Slide 1

Activity and the Systematic Treatment of Equilibrium2Effect of Ion Strength on Solubility of Salts

The region surrounding ions in solution is referred to as the ionic atmosphere.The greater the ionic strength of a solution, the higher the charge in the ionic atmosphere. Each ion-plus-its ionic atmosphere contains less NET charge, so there is less attraction between any particular cation and anion. This leads to a decreased tendency to come together and increases solubility.

* Increasing ionic strength promotes dissociation of salts into ions. 3Effect of Ionic Strength on the Solubility of Salts

The ionic strength of a solution increases as salt is added to the solution.Adding an inert salt (one whose ions do not react with the compound of interest) to a solution of a sparingly soluble salt increases the solubility of the sparingly soluble salt.The presence of other ions in solution decreases the attraction between the ions of the sparingly soluble salt because the oppositely charged ions of the inert salt surround the ions of the sparingly soluble salt and decrease its apparent charge.4What is Meant by Ionic Strength

Ionic strength, , is a measure of the total concentration of ions in solution.*Multiply charged ions increase ionic strength more than singly charged ions. = (c1z12 + c2z22 + . . .) = cizi2

where ci is the concentration of the ith species and zi is its charge. The summation is over all ions in solution.Calculation of Ionic StrengthWhat is the ionic strength of a 0.1 M solution of Na2SO4 = (cizi2)

Na2SO4 2 Na+ + SO42- [Na] = 2 x 0.1 = 0.2 M[SO42-] = 0.1 M= [(0.2)(1)2 + (0.1)(2)2] = 0.3 MLimitation: may overestimate the true ionic strength due to ion pair formation6What is Meant by Ionic StrengthFor 1:1 electrolytes (i.e., electrolytes involving singly charged cation and anions), the ionic strength equals the molarity ( = M). For other electrolytes, the ionic strength is greater than the molarity.The ionic strength of solutions containing salts whose ions are multiply charged is very difficult to quantify because these salts are not completely dissociated and generally exist as a soluble ion pair M2+X2-(aq). To calculate ionic strength you would need to know the extent of dissociation (or the concentration of ion pairs).7Activity According to the simple form of the law of mass action for the following reaction equilibriaaA + bB cC + dDThe equilibrium constant is given byK = [C]c[D]d/[A]a[B]bHowever, the equilibrium constant is not truly constant and varies with the concentration of electrolytes in solution. In order for this expression for the equilibrium constant to hold we must replace concentrations by activities. K = ACcADd /AAaABb8Activity Coefficients

The expression for the equilibrium constant does not predict any effect of ionic strength of a chemical reaction. To account for the effect of ionic strength, concentrations must be replaced by activitiesactivityAC = C[C]

The activity coefficient, c, is a measure of the deviation of behavior from ideality.

9Activity Coefficients

The accurate form of the equilibrium constant

K = ACc ADd/ AAa ABb = Cc[C]c Dd[D]d / Aa[A]a Bb[B]bAt low ionic strength, s 1 and K can be written fairly accurately in terms of concentrations.10Activity Coefficients of Ions

Activity coefficients are related to ionic strength by the extended Debye-Hckel equation

log = (-0.51 z2 ) /(1 + ((/305)) at 25C

z is the charge of the ion, and is the size of the ion in pm (and is the effective radius of the hydrated ion)Smaller highly charged ions bind solvent molecules more tightly and have larger hydrated radii than larger or less highly charged ions.1011Effect of Ionic Strength, Ion charge, and Ion Size on the Activity Coefficient

Over the range of ionic strengths from 0-0.1 M

1. As ionic strength, , the activity coefficient, , . 1 as 0.2. As the charge, z, of an ion , the deviation of from 1 increases. i.e. Activity corrections are larger for more highly charged ions, but do not depend upon the sign of the charge. ( - or +)3. The smaller the hydrated radius, , of the ion, the more important the activity effects become.12Effect of Ionic Strength, Ion charge, and Ion Size on the Activity Coefficient

13Activity Coefficients of Nonionic Compounds The activity coefficients of neutral molecules are ~1 in solutions with an ionic strength of < 0.1 M. * We will assume that the activity of a neutral molecule is equal to its concentration AC = [C]. 14Activity Coefficients of Nonionic Compounds For gases, the activity is refereed to its fugacity.The fugacity is given by AC = CPC where C is its fugacity coefficient.When a gas behaves ideally (i.e., at pressures 1 bar), C = 1.

We will assume that the fugacity of a gas is equal to its partial pressure AC = PC. 15pH Revisited The definition of pH given previously is not correct. The accurate definition is in terms of activity.pH = -logAH+ = -log{H+[H3O+]}

A pH meter measures activity not concentration.As for other reactions, the extent of ionization of H2O is influenced by the ionic strength of the solution. *Thus pH depends upon the concentration(s) of other electrolytes in solution. 16Systematic Treatment of Equilibrium Many chemical systems are exceedingly complex because of the large number of reactions and species involved.Need to treat equilibria systematically in order to solve problems where many reactions/species are involved. Many equations in many unknowns that are derived fromThe equilibrium expressions for the reactions involvedCharge balanceMass balance(s)17Charge Balance The charge balance is derived from the fact that solutions have no net charge.

The sum of the (+) charges in solution = the sum of the (-) charges in solution.If we know all of the charged species that exist in solution, then it is straightforward to write the charge balance equation.18Charge Balance * The coefficient in front of each species always equals the magnitude of the charge on the ion.

In general terms, charge balance is given by

p1[C1] + p2[C2] + . . . = n1[A1] + n2[A2] + . . .or Spi[Ci] = Sni[Ai]

where [Ci] = concentration of cation i pi = the (+) charge on cation i [Ai] = concentration of anion i ni = the (-) charge on anion iCharge Balance Write a charge balance equation for the dissociation of sulfuric acid in aqueous solutionH2SO4 H+ + HSO4- + SO42-H2O H+ + OH-The sum of the (+) charges in solution = the sum of the (-) charges in solution[H+] = [HSO4-] + 2[SO42-] + [OH-]

20Example: Write the charge balance equation for a solution that contains the following ionic species, H+, OH-, Na+, HSO4-, and SO42-.

Total (+) charge in solution = Total (-) charge in solution[H+] + [Na+] = [OH-] + [HSO4-] + 2 [SO42-] .

21Example: Write the charge balance equation for a solution that contains 0.100 M H3PO4.

To determine all of the ionic species in solution we need to recognize that . . .2 H2O() H3O+(aq) + OH-(aq)H3PO4(aq) + H2O() H3O+(aq) + H2PO4-(aq)H2PO4-(aq) + H2O() H3O+(aq) + HPO42-(aq)HPO42-(aq) + H2O() H3O+(aq) + PO43-(aq)Total (+) charge in solution = Total (-) charge in solution[H3O+] = [OH-] + [H2PO4-] + 2 [HPO42-] + 3[PO43-].

22Mass Balance The mass balance, also referred to as the material balance, arises because we must have conservation of matter in chemical reactions.i.e., matter is neither created nor destroyed in a chemical reaction, it merely changes form.The quantity of all species in a solution containing a particular atom (or group of atoms) must equal the amount of the atom (or group) placed in solution.23Example: The fizz (carbonation) in soft drinks in maintained by complex equilibria involving carbonic acid, H2CO3. Consider a solution of 0.050 M H2CO3. Write a mass balance equation for the CO2 group.H2CO3(aq) + H2O() HCO3-(aq) + H3O+(aq)HCO3-(aq) + H2O() CO32-(aq) + H3O+(aq)HCO3-(aq) CO2(aq) + OH-(aq)H3O+(aq) + OH-(aq) 2 H2O()mass balance for CO2 group is given by0.050 M = [H2CO3] + [HCO3-] + [CO32-] + [CO2]When in a closed container, soft drinks are able to maintain their fizz because CO2 cannot escape solution, but pop loses its fizz when opened to the atmosphere because the CO2 evaporates and the above equilibria for the first three reactions are pushed to completion by Le Chteliers principle.Mass BalanceCe(OH)3 Ce3+ + 3 OH-

Mass Balance[OH-] = 3 [Ce3+]In words: the concentration of hydroxide ions is 3 times the concentration of ceriumSolubility Using a Mass BalanceDetermine the solubility of tributylammonium bromide in a saturated solution with the pH fixed to 9.50R3NHBr R3NH++ Br-Ksp = 4.0 x 10-8R3NH+ R3N + H+Ka = 2.3 x 10-9Ksp = [R3NH+][Br-]Ka = [R3N][H+] / [R3NH+]Two equations with three unknowns? Solubility Using a Mass BalanceCannot use a charge balance because pH is fixed by an external mechanismMass Balance[Br-] = [R3NH+] + [R3N]

From the equation for the acid dissociation constant[R3N] = Ka [R3NH+] / [H+] = 2.3 x 10-9[R3NH+] / 1 x 10-9.5 [R3N] = 7.273 [R3NH+]

Substitute into mass balance equation [Br-] = [R3NH+] + 7.273 [R3NH+] [Br-] = 8.273 [R3NH+]Solubility Using a Mass BalanceSubstitute for [Br-] and [R3NH+] in Ksp relationship([R3NH+])(8.273 [R3NH+]) = 4.0 x 10-8 [R3NH+] = (4.0 x 10-8) / 8.273 = 7.0 x 10-5

From stoichiometry solubility = [Br-][Br-] = [R3NH+] + [R3N]= 7.0 x 10-5 + (7.273 x 7.0 x 10-5)= 5.79 x 10-4 MCorrect for significant figures Solubility of R3NHBr = 5.8 x 10-4 M28Example:Calculate the concentrations of all species present in a solution containing 1.0M HCl and 0.010M Cd(NO3)2Because the solution is strongly acidic, hydrolysis of the metal ion is negligible (i.e., we can ignore reactions such as Cd2+ + 2x H2O = Cd(OH)x2-x + x H3O+). equilibria presentCd2+(aq) + Cl-(aq) CdCl+(aq)K1 = [CdCl+] / [Cd2+][Cl-] = 21CdCl+(aq) + Cl-(aq) CdCl2(aq)K2 = [CdCl2] / [CdCl+][Cl-] = 7.9CdCl2(aq) + Cl-(aq) CdCl3-(aq)K3 = [CdCl3-] / [CdCl2][Cl-] = 1.23

29Example:Calculate the concentrations of all species present in a solution containing 1.0M HCl and 0.010M Cd(NO3)2Because the solution is strongly acidic, hydrolysis of the metal ion is negligible (i.e., we can ignore reactions such as Cd2+ + 2x H2O = Cd(OH)x2-x + x H3O+). equilibria presentCdCl3-(aq) + Cl-(aq) CdCl42-(aq)K4 = [CdCl42-] / [CdCl3-][Cl-] = 0.35mass balances[Cl-] + [CdCl+] + 2[CdCl2] + 3[CdCl3-] + 4[CdCl42-] = 1.00[Cd2+] + [CdCl+] + [CdCl2] +[CdCl3-] + [CdCl42-] = 0.01030Example:Calculate the concentrations of all species present in a solution containing 1.0M HCl and 0.010M Cd(NO3)2Because the solution is strongly acidic, hydrolysis of the metal ion is negligible (i.e., we can ignore reactions such as Cd2+ + 2x H2O = Cd(OH)x2-x + x H3O+). equilibria presentCdCl3-(aq) + Cl-(aq) CdCl42-(aq)K4 = [CdCl42-] / [CdCl3-][Cl-] = 0.35mass balances[Cl-] + [CdCl+] + 2[CdCl2] + 3[CdCl3-] + 4[CdCl42-] = 1.00[Cd2+] + [CdCl+] + [CdCl2] +[CdCl3-] + [CdCl42-] = 0.01031Example:Calculate the concentrations of all species present in a solution containing 1.0 M HCl and 0.010 M Cd(NO3)2Because the solution is strongly acidic, hydrolysis of the metal ion is negligible (i.e., we can ignore reactions such as Cd2+ + 2x H2O Cd(OH)x2-x + x H3O+). Assume that most of the chlorine present is as free Cl-. This should be reasonable because the total Cl- concentration is 100 times as large as that of Cd.[Cl-] = 1.0from reaction 121 = [CdCl+]/[Cd2+][Cl-]or [CdCl+] = 21[Cd2+]32Example:Calculate the concentrations of all species present in a solution containing 1.0 M HCl and 0.010 M Cd(NO3)2Because the solution is strongly acidic, hydrolysis of the metal ion is negligible (i.e., we can ignore reactions such as Cd2+ + 2x H2O Cd(OH)x2-x + x H3O+). from reaction 27.9 = [CdCl2]/[CdCl+][Cl-]or [CdCl2] = 7.9[CdCl+] = 7.9(21)[Cd2+] = 166[Cd2+]from reaction 31.23 = [CdCl3-]/[CdCl2][Cl-]or [CdCl3-] = 1.23[CdCl2] = 1.23(166)[Cd2+] = 204[Cd2+]33Example:Calculate the concentrations of all species present in a solution containing 1.0 M HCl and 0.010 M Cd(NO3)2Because the solution is strongly acidic, hydrolysis of the metal ion is negligible (Cd2+ + 2x H2O Cd(OH)x2-x + x H3O+). from reaction 40.35 = [CdCl42-]/[CdCl3-][Cl-]or [CdCl42-] = 0.35[CdCl3-] = 0.35(204)[Cd2+] = 71[Cd2+]substitute into mass balance for Cd2+[Cd2+] + 21[Cd2+] + 166[Cd2+] + 204[Cd2+] + 71[Cd2+] = 0.010[Cd2+] = 2.16 x 10-5M 2.2 x 10-5M

Example:Calculate the concentrations of all species present in a solution containing 1.0M HCl and 0.010M Cd(NO3)2Because the solution is strongly acidic, hydrolysis of the metal ion is negligible (Cd2+ + 2x H2O Cd(OH)x2-x + x H3O+). substitute into other equations to derive all concentrations

[CdCl+] = 4.53 x 10-4 M 4.5 x 10-4 M[CdCl2] = 3.58 x 10-3 M 3.6 x 10-3 M[CdCl3-] = 4.41 x 10-3 M 4.4 x 10-3 M[CdCl42-] = 1.53 x 10-3 M 1.5 x 10-3 MThe formation of coordination complexes can have a large effect on the solubility of a compound in water by consuming one of the ions through complex formation.35Example:Calculate the concentrations of all species present in a solution containing 1.0M HCl and 0.010M Cd(NO3)2Because the solution is strongly acidic, hydrolysis of the metal ion is negligible (Cd2+ + 2x H2O Cd(OH)x2-x + x H3O+). Because the question asks for the concentrations of all species, the concentrations of H3O+, OH-, and NO3- should also be determined as these species will be present in the solution

[H3O+] = 1.0 M (because HCl(aq) + H2O() H3O+(aq) + Cl-(aq))[OH-] = Kw/[H3O+] = (1.0 x 10-14/1.0)M = 1.0 x 10-14 M [NO3-] = 2(0.010 M) = 0.020 M (because Cd(NO3)2(aq) Cd2+(aq) + 2 NO3-(aq) in H2O)