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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.

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Presentation on theme: "Activity Introduction 1.)Hydration  Ions do not act as independent particles in solvent (water)  Surrounded by a shell of solvent molecules Oxygen has."— Presentation transcript:

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.)H 2 O 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 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 Activity Each ion see less of the other ions charge and decreases the attraction Each ion-plus-atmosphere contains less net charge and there is less attraction between any particular cation and anion

5 Activity Effect of Ionic Strength on Solubility 2.)Ionic Strength (  )  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 (  )  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: C i is the concentration of the ith species and z i is its charge

7 Activity Effect of Ionic Strength on Solubility 2.)Ionic Strength (  )  Example: What is the ionic strength of a M KOH and M La(IO 3 ) 3 solution? Assume complete dissociation and no formation of LaOH 2+

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: K sp = 1.3x If Hg 2 (IO 3 ) 2 is placed in pure water, up to 6.9x10 -7 M will dissolve. If M KNO 3 is added, up to 1.0x10 -6 M Hg 2 (IO 3 ) 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: A A, A B, A C, A D 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 (  ) 4.)“Real” Equilibrium Constant Using Activity Coefficients where: A C is activity of C [C] is concentration of C  C is activity coefficient of C

11 Activity Equilibrium Constant and Activity 4.)“Real” Equilibrium Constant Using Activity Coefficients   is always ≤ 1  Activity coefficient measures the deviation from ideal behavior - If  =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 (  ) 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-H ϋ ckel Equation  Only valid for concentrations ≤ 0.1M  In theory,  is the diameter of hydrated ion where:  is the activity coefficient  is ion size (pm) z is the ion charge  is the ionic strength

13 Activity Equilibrium Constant and Activity 5.)Activity Coefficients of Ions  In practice  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  : Li + > Na + > K + > Rb + Ideal behavior when  = 1 - low ionic strength - low concentration - low charge/large 

14 Activity Activity Coefficients from Debye-Hϋckel Equation

15 Activity Equilibrium Constant and Activity 6.)Example 1: What is the activity coefficient of Hg 2 2+ in a solution of M Hg 2 (NO 3 ) 2 ? Solution: Step 1 – Determine 

16 Activity Equilibrium Constant and Activity 6.)Example 1: What is the activity coefficient of Hg 2 2+ in a solution of M Hg 2 (NO 3 ) 2 ? Solution: Step 2 – Identify Activity Coefficient from table at corresponding ionic strength.  = at  = 0.10 M

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

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

19 Activity Equilibrium Constant and Activity 6.)Example 2: What is the activity coefficient for H + at  = M? b.) Interpolation Use values for  H+ given at  = 0.01 and 0.05  from table and assume linear change in  with  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 -  ≈ 1 for  ≤ 0.1 M -or A c = [C]  For gases, -  ≈1 for pressures ≤ 1 atm -or A ≈ P, where P is pressure in atm 8.)Limitation of Debye-Hϋckel Equation  Debye-Hϋckel predicts  decreases as  increases -true up to  = 0.10 M  At higher , the equation is no longer accurate - at  ≥ 0.5 M, most ions actually show an increase in  with an increase in  -at higher , 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 M KClO 4 ?

23 Activity Using Activity Coefficients 1.) Activity Coefficients Need to be Considered for Accurate Answers Involving Equilibrium Constants  Example #1: What is the [Hg 2 2+ ] in a saturated solution of Hg 2 Br 2 with M KCl, where and KCl acts as an “inert salt”? K sp = 5.6x10 -23

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

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 Initial Concentrationsolid00 Final Concentrationsolidxx 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  Li+,  F- depend on the final amount of LiF dissolved in solution To solve, use the method of successive of approximation Solution: Assume  Li+ =  F- = 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  Values. Obtained by using  =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 3 use the calculated values for  F and  Li 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  F,  Li [F - ], [Li + ] Use  to calculate new concentrations. Use concentrations to calculate new  and 

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. 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 ( positive charge ) ( negative charge ) 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 +,H 2 PO 4 -,HPO 4 2- and PO 4 3-, 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 mol of KH 2 PO 4 and mol of KOH in 1.00L: [H + ] = 5.1x M[H 2 PO 4 - ] = 1.3x10 -6 M [K + ] = M[HPO 4 2- ] = M [OH - ] = M[PO 4 3- ] = M 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 acidAcetate Mass balance for M in water: Include ALL products in mass balance: H 3 PO 4  H 2 PO 4 -,HPO 4 2-, PO 4 3-

33 Equilibrium Systematic Treatment of Equilibrium 3.)Mass Balance  Example #1: Write the mass balance for a saturated solution of the slightly soluble salt Ag 3 PO 4, which produces PO 4 3- and Ag + when it dissolves. Solution: If phosphate remained as PO 4 3-, then but, PO 4 3- reacts with water

34 Equilibrium Systematic Treatment of Equilibrium 3.)Mass Balance  Example #2: Write a mass balance for a solution of Fe 2 (SO 4 ) 3, if the species are Fe 3+, Fe(OH) 2+, Fe(OH) 2 +, Fe 2 (OH) 2 4+, FeSO 4 +, SO 4 2- and HSO 4 -.

35 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 Equilibrium

36 Applying the Systematic Treatment of Equilibrium 1.)Example #1:  Ionization of water KwKw K w = 1.0x at 25 o C Step 1: Pertinent reactions: :[H 2 O], [H + ], [OH - ] determined by K w Not True! Step 2:Charge Balance: Step 3:Mass Balance

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) (1) (2) Two unknowns:

38 Equilibrium Applying the Systematic Treatment of Equilibrium 1.)Example #1:  Ionization of water Step 6: Solve: Ionic strength (  ) of pure water is very low,  H+ and  OH- ~ 1 substitute

39 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 CaSO 4 Step 1: Pertinent reactions: KwKw K w = 1.0x K base K acid K ion pair K sp K base = 9.8x K acid = 2.0x K ion pair = 5.0x10 -3 K sp = 2.4x10 -5 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 CaSO 4 Step 2: Charge Balance: Step 3: Mass Balance: Doesn’t matter what else happens to these ions!

41 Step 4: Equilibrium constant expression (one for each reaction): 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 CaSO 4

42 Step 5: Count equations and unknowns: 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 CaSO 4 Seven Equations: Seven Unknowns: (1) (2) (3) (4) (5) (6) (7) (CB) (MB)

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 CaSO 4 Step 6: Solve (Not Easy!): - don’t know ionic strength  don’t know activity coefficients - where to start with seven unknowns Make Some Initial Assumptions:  At first, set all activities to one to calculate ionic strength  [H + ]=[OH - ]=1x10 -7, remaining chemical reactions are independent of water  At first, ignore equations with small equilibrium constants

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 CaSO 4 Step 6: Solve (Not Easy!): Assumptions Reduce Number of Equations and Unknowns:  Three unknowns:  Three equations Mass balance and charge balance reduces to: Low concentrations  small equilibrium constant [H + ] = [OH - ] Charge balance: Mass balance: Simple Cancellation Low concentrations  small equilibrium constant

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 CaSO 4 Step 6: Solve (Not Easy!): So, [CaSO 4 ] is known substitute and Therefore, only two equations and two unknowns:

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 CaSO 4 Step 6: Solve (Not Easy!): From table Determine Ionic Strength: Determine Activity Coefficients: Given:

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 CaSO 4 Step 6: Solve (Not Easy!): Use activity coefficients and K sp 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 CaSO 4 Step 6: Solve (Not Easy!): Repeat process until calculated numbers converge to a constant value: Iteration  Ca 2+  SO 4 2- [Ca 2+ ] (M)  (M) 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 CaSO 4 Step 7: Check Assumptions: With: Both [HSO 4 - ] and [CaOH + ] are ~ 5 times less than [Ca 2 +] and [SO 4 2- ]  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: KwKw K w = 1.0x K1K1 K sp K 1 = 3.8x10 2 K sp = 7.1x Step 2: Charge Balance:

51 [OH - ] = 2[Mg 2+ ] : 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 3: Mass Balance (tricky): But, two sources of OH -, [OH - ] = [H + ]: Account for both sources of OH - : : Species containing OH - Species containing Mg +

52 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 4: Equilibrium constant expression (one for each reaction): Proper to write equilibrium equations using activities, but complexity of manipulating activity coefficients is a nuisance. Most of the time we will omit activity coefficients

53 Step 5: Count equations and unknowns: Four equations: Four unknowns: 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 CB=MB (1) (2) (3) (4)

54 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 6: Solve (Not Easy!): Assumption to Reduce Number of Equations and Unknowns:  Solution is very basic [OH - ] >> [H + ], neglect [H + ] CB=MB Rearrange K 1 (ignore activity coefficients):

55 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 6: Solve (Not Easy!): Substitute K 1 into Mass or Charge Balance: Solve for [Mg 2+ ]:

56 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 6: Solve (Not Easy!): Substitute [Mg 2+ ] into K sp equation: Reduces to a single equation with a single variable: Solve using spreadsheet, vary [OH - ] until obtain correct value for K sp (7.1x )

57 Excel Demo of Goal Seek


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