can be written directly considering that the concentration of the charge on Al 3+ is three times aluminum concentration, therefore we have: [I - ] + [OH - ] = [H + ] + [K + ] + 3[Al 3+ ]

53 Equilibrium Calculations Using the Systematic Approach This section will only be briefly presented since it will not be our approach in solving equilibrium problems. T depends on building up relations as the equilibrium constant, solubility product, autoprotolysis, mass balance, charge balance, etc.. so as to produce equations at least equaling the number of unknowns. The systematic approach is not easily presented in a part of a lecture but needs enough time for mastering in order to be able to solve equilibrium problems effectively. Therefore, we are justified to keep it as simple as possible in our discussion.

54 Example Use the systematic approach to calculate the equilibrium concentrations of A, B, and AB in a 0.10 M AB solution. The equilibrium constant is 3.0x AB  A + B Solution We do not have charges in the chemical equation; therefore do not attempt to write a charge balance equation.

55 The equilibrium constant is K = [A][B]/[AB] The mass balance equation can be formulated as: C AB = [AB] + [A] 0.10 = [AB] + [A] From the value of the equilibrium constant we know that [AB] >> [A] Therefore, [AB] = 0.10

56 Also, AB dissociates into A and B only, therefore [A] = [B] Substitution in the equilibrium constant expression gives 3.0x10 -6 = [A] 2 /0.10 [A] = 5.5x10 -4 M The relative error = (5.5x10 -4 / 0.10) x 100 = 0.55% which means that A is really much smaller than AB [A] = [B] = 5.5x10 -4 M [AB] = x10 -4 ~ 0.10 M

57 Example Using the systematic approach Calculate the equilibrium concentrations of A, B, and AB in a 0.10 M AB (equilibrium constant is 3.0x10 -6 ) in presence of 0.20 M MB (strong electrolyte)where the equilibrium below takes place. AB  A + + B - MB  M + + B -

58 Mass balance for B - C B - = [B - ] from strong electrolyte + [B - ] from dissociation of AB [B - ] from dissociation of AB = [A + ] C B - = [B - ] from strong electrolyte + [A + ] A + concentration is negligible as compared to amount of B - from strong electrolyte due to very small K and common ion effect seen earlier. Therefore we have [B - ] = 0.20 M We also have from mass balance that C AB = [AB] + [A + ]

59 The same argument can also be used where the very low K suggests that [AB] >> [A + ], therefore: 0.10 = [AB] Now, substitution in the equilibrium constant expression ew get: K = [A + ][B - ]/[AB] 3.0x10 -6 = [A + ] * 0.20/0.10 [A + ] = 1.5x10 -6 M

60 The relative error = (1.5x10 -6 /0.10) x 100 =1.5x10 -3 % Therefore, the concentrations of the different species are: [A + ] = 1.5x10 -6 M [B] = x10 -6 ~ 0.20 M [AB] = 0.10 – 1.5x10 -6 ~ 0.10 M

What is the titer of a 1.58 g/L KMnO 4 (FW = 158 mg/mmol) in terms of mg Na 2 O 2 (FW = 34 mg/mmol). The equation is: 5Na 2 O 2 + 2KMnO H +  2Mn O 2 + 8H 2 O + 10Na + + 2K + 61 Name: Student Number:

Lecture 16 Equilibrium Calculations, Cont…. Activity and Activity Coefficients

63 Activity and Activity Coefficients In presence of strong electrolytes of diverse ions, weak electrolytes tend to dissociate more. This behavior is attributed to the shielding effects exerted by the ions of the strong electrolyte on the ionic species of opposite signs from the weak electrolyte. The concentration of an ion, from a weak electrolyte, in presence of strong electrolyte is called the activity of the ion and is defined by the relation: a i = C i f i Where a i is the activity of the ion i, C i is the concentration of ion i, and f i is the activity coefficient of ion i.

65 Properties of the activity Coefficient 1.The activity coefficient of uncharged species is unity. 2.The activity coefficient of ions in dilute solutions (less than M) is very close to unity and in fact can be approximated to unity. 3.As the concentration of electrolyte (diverse ions) increases, the activity coefficient decreases and activity becomes less than concentration 4.The activity coefficient is highly dependent on the charge of diverse ions where as the charge increases, the activity coefficient decreases significantly

66 Ionic Strength A property which represents the strength of ions in solution can be seen using the term ionic strength (  ) which can be defined as:  = 1/2  C i Z i 2 Z i is the charge on the ion i. If we look carefully to properties of activity coefficients listed above (3 and 4) it is clear that the term ionic strength combines the effects of concentration and charge of ions and thus will be advantageously used as an important factor affecting f i.

67 Example Calculate the ionic strength of a 0.2 M KNO 3 solution. Solution  = 1/2  C i Z i 2  = 1/2 (C K+ Z 2 K+ + C NO3 - Z 2 NO3 - )  = 1/2 (0.2 x x 1 2 ) = 0.2

68 Calculate the ionic strength of a 0.2 M K 2 SO 4 solution. Solution  = 1/2  C i Z i 2  = 1/2 (C K + Z 2 K + + C SO4 2- Z 2 SO4 2- )  = 1/2 (2 x 0.2 x x 2 2 ) = 0.6 Compare between the value of the ionic strength in this and previous example and see the effect of charge on the answer.

69 Calculate the ionic strength of the solution containing 0.3 M KCl and 0.2 M K 2 SO 4.  = 1/2  C i Z i 2  = 1/2 (C K + Z 2 K + + C Cl - Z 2 Cl - + C SO4 2- Z 2 SO4 2- )  = 1/2 (0.7 x x x 2 2 ) = 0.9 We have substituted 0.7 for C K + since this is the overall potassium ion concentration but we could make it the sum of two terms: { C K + Z 2 K + } from KCl + { C K + Z 2 K + } from K2SO4 which will also give the same result.

70 Calculation of Activity coefficients The activity coefficients of an ion i can be calculated provided that three pieces of information are known. These are: 1.The ionic strength of the solution 2.The charge of the ion 3. The effective diameter of the hydrated ion (  i ) in angstrom (A o ) The Debye-Huckel equation can be used for such a calculation where: - log f i = (0.51 Z i 2  1/2 )/(  i  1/2 )

72 In situations where the effective diameter of the ion is 3 A o (you can assume this situation to solve assigned problems) we can write - log f i = (0.51 Z i 2  1/2 )/(1 +  1/2 ) However the Debye-Huckel equation gives an approximate value for the activity coefficient. A better relation is the Davies modification of the Debye-Huckel equation. The equation is: - log f i = (0.51 Z i 2  1/2 )/(  i  1/2 ) – 0.10 Z i 2  In our calculations we will only use the Debye-Huckel equation for estimation of activity coefficients.

73 Example Calculate the activity coefficients of K + and SO 4 2- in a M K 2 SO 4 assuming an effective diameter of 3 A o for both ions. Solution To apply Debye-Huckel equation we need to determine the ionic strength of the solution  = 1/2  C i Z i 2  = 1/2 (0.002 x 2 x x 2 2 ) = 0.006

74 - log f i = (0.51 Z i 2  1/2 )/(1 +  1/2 ) - log f K + = (0.51 x /2 )/( /2 ) = f K + = = For sulfate we have - log f SO4 2- = (0.51 x 2 2 x /2 )/( /2 ) = f SO4 2- = = 0.71 If we repeat the calculation for a more concentrated solution (0.02 M K 2 SO 4 ) the ionic strength will be 0.06 and the activity coefficients for K + and SO 4 2- will be 0.79 and 0.38, respectively. Therefore, you should appreciate the effect of concentration on both ionic strength and activity coefficients as well.

75 Thermodynamic Equilibrium Constant The equilibrium constant expressions which use concentrations are called concentration equilibrium constants. For example, in the reaction: AB  A + B We can write K = [A][B]/[AB] This is the so called concentration equilibrium constant in order to differentiate it from the thermodynamic equilibrium constant which uses activities rather than concentrations.

76 K Th = a A a B /a AB Since a = C f, we can write K Th = [A] f A [B] f B /[AB] f AB K Th = ([A][B]/[AB]) (f A f B /f AB ) Substituting K for [A][B]/[AB] we get K Th = K (f A f B /f AB )

Since f i = 1 for very dilute solutions (< M) then we can say that K approximates K Th as the ionic strength of the solution approaches zero. Therefore, we should calculate K for any equilibrium especially when the ionic strength has a value away from zero. You should also know that equilibrium constants listed in appendices are all thermodynamic equilibrium constants, i.e. measured at very low concentrations where the ionic strength approaches zero.

78 Calculate the concentration equilibrium constant for the dissociation of AB if f A +, f B - are 0.6 and 0.7, respectively. The thermodynamic equilibrium constant is 2.0 x Solution K Th = K (f A + f B - /f AB ) Substitution for f A +, f B - and remembering that f AB = 1 since AB is not charged, we get 2 x = K (0.6 * 0.7)/1 K = 5 x 10 -8

79 Calculate the percent dissociation of a 1.0x10 -4 M AB in water. The thermodynamic equilibrium constant is 2.0 x AB  A + + B -

80 The equilibrium constant is very small, therefore we can assume that 1.0x10 -4 >> x. Substitution in the equilibrium constant expression gives: 2.0x10 -8 = x 2 /1.0*10 -4 x = 1.4x10 -6 M The relative error = (1.4x10 -6 /1.0x10 -4 ) x 100 = 1.4% Therefore the assumption is valid % dissociation = (amount dissociated/initial amount) x 100 = (1.4x10 -6 /1.0x10 -4 ) x 100 = 1.4%

81 Example Calculate the percent dissociation of a 1.0x10 -4 M AB in presence of diverse ions where the ionic strength is 0.1 and f A +, f B - are 0.6 and 0.7, respectively. The thermodynamic equilibrium constant is 2.0 x AB  A + + B -

82 We first calculate the equilibrium constant since we can not use the thermodynamic equilibrium constant in presence of diverse ions: K Th = K (f A + f B - /f AB ) Substitution for f A +, f B - and remembering that f AB = 1 since AB is not charged, we get 2 x = K (0.6 * 0.7)/1 K = 5 x Now we can substitute in the equilibrium constant expression where we have

83 5x10 -8 = x 2 /1.0*10 -4 x = 2. 2 x10 -6 M The relative error = (2. 2 x10 -6 /1.0x10 -4 ) x 100 = 2. 2 % Therefore the assumption is valid % dissociation = (amount dissociated/initial amount) x 100 = (2. 2 x /1.0x10 -4 ) x 100 = 2. 2 % If you compare the results in this and previous examples you will see that the percent dissociation increased from 1.4 to 2.2% which represents an increase of: {(2.2 – 1.4)/1.4}x 100 = 57%

84 Therefore, it is clear that dissociation can significantly increase in presence of diverse ions. However, I would like to emphasize that we will use concentrations rather than activities throughout this course and will neglect the effects of diverse ions except in situations where I ask you to work differently.

In the equilibrium: A + B  C + D If 0.50 mol of A is mixed with 0.20 mol of B in a 1 L solution, find the concentrations of all species at equilibrium. K=4* Name: Student Number: