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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

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Activity Introduction 2.)H 2 O exchanges rapidly between bulk solvent and ion-coordination sites

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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

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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

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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

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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

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

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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 = 1.3x10 -18 If Hg 2 (IO 3 ) 2 is placed in pure water, up to 6.9x10 -7 M will dissolve. If 0.050 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

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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

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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

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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.

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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

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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

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Activity Activity Coefficients from Debye-Hϋckel Equation

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Activity Equilibrium Constant and Activity 6.)Example 1: What is the activity coefficient of Hg 2 2+ in a solution of 0.033 M Hg 2 (NO 3 ) 2 ? Solution: Step 1 – Determine

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Activity Equilibrium Constant and Activity 6.)Example 1: What is the activity coefficient of Hg 2 2+ in a solution of 0.033 M Hg 2 (NO 3 ) 2 ? Solution: Step 2 – Identify Activity Coefficient from table at corresponding ionic strength. = 0.355 at = 0.10 M

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Activity Equilibrium Constant and Activity 6.)Example 2: What is the activity coefficient for H + at = 0.025 M? Note: Values for at = 0.025 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

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Activity Equilibrium Constant and Activity 6.)Example 2: What is the activity coefficient for H + at = 0.025 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

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Activity Equilibrium Constant and Activity 6.)Example 2: What is the activity coefficient for H + at = 0.025 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.

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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

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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

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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 KClO 4 ?

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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 0.00100M KCl, where and KCl acts as an “inert salt”? K sp = 5.6x10 -23

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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 0.00100M KBr?

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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

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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.

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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

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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

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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

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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!

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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 0.0250 mol of KH 2 PO 4 and 0.0300 mol of KOH in 1.00L: [H + ] = 5.1x10 -12 M[H 2 PO 4 - ] = 1.3x10 -6 M [K + ] = 0.0550 M[HPO 4 2- ] = 0.0220M [OH - ] = 0.0020M[PO 4 3- ] = 0.0030M Charge balance:

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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 0.050 M in water: Include ALL products in mass balance: H 3 PO 4 H 2 PO 4 -,HPO 4 2-, PO 4 3-

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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

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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 -.

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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

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Applying the Systematic Treatment of Equilibrium 1.)Example #1: Ionization of water KwKw K w = 1.0x10 -14 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

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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:

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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

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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 = 1.0x10 -14 K base K acid K ion pair K sp K base = 9.8x10 -13 K acid = 2.0x10 -13 K ion pair = 5.0x10 -3 K sp = 2.4x10 -5 This information is generally given:

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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!

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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

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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)

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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

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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

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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:

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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:

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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:

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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) 1110.00490.020 20.6280.6060.00790.032 30.5700.5420.00880.035 40.5560.5260.00910.036 50.5510.5200.00920.037 60.5470.5150.00920.037 Stop, concentrations converge

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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

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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 = 1.0x10 -14 K1K1 K sp K 1 = 3.8x10 2 K sp = 7.1x10 -12 Step 2: Charge Balance:

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[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 +

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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

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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)

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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):

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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+ ]:

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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 (7.1x10 -12 )

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Excel Demo of Goal Seek

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