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Activity. Introduction 1.) Hydration Ions do not act as independent particles in solvent (water) Surrounded by a shell of solvent molecules. Oxygen has a partial negative charge and hydrogen partial positive charge. Oxygen binds cations. Hydrogen binds anions. Activity. Introduction

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Activity

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  1. Activity • Introduction • 1.)Hydration • Ions do not act as independent particles in solvent (water) • Surrounded by a shell of solvent molecules Oxygen has a partial negative charge and hydrogen partial positive charge Oxygen binds cations Hydrogen binds anions

  2. Activity • Introduction • 2.)H2O exchanges rapidly between bulk solvent and ion-coordination sites

  3. Activity • Introduction • 3.)Size of Hydration • Size and charge of ion determines number of bound waters • Smaller, more highly charged ions bind more water molecules Activity – is related to the size of the hydrated species Small Ions bind more water and behave as larger species in solution

  4. Activity • Effect of Ionic Strength on Solubility • 1.)Ionic Atmosphere • Similar in concept to hydration sphere • Cation surrounded by anions and anions are surrounded by cations - Effective charge is decreased • - Shields the ions and decreases attraction • Net charge of ionic atmosphere is less than ion - ions constantly moving in/out of ionic atmosphere Each ion-plus-atmosphere contains less net charge and there is less attraction between any particular cation and anion Each ion see less of the other ions charge and decreases the attraction

  5. Activity • Effect of Ionic Strength on Solubility • 2.)Ionic Strength (m) • Addition of salt to solution increases ionic strength - Added salt is inert  does not interact or react with other ions • In general, increasing ionic strength increases salt solubility - Opposite of common ion effect The greater the ionic strength of a solution, the higher the charge in the ionic atmospheres More ions added, more ions can be present in ionic atmospheres

  6. Activity • Effect of Ionic Strength on Solubility • 2.)Ionic Strength (m) • Measure of the total concentration of ions in solution - More highly charged an ion is the more it is counted - Sum extends over all ions in solution where: Ci is the concentration of the ith species and zi is its charge

  7. Activity • Effect of Ionic Strength on Solubility • 2.)Ionic Strength (m) • Example: What is the ionic strength of a 0.0087 M KOH and 0.0002 M La(IO3)3 solution? Assume complete dissociation and no formation of LaOH2+

  8. Activity • Effect of Ionic Strength on Solubility • 3.)Equilibria Involving Ionic Compounds are Affected by the Presence of All Ionic Compounds in the Solution • Knowing the ionic strength is important in determining solubility • Example: Ksp = 1.3x10-18 If Hg2(IO3)2 is placed in pure water, up to 6.9x10-7M will dissolve. If 0.050 M KNO3 is added, up to 1.0x10-6M Hg2(IO3)2 will dissolve. Occurs Due to Changes in the Ionic Strength & Activity Coefficients

  9. Activity • Equilibrium Constant and Activity • 1.)Typical Form of Equilibrium Constant • However, this is not strictly correct • Ratio of concentrations is not constant under all conditions • Does not account for ionic strength differences 2.) Activities, instead of concentrations should be used • Yields an equation for K that is truly constant where: AA, AB, AC, AD is activities of A through D

  10. Activity • Equilibrium Constant and Activity 3.) Activities account for ionic strength effects • Concentrations are related to activities by an activity coefficient (g) 4.) “Real” Equilibrium Constant Using Activity Coefficients where: AC is activity of C [C] is concentration of C gC is activity coefficient of C

  11. Activity • Equilibrium Constant and Activity 4.) “Real” Equilibrium Constant Using Activity Coefficients • g is always ≤ 1 • Activity coefficient measures the deviation from ideal behavior - If g =1, the behavior is ideal and typical form of equilibrium constant is used • Activity coefficient depends on ionic strength - Activity coefficient decrease with increasing ionic strength - Approaches one at low ionic strength Activity depends on hydrated radius (a) of the ion. This includes the ion itself and any water closely associated with it.

  12. Activity • Equilibrium Constant and Activity 5.) Activity Coefficients of Ions • Extended Debye-Hϋckel Equation • Only valid for concentrations ≤ 0.1M • In theory, a is the diameter of hydrated ion where: g is the activity coefficient a is ion size (pm) z is the ion charge m is the ionic strength

  13. Activity • Equilibrium Constant and Activity 5.) Activity Coefficients of Ions • In practice, a is an empirical value, provide agreement between activity and ionic strength - sizes can not be taken literally - trends are sensible  small, highly charged ions have larger effective sizes a: Li+ > Na+ > K+ > Rb+ Ideal behavior when g = 1 - low ionic strength - low concentration - low charge/large a

  14. Activity Activity Coefficients from Debye-Hϋckel Equation

  15. Activity • Equilibrium Constant and Activity 6.) Example 1: What is the activity coefficient of Hg22+ in a solution of 0.033 M Hg2(NO3)2? Solution: Step 1 – Determine m

  16. g = 0.355 at m = 0.10 M Activity • Equilibrium Constant and Activity 6.) Example 1: What is the activity coefficient of Hg22+ in a solution of 0.033 M Hg2(NO3)2? Solution: Step 2 – Identify Activity Coefficient from table at corresponding ionic strength.

  17. Activity • Equilibrium Constant and Activity 6.) Example 2: What is the activity coefficient for H+ at m = 0.025 M? Note: Values for g at m = 0.025 are not listed in the table. There are two possible ways to obtain g in this case: a.) Direct Calculation (Debye-Hϋckel) zH+ m a for H+ from table

  18. Activity • Equilibrium Constant and Activity 6.) Example 2: What is the activity coefficient for H+ at m = 0.025 M? b.) Interpolation Use values for gH+ given at m = 0.01 and 0.05 m from table and assume linear change in g with m. To solve for gH+ at m = 0.025: Fract. Of Interval Between 0.01 and 0.05 Diff. in g values at 0.01 and 0.05 gH+ at m = 0.01

  19. Activity • Equilibrium Constant and Activity 6.) Example 2: What is the activity coefficient for H+ at m = 0.025 M? b.) Interpolation Use values for gH+ given at m = 0.01 and 0.05 m from table and assume linear change in g with m. Note: This value is slightly different from the calculated value (0.88) since it is only an estimate.

  20. Activity • Equilibrium Constant and Activity 7.) Activity Coefficients of Gasses and Neutral Molecules • For nonionic, neutral molecules - g ≈ 1 form ≤ 0.1 M - or Ac = [C] • For gases, - g ≈1 for pressures ≤ 1 atm - or A ≈ P, where P is pressure in atm 8.)Limitation of Debye-Hϋckel Equation • Debye-Hϋckel predicts g decreases as m increases - true up to m = 0.10 M • At higher m, the equation is no longer accurate - at m ≥ 0.5 M, most ions actually show an increase in g with an increase in m - at higher m, solvent is actually a mixture instead of just water Hydration sphere is mixture of water and salt at high concentration

  21. Activity • pH 1.) When we measure pH with a pH meter, we are measuring the negative logarithm of the hydrogen ion activity • Not measuring concentration 2.)Affect of pH with the Addition of a Salt • Changes ionic strength  Changes H+ and OH- activity

  22. Activity • pH 2.)Affect of pH with the Addition of a Salt • Example: What is the pH of a solution containing 0.010M HCl plus 0.040 M KClO4?

  23. Activity • Using Activity Coefficients 1.) Activity Coefficients Need to be Considered for Accurate Answers Involving Equilibrium Constants • Example #1: What is the [Hg22+] in a saturated solution of Hg2Br2 with 0.00100M KCl, where and KCl acts as an “inert salt”? Ksp = 5.6x10-23

  24. Activity • Using Activity Coefficients 1.) Activity Coefficients Need to be Considered for Accurate Answers Involving Equilibrium Constants • Example #2: What is the [Hg22+] in a saturated solution of Hg2Br2 with 0.00100M KBr? Note: KBr is not an inert salt, since Br- is also present in the Ksp reaction of Hg2Br2

  25. Activity • Using Activity Coefficients 1.) Activity Coefficients Need to be Considered for Accurate Answers Involving Equilibrium Constants • Example #3: What is the true concentration of Li+ and F- in a saturated solution of LiF in water? Note: Only LiF is present in solution. Ionic strength is only determined by the amount of LiF that dissolves Solution:Set-up the equilibrium equation in terms of activities

  26. Activity • Using Activity Coefficients 1.) Activity Coefficients Need to be Considered for Accurate Answers Involving Equilibrium Constants • Example #3: Note: Both x and gLi+,gF- depend on the final amount of LiF dissolved in solution To solve, use the method of successive of approximation Solution:Assume gLi+ = gF- = 1. Solve for x.

  27. Activity • Using Activity Coefficients 1.) Activity Coefficients Need to be Considered for Accurate Answers Involving Equilibrium Constants • Example #3: Solution:Step 2 use the First Calculated Value of [Li+] and [F-] to Estimate the Ionic Strength and g Values. Obtained by using m=0.041 and interpolating data in table

  28. Activity • Using Activity Coefficients 1.) Activity Coefficients Need to be Considered for Accurate Answers Involving Equilibrium Constants • Example #3: Solution:Step 3use the calculated values for gF and gLi to re-estimate [Li+] and [F-]. substitute

  29. Activity • Using Activity Coefficients 1.) Activity Coefficients Need to be Considered for Accurate Answers Involving Equilibrium Constants • Example #3: Solution:Repeat Steps 2-3 Until a Constant Value for x is obtained For this example, this occurs after 3-4 cycles, where x = 0.050M gF, gLi Use g to calculate new concentrations. Use concentrations to calculate new m and g. [F-], [Li+]

  30. Equilibrium • Systematic Treatment of Equilibrium 1.) Help Deal with Complex Chemical Equilibria • Set-up general equations • Simplify using approximations • Introduce specific conditions  number of equations = number of unknowns 2.) Charge Balance • The sum of the positive charges in solution equals the sum of the negative charges in solution. (positive charge) (negative charge) where [C] is the concentration of a cation n is the charge of the cation [A] is the concentration of an anion m is the charge of the anion A solution will not have a net charge!

  31. Equilibrium • Systematic Treatment of Equilibrium 2.) Charge Balance • If a solution contains the following ionic species: H+, OH-,K+,H2PO4-,HPO42- and PO43-, the charge balance would be: The coefficient in front of each species always equals the magnitude of the charge on the ion. For a solution composed of 0.0250 mol of KH2PO4 and 0.0300 mol of KOH in 1.00L: [H+] = 5.1x10-12M [H2PO4-] = 1.3x10-6 M [K+] = 0.0550 M [HPO42-] = 0.0220M [OH-] = 0.0020M [PO43-] = 0.0030M Charge balance:

  32. Equilibrium • Systematic Treatment of Equilibrium 3.) Mass Balance • Also called material balance • Statement of the conservation of matter • The quantity of all species in a solution containing a particular atom must equal the amount of that atom delivered to the solution Acetic acid Acetate Mass balance for 0.050 M in water: Include ALL products in mass balance: H3PO4 H2PO4-,HPO42-, PO43-

  33. Equilibrium • Systematic Treatment of Equilibrium 3.) Mass Balance • Example #1: Write the mass balance for a saturated solution of the slightly soluble salt Ag3PO4, which produces PO43- and Ag+ when it dissolves. Solution:If phosphate remained as PO43-, then but, PO43- reacts with water

  34. Equilibrium • Systematic Treatment of Equilibrium 3.) Mass Balance • Example #2: Write a mass balance for a solution of Fe2(SO4)3, if the species are Fe3+, Fe(OH)2+, Fe(OH)2+, Fe2(OH)24+, FeSO4+, SO42- and HSO4-.

  35. Equilibrium • Systematic Treatment of Equilibrium 1.) Write all pertinent reactions. 2.) Write the charge balance equation. • Sum of positive charges equals the sum of negative charges in solution 3.) Write the mass balance equations. There may be more than one. • Conservation of matter • Quantity of all species in a solution containing a particular atom must equal the amount of atom delivered to the solution 4.) Write the equilibrium constant expression for each chemical reaction. • Only step where activity coefficients appear 5.) Count the equations and unknowns • Number of unknowns must equal the number of equations 6.) Solve for all unknowns 7.) Verify any assumptions

  36. Equilibrium • Applying the Systematic Treatment of Equilibrium 1.) Example #1: • Ionization of water Kw Kw= 1.0x10-14 at 25oC Step 1:Pertinent reactions: Step 2: Charge Balance: Step 3: Mass Balance :[H2O], [H+], [OH-] determined by Kw Not True!

  37. Equilibrium • Applying the Systematic Treatment of Equilibrium 1.) Example #1: • Ionization of water Step 4:Equilibrium constant expression: Step 5:Count equations and unknowns: Two equations: (1) (2) Two unknowns: (1) (2)

  38. Equilibrium • Applying the Systematic Treatment of Equilibrium 1.) Example #1: • Ionization of water Step 6:Solve: Ionic strength (m) of pure water is very low, gH+ and gOH- ~ 1 substitute

  39. Ksp Ksp= 2.4x10-5 Kion pair Kion pair= 5.0x10-3 Kacid Kacid= 2.0x10-13 Kbase Kbase= 9.8x10-13 Kw Kw= 1.0x10-14 Equilibrium • Applying the Systematic Treatment of Equilibrium 2.) Example #2: • Solubility of Calcium Sulfate • Find concentrations of the major species in a saturated solution of CaSO4 Step 1:Pertinent reactions: This information is generally given:

  40. Equilibrium • Applying the Systematic Treatment of Equilibrium 2.) Example #2: • Solubility of Calcium Sulfate • Find concentrations of the major species in a saturated solution of CaSO4 Step 2:Charge Balance: Step 3:Mass Balance: Doesn’t matter what else happens to these ions!

  41. Equilibrium • Applying the Systematic Treatment of Equilibrium 2.) Example #2: • Solubility of Calcium Sulfate • Find concentrations of the major species in a saturated solution of CaSO4 Step 4:Equilibrium constant expression (one for each reaction):

  42. Equilibrium • Applying the Systematic Treatment of Equilibrium 2.) Example #2: • Solubility of Calcium Sulfate • Find concentrations of the major species in a saturated solution of CaSO4 Step 5:Count equations and unknowns: Seven Equations: (1) (CB) (2) (MB) (3) (4) (6) (5) (7) Seven Unknowns:

  43. Equilibrium • Applying the Systematic Treatment of Equilibrium 2.) Example #2: • Solubility of Calcium Sulfate • Find concentrations of the major species in a saturated solution of CaSO4 • Step 6:Solve (Not Easy!): • - don’t know ionic strength  don’t know activity coefficients • - where to start with seven unknowns • [H+]=[OH-]=1x10-7, remaining chemical reactions are independent of water • At first, ignore equations with small equilibrium constants • Make Some Initial Assumptions: • At first, set all activities to one to calculate ionic strength

  44. Equilibrium • Applying the Systematic Treatment of Equilibrium 2.) Example #2: • Solubility of Calcium Sulfate • Find concentrations of the major species in a saturated solution of CaSO4 Step 6:Solve (Not Easy!): • Assumptions Reduce Number of Equations and Unknowns: • Three unknowns: • Three equations Mass balance and charge balance reduces to: Charge balance: Mass balance: [H+] = [OH-] Low concentrations  small equilibrium constant Low concentrations  small equilibrium constant Simple Cancellation

  45. Equilibrium • Applying the Systematic Treatment of Equilibrium 2.) Example #2: • Solubility of Calcium Sulfate • Find concentrations of the major species in a saturated solution of CaSO4 Step 6:Solve (Not Easy!): So, [CaSO4] is known Therefore, only two equations and two unknowns: substitute and

  46. Equilibrium • Applying the Systematic Treatment of Equilibrium 2.) Example #2: • Solubility of Calcium Sulfate • Find concentrations of the major species in a saturated solution of CaSO4 Step 6:Solve (Not Easy!): Given: Determine Ionic Strength: Determine Activity Coefficients: From table

  47. Equilibrium • Applying the Systematic Treatment of Equilibrium 2.) Example #2: • Solubility of Calcium Sulfate • Find concentrations of the major species in a saturated solution of CaSO4 Step 6:Solve (Not Easy!): Use activity coefficients and Ksp equation to calculate new concentrations: Use new concentrations to calculate new ionic strength and activity coefficients:

  48. Equilibrium • Applying the Systematic Treatment of Equilibrium 2.) Example #2: • Solubility of Calcium Sulfate • Find concentrations of the major species in a saturated solution of CaSO4 Step 6:Solve (Not Easy!): Repeat process until calculated numbers converge to a constant value: Stop, concentrations converge

  49. Equilibrium • Applying the Systematic Treatment of Equilibrium 2.) Example #2: • Solubility of Calcium Sulfate • Find concentrations of the major species in a saturated solution of CaSO4 Step 7:Check Assumptions: With: Both [HSO4-] and [CaOH+] are ~ 5 times less than [Ca2+] and [SO42-]  assumption is reasonable

  50. Equilibrium • Applying the Systematic Treatment of Equilibrium 2.) Example #3: • Solubility of Magnesium Hydroxide • Find concentrations of the major species in a saturated solution of Mg(OH)2 Step 1:Pertinent reactions: Ksp Ksp= 7.1x10-12 K1 K1= 3.8x102 Kw Kw= 1.0x10-14 Step 2:Charge Balance:

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