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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 • 2.)H2O exchanges rapidly between bulk solvent and ion-coordination sites**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**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**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**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**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+**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**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**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**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.**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**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**Activity**Activity Coefficients from Debye-Hϋckel Equation**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**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.**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**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**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.**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**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**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?**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**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**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**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.**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**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**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+]**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!**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:**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-**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**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-.**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**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!**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)**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**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:**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!**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):**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:**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**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**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**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**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:**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**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**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: