1 Example 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 Solution 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.

2 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 = M

3 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

4 For this type of equation, we can use the quadratic equation to get a solution for x The quadratic equation solution of the 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

5 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:

6 Example 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.

7 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:

8 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 M Let us calculate the relative error to check for the validity of our assumption

9 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

10 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).

11 Example 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 - ]

12 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 - ]

13 Example 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 -

14 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.

15 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

16 Example 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 -

17 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- ]

18 Example Write a mass and charge balance equations for CdS in solution. 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 -

19 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 and 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- ]