8.00 Activity and Systematic Treatment of Equilibrium

Size: px

Start display at page:

Download "8.00 Activity and Systematic Treatment of Equilibrium"

James Barker
5 years ago
Views:

1 8.00 Activity and Systematic Treatment of Equilibrium Recommended Problem Set: 7.7, 7.12, 7.18, 7.19, 7.27 Dr. Fred Omega Garces Chemistry 251 Miramar College 1

2 Effect of Electrolyte on Equilibrium The position of solution equilibrium depends on the concentration of the electrolyte. Even when the electrolyte is not part of the mass action expression. Consider three systems at equilibrium at various concentration of NaCl: A) 2 H 2 O (l) D H 3 O + (aq) + OH - (aq) g B) BaSO 4 (s) D Ba 2+ (aq) + SO 4 2- (aq) g K sp C) HCH 3 CO 2 (aq) D H 3 O + (aq) + CH 3 CO 2 - (aq) g K a The equilibrium constant increases as the concentration of NaCl electrolyte increases. K eq = [Product]i [Reactant] j The increase in K eq an increase in solubility or the decrease of product ion formation back to the reactant. 2

3 Formation of Fe(SCN)2+ Ions and molecules in solution are surrounded by solvent molecules that can impede reaction process. Consider the formation of Fe(SCN)2+ : Fe+3 + SCN- D Fe(SCN)2+ The graph shows the Kf (equilibrium constant ) is influence by the concentration of KNO3, a supposedly inert reagent. The higher the KNO3 conc. The less efficient the formation of Fe(SCN)2+. The concentration of KNO3 changes the ionic strength of the solution which influences the formation of the complex, Fe(SCN)2+ The reaction: CaSO4 (s) D Ca+2 + SO42- is also influence by the concentration of KNO3 in solution. Addition of KNO3 increases the solubility of the CaSO4 salt. This is due to the K+ and NO3- formation of ionic atmosphere that prevents effective attraction between Ca+2 and SO42-, thus preventing the formation of the CaSO4 precipitate (increasing solubility). The more ions K+ and NO3- in solution, the greater the solubility, the less likely Ca+2 ions and SO42- ions want to combine. The amount of ions in solution is called the ionic strength µ. (Not to be confused with population mean µ) 3

4 The Explanation The electrolyte effect (salt effect) results from electrostatic attractive and repulsive force between the electrolyte ions and other ions involve in the equilibrium. BaSO 4 (s) Ba 2+ + SO 2- Cl - 4 Cl - Cl - Na + Cl - Na + Cl - Na + Na + Na + The charge layer around each ion decreases the overall attraction between Ba 2+ and SO 2-4 resulting in an increase in solubility or an increase in the K sp. This trend increases as the number of electrolyte become larger; the effective concentration of barium ions and of sulfate ion (unsolvated ions of barium and sulfate) decrease as the ionic strength of the medium increases. 4

5 Ionic strength Magnitude of equilibrium constant is dependent highly on charges of the chemical specie involved in the equilibrium. Neutral species do not affect the equilibrium from ideal behavior. Ionic specie affects the equilibrium constant as a function of the ions charge. The higher the charge of the species, the less ideal the solution behaves in terms of the mass action expression. Graph shows deviation of BaSO 4 compared to AgCl. Ionic strength µ is the total concentration of ions in solution. ionic strength= µ = 1 ( 2 [C ]Z [C 2 ]Z 2 2 +[C 3 ]Z ) where, [C 1 ],[C 2 ],[C 3 ]... represent the speice molar concentration [Z A ],[Z B ],[Z C ]... represent their charges 5

Ionic strength, calculations Ionic strength calculations: i) Calculate the ionic strength (µ) of 0.10M Na 2 SO 4 µ =1/2 { [Na + ] 1 2 + [SO 2-4 ] 2 2 } ; µ =1/2 { [2x.10] 1 2 + [.10] 2 2 } = 0.

6 Ionic strength, calculations Ionic strength calculations: i) Calculate the ionic strength (µ) of 0.10M Na 2 SO 4 µ =1/2 { [Na + ] [SO 2-4 ] 2 2 } ; µ =1/2 { [2x.10] [.10] 2 2 } = 0.30 M ii) Calculate the ionic strength (µ) of 0.05 M KNO 3 and 0.10M Na 2 SO 4 µ =1/2 { [C K ] K + chg 2 + [C KNO3 ] NO 3 - chg 2 + [C Na2 ] Na + chg 2 + [C SO4 ] SO 4-2 chg2 } µ =1/2([.05] [.05] [.20] [.10] 2 2 ) = 0.35 M In general, µ = i Total * C, Where, i Total is the total ions (electrolyte) and C is the analytical molar (F) concentration. Ionic strength greatest for chemical with largest charge. 6

Ionic strength 9.7c Calculate the ionic strength of a solution that contains 0.10 M FeCl 3 and 0.20 M FeCl 2. ionic strength = µ = 1 ( 2 [C ]Z 2 + [C 1 1 2 ]Z 2 2 + [C 3 ]Z 2 3 +.

7 Ionic strength 9.7c Calculate the ionic strength of a solution that contains 0.10 M FeCl 3 and 0.20 M FeCl 2. ionic strength = µ = 1 ( 2 [C ]Z 2 + [C ]Z [C 3 ]Z ) ions formed: FeCl 3 (aq) Fe +3 (aq) + 3Cl - (aq) Conc of ions: = 0.30 ions formed: FeCl 2 (aq) Fe +2 (aq) + 2Cl - (aq) Conc of ions: = 0.40 µ = 1 2 [.10M] (3) 2 + [.30M](1) 2 Fe+3 Cl- + [.20M](2) 2 2 ( Fe+2 + [.40M](1) Cl- ) µ = 1.2 7

8 Activity Coefficients The effect of electrolyte on the equilibrium is taken into account by the activity,a. ax = [C] g x Where, ax is the activity of species X [C] is the molar concentration g x is the dimensionless activity coefficient Thus, for X m Y n, (X m Y n D mx + ny) concentration solubility product; K sp = [X] m [Y] n Classic Mass Action Thermodynamic solubility product; K sp = a m x a n y or ; K sp = [X] m [Y] n g m x g n y = K sp g m x g n y 8

9 Properties of Activity Coefficient (g). 1. a is a measure of effectiveness. In dilute solutions, µ (ionic strength) is small, g ~ 1,thus a ~ [X] (concentration). See figure below. 2. Solution not too concentration g (activity coefficient) is independent of type electrolyte and depends only in µ. 3. g departs from unity as the charge of species increase. 4. At any µ, g x = g y if X and Y have the same charge Dilute Solution 5. g = 1 of for neutral chemicals regardless of µ. 6. g x applies to all equilibria in which that ion participates. 9

for X Z x = charge of X µ = ionic strength of solution a x = effective diameter of the hydrated ion X (pm) When µ is less than 0.

10 Debye-Huckel: Equation for g Model based on ionic atmosphere Equation to calculate activity coefficient, g log γ = 0.51 Z 2 µ x x α x µ or Z log γ = x µ 1 + (α x µ / 305) Where, g x = activity coeff. for X Z x = charge of X µ = ionic strength of solution a x = effective diameter of the hydrated ion X (pm) When µ is less than 0.01, 1 + (µ) 1/2 ~ 1 (denominator) and the eqn. reduces to logg x = Z x2 (µ) 1/2 This is known as the: (DHLL) Debye-Huckel Limiting Law and applies at very low µ 10

11 Table: Activity Coefficient (g) for aqueous soln Be sure you also know how to estimate between u values. Linear interpolation is sufficient in many cases. 11

Debye-Huckel Interpolate for g To find activity coefficient for an ionic strength between values, use the Debye-Huckel Equation or you can interpolate values from the table.

12 Debye-Huckel Interpolate for g To find activity coefficient for an ionic strength between values, use the Debye-Huckel Equation or you can interpolate values from the table. Or linear interpolation can be applied. If values are between two entries of a table, then assume a linear relationship. For example, consider a table in which y = 0.67 when x = 10 and y = 0.83 when x = 20. What is the value of y when x = 16? For x = 16, our estimate of y is

13 Mean Activity Coefficients The Mean activity coefficient of electrolyte A m B n is defined as g + = mean activity coefficient = (g Am g Bn ) 1/(m+n) The contributor to the mean activity coefficient are impossible to measure individually since g Am and g Bn are specific to an ion. Experimentally a cation ion cannot exist without an anion so one cannot be independently measured without the other. The activity coefficient measures the deviation of behavior from ideal. The activity coefficient allows the correct form of the mass action expression to be used in equilibrium equations. K = α c d α C D α a A α = [C ] c c γ C b B [A] a a γ A [D] d γ D d [B] B γ B b 13

Calculation of Activity Coefficient. 9.8d Determine the activity coefficient g of Sn 4+ at µ = 0.060 log γ = 0.

14 Calculation of Activity Coefficient. 9.8d Determine the activity coefficient g of Sn 4+ at µ = log γ = 0.51 Z x 2 µ 1 + (α x µ / 305) at 25 Chemical Z µ (µ) 1/2 a* log g g Sn *Table

15 Equilibrium Calculations and Activity Coefficient Calculate the % relative error in solubility (s) by using concentration instead of activities for Ba(IO3)2 in M Mg(IO3)2, Ksp for Ba(IO3)2 is (ApxF, AP10) Factoring in activities: K sp (Thermodynamics) = MgIO 3! 1Mg IO-3 ; "Mg2+ $ = 0.33M, "IO-3 $ = 0.066M # % # % 1 µ = [0.033] 22Mg + [0.066] 12IO3 = thus, using table 7.1 γ +2 =.38, γ =.775 ( ) Ba IO K sp = a K sp = γ Ba +2 Ba +2 (a (γ 3 IO )2 = γ 3 IO 3 Ba +2 [Ba +2] (γ )2 ([Ba +2] [IO-3]2 ) IO 3 [IO-3] ) i solid Δ [Eq] -s constant +2 Ba (aq) + 0 +s s 2 IO(aq) s 2s s = Molar solubility, s= M = ([Ba +2] [IO-3]2 ), note that IO-3 = {from Mg(IO 3 ) 2 } + 2[Ba +2] 0.066M Ba(IO 3 )(s) 2 K sp = = [Ba +2][IO 3]2 = s (0.066)2 K sp = = ([Ba +2] [IO-3]2 ) Neglecting activities, 2 = [Ba +2]=s = Relative Error: relative error = ( ) = 77%

16 Equilibrium Calculations and Activity Coefficient Calculate the % relative error in solubility (s) by using concentration instead of activities for Ba(IO 3 ) 2 in M Mg(IO 3 ) 2, K sp for Ba(IO 3 ) 2 is (ApxF, AP10) Factoring in activities: K sp (Thermodynamics) = MgIO 3 1Mg IO 3 - µ = 1 2 [0.033] ( + [0.066] 1 Mg IO3 ) = 0.10 thus, using table 8.1 γ Ba +2 =.38, γ IO 3 =.775 K sp = a Ba +2 (a IO 3 )2 = γ Ba +2 [Ba+2 ] (γ IO3 [IO 3 K sp = γ Ba +2 (γ IO 3 )2 ([Ba +2 ] [IO 3 - ] 2 ) ; # Mg $ 2+ % & = 0.33M, # IO - % $ 3& = 0.066M - ] ) 2 K sp = = ([Ba +2 ] [IO 3 - ] 2 ) = 2 ([Ba+2 ] [IO - 3 ] 2 ), note that IO 3 - = {from Mg(IO 3 ) 2 } + 2[Ba +2 ] 0.066M = [Ba+2 ]=s = Neglecting activities, Ba(IO 3 ) 2(s) Ba +2 (aq) + 2 IO 3 (aq) i solid Δ -s +s +2s [Eq] constant s 2s K sp = = [Ba +2 ][IO 3 ] 2 = s (0.066) 2 s = Molar solubility, s= M Relative Error: relative error = ( ) = 77% 16

17 Equilibrium Calculations and Activity Coefficient Calculate the % relative error in solubility (s) by using concentration instead of activities for the following compound in M KNO3. Ksp for Fe(OH)3 is Factoring in activities: K sp (Thermodynamics) = µ= 1 Neglecting activities, Fe(OH)(s) 3 ([0.0500]1 + [.05000]1 ) = thus, γ +3 =.245, γ Fe K sp = a K sp = γ Fe+3 Fe (a (γ +3 OH 2 2 K NO OH s = s constant + 3OH(aq) 0 +s 0 +3s s 3s s = )3 = γ Fe+3 +3 [Fe +3] (γ OH [OH- ] )3 Molar solubility, s= M OH ([Fe ] [OH ] ) = Δ +3 Fe(aq) K sp = = [Fe +3][OH- ]3 = s (3s )3 = 27s 4 =.81 )3 ([Fe ] [OH- ]3 ) solid [Eq] 3 K sp = = ([Fe +3] [OH- ]3 ) +3 i Relative Error:, note that s = [Fe ], 3s=[OH ] +3 - relative error = ( ) = 0.399% = = s 3s3 = 27s =

76 = 1 10-14 = ([H + ] 0.83) ( [OH- ] 0.76) 1 10-14 = ([x] 0.83) ([x] 0.76) (.83) (.76) = x2 x = 1.

18 ph Revisited ph = - log a H+ = - log [H + ] g H+ What is the true ph of pure water in 0.10 M KCl? = = [H + ] γ H + [OH- ] γ OH let [H+ ]=[OH - ] = x for µ = 0.1, γ H + = 0.83, γ OH = 0.76 = = ([H + ] 0.83) ( [OH- ] 0.76) = ([x] 0.83) ([x] 0.76) (.83) (.76) = x2 x = M = [H + ] = [OH - ] ph = -log α H + α H + = [H + ]γ H + = = ph = -log =

19 Summary: Omitting Activity Coefficient For most problems activity coefficients can be ignored and molar concentration is applied for equilibrium problems. This approximation simplifies the calculations. The activity coefficient will provide the necessary information if neglecting the activities will result in large errors. 19

20 Systematic Treatment of Equilibrium The systematic treatment of equilibrium is the best way to deal with all types of equilibrium problems. The steps are- 1 Balance Equation for all pertinent equilibria 2. Goal of problem in terms of equilibrium concentrations 3. Law of Mass action 4. Charge-balance expression 5. Mass-balance expression 6. Check for unknown to # of equations 7a Approximation method if possible 7b. Solve simultaneous equation for exact soln. 8. Solve manually the algebraic equation 9. Check approximation. 20

21 Charge and Mass Balance Electrical neutrality for all samples must be obeyed. The sum of the positive charges in solution equals the sum of negative charges. S (positive) = S (negative) Law of Conservation of Mass must be obeyed. Material or mass balance is the statement that matter is neither created or destroyed. The quantity of all species in a solution containing a particular atom (or group of atoms) must equal the amount of atoms (or group of atoms) delivered to the solution. S (reactant) = S (products) 21

Charge Balance: KH2PO4 and KOH Charge balance is an algebraic statement of electro neutrality. The sum of the positive charges in solution equals the sum of negative charges.

22 Charge Balance: KH2PO4 and KOH Charge balance is an algebraic statement of electro neutrality. The sum of the positive charges in solution equals the sum of negative charges. S (positive) = S (negative) Consider a solution containing KH2PO4 and KOH. [H+] + [K+] = [H2PO4-] + 2 [HPO4-2] + 3 [PO4-3] + [OH-] Example: Write the charge balance for a solution containing mol KH2PO4 and mol KOH diluted to 1 L. * [H+] + [K+] = [H2PO4-] + 2 [HPO4-2] + 3 [PO4-3] + [OH-] [ M] + [.0550M] = [ M] + 2 [.0020M] + 3 [.0030M] + [.0020M] [.0550M] = [.0550M] 22 * Conc. is to be calculated after studying acid/base.

Mass Balance Mass balance is a statement on the concentration of the solution.

S(reactant) = S (products) S Consider a solution containing 0.050 M acetic acid in 1.0 L water.

050M] = [CH 3 CO 2 H] + [CH 3 CO 2- ] Example: Write the mass balances for potassium and for

23 Mass Balance Mass balance is a statement on the concentration of the solution. The concentration of the ions in solution is equal sum of the concentration of solution. S(reactant) = S (products) S Consider a solution containing M acetic acid in 1.0 L water. CH 3 COOH = CH 3 CO 2- + H + [0.050M] = [CH 3 CO 2 H] + [CH 3 CO 2- ] Example: Write the mass balances for potassium and for phosphate in solution by mixing M KH 2 PO 4 and mol KOH. Total [K + ] = [0.0250M] + [0.030] = [0.0550M] PO 4 ] + [H 2 PO 4- ] + [HPO -2 4 ] + [PO -3 4 ] = [0.0250M] 23

24 Applying Systematic Treatment of Equilibrium 7.26 Following the example of ammonia in Section 7.5 (Harris 8 th ed), write the equilibria and charge and mass balances needed to find the composition of 0.01 M NaC 2 H 3 O 2, (NaAc) Include activity when appropriate. Pertinent Equation: Remember that NaAc dissociates to Na + + Ac - Eq1 K b reaction: Ac - (aq ) K + H 2 O b (l) ## HAc (aq) + OH - (aq) K b = = [OH- ] γ OH - [HAc] γ HAc [Ac - ] γ Ac - K Eq2 reaction: 2 H 2 O w (l) ## H 3 O + (aq) + OH - (aq) = = [OH - ] γ OH - O+ ] γ H3 O + Eq3 Charge Bal: [Na + ] + O + ] = [OH - ] + [Ac - ] Eq4 Mass Bal 1: 0.01 M = [Na + ] Eq5 Mass Bal 2: 0.01 M = [HAc] + [Ac - ] Unknowns: [HAc], [Ac - ], [OH - ], O + ], [Na + ] Neglecting activity, set 0.01M = F, reduce the equations from part a to one equation in terms of O + ]. From Eq2: [OH - ] = O + ] From Eq1: K b = [OH- ] [HAc] [Ac - ] [HAc] = K b [Ac - ] [OH - ] From Eq5: 0.01 M= F = [HAc] + [Ac - ] [Ac - ] = F - [HAc] = F - K b [Ac - ] [Ac - ](1 + K b O + ] ) = F [Ac - ] = F 1 + K b O + ] [OH - ] [Ac - [Ac - ] ] + K b # K & % w ( % $ O + ]( ' = [Ac - ] + K b [Ac - ] O + ] = F Sub into Eq3: [Na + ] + O + ] = [OH - ] + [Ac - ] F + O + ] = O + ] + F 1 + K b O + ] One Equation one unknown: F = O + ] + F 1 + K b O + ] - O + ] 0 = O + ] + F 1 + K b O + ] - O + ] - F 24

25 Applying Systematic Treatment of Equilibrium 7.26 Following the example of ammonia in Section 7.5 (Harris 8 th ed), write the equilibria and charge and mass balances needed to find the concentration of specie for 0.01 M NaC 2 H 3 O 2, (NaAc) Include activity when appropriate. Pertinent Equation: Remember that NaAc dissociates to Na + + Ac - Eq1 K b reaction: Ac - (aq ) K + H 2 O b (l) ## HAc (aq) + OH - (aq) K b = = [OH- ] γ OH - [HAc] γ HAc [Ac - ] γ Ac - K Eq2 reaction: 2 H 2 O w (l) ## H 3 O + (aq) + OH - (aq) = = [OH - ] γ OH - O+ ] γ H3 O + Eq3 Charge Bal: [Na + ] + O + ] = [OH - ] + [Ac - ] Eq4 Mass Bal 1: 0.01 M = [Na + ] Eq5 Mass Bal 2: 0.01 M = [HAc] + [Ac - ] Unknowns: [HAc], [Ac - ], [OH - ], O + ], [Na + ] Neglecting activity, set 0.01M = F, reduce the equations from part a to one equation in terms of O + ]. From Eq2: [OH - ] = O + ] From Eq1: K b = [OH- ] [HAc] [Ac - ] [HAc] = K b [Ac - ] [OH - ] From Eq5: 0.01 M= F = [HAc] + [Ac - ] [Ac - ] = F - [HAc] = F - K b [Ac - ] [Ac - ](1 + K b O + ] ) = F [Ac - ] = F 1 + K b O + ] [OH - ] F = [Ac - [Ac - ] ] + K b # K & % w ( % $ O + ]( ' = [Ac - ] + K b [Ac - ] O + ] = F Sub into Eq3: [Na + ] + O + ] = [OH - ] + [Ac - ] F + O + ] = O + ] + F 1 + K b O + ] One Equation one unknown: F = O + ] + F 1 + K b O + ] - O + ] 0 = O + ] + F 1 + K b O + ] - O + ] - F 25

Homework, 7.7, 7.12 7.7 Calculate the activity coefficient of Al 3+, µ = 0.

18 ] =.065 Dµ [.05.083 ] =.033 Al 3+ Dg [.245.083 ] = x x =.065 (.033/.050) = 0.

012 M LiNO 3. What would be the ph if you neglect activities. µ = 1 2 [0.

010 (1)2 + 0.010 (1) 2 + 0.012 (1) 2 + 0.012 (1) 2 ] =.022 $ γ H + =.914 -.0540.

0 10-14 = [H + ] γ H + + [OH - ] γ OH - = [H + ] 0.8978 [0.010] 0.873 = 0.900 -.

26 Homework, 7.7, Calculate the activity coefficient of Al 3+, µ = M by linear interpolation in Table 8-1. Dµ [.05.1 ] =.050 Al 3+ Dg [ ] =.065 Dµ [ ] =.033 Al 3+ Dg [ ] = x x =.065 (.033/.050) = g Al 3+ (µ =.083) = = γ Al 3+ = # & $ % ' ( ( ) = Using activities, calculate the ph of a solution containing M NaOH plus M LiNO 3. What would be the ph if you neglect activities. µ = 1 2 [0.010 (Na+ ) (OH - ) (Li + ) (NO 3 - ) 2 ] µ = 1 2 [0.010 (1) (1) (1) (1) 2 ] =.022 $ γ H + = ' $ & ) %&.040() = = γ.012 ' = &.090 ) OH %&.040(). = = [H + ] γ H + + [OH - ] γ OH - = [H + ] [0.010] = = [H + ] = [0.010] = ; ph = -log α H + ph = -log [ ] = -log [ ] =

Homework, 7.18, 7.19, 7.27 7.18 Write a charge balance for a solution containing: H +, OH -, Ca 2+, HCO 3-, CO -2 3, Ca(HCO 3 ) +, Ca(OH) +, K +, ClO 4-.

19 Write a charge balance for a solution containing: H 2 SO 4 [H + ] + = [HSO 4- ] + 2 [HSO 4-2 ] + [OH - ] 7.XX What is the ph of KOH solution that is 1.0 10-8 M Pertinent Equations: H 2 O!

27 Homework, 7.18, 7.19, Write a charge balance for a solution containing: H +, OH -, Ca 2+, HCO 3-, CO -2 3, Ca(HCO 3 ) +, Ca(OH) +, K +, ClO 4-. [H + ] + 2[Ca +2 ] + [Ca(HCO 3 ) + ] + [Ca(OH) + ] + [K + ] = [OH - ] + [HCO 3- ] + 2 [CO -2 3 ] + [ClO 4- ] 7.19 Write a charge balance for a solution containing: H 2 SO 4 [H + ] + = [HSO 4- ] + 2 [HSO 4-2 ] + [OH - ] 7.XX What is the ph of KOH solution that is M Pertinent Equations: H 2 O! H + + OH - Charge Balance: [K + ] + [H + ] = [OH - ] Three equations 3 unknown [K + ] + [H + ] = [OH - ] [K + ] = M = = [H + ] γ + [OH- ] γ H + OH - Mass balance: [K + ] + [H + ] = [OH - ] Already a charge balance [K + ] = M Solve : [ M] + [H + ] = [H + ] Equilibrium Constant: = = [H + ] γ H + + [OH- ] γ OH - µ = M γ = 1 [ M] [H + ] + [H + ] 2 = = Solving quadratic; [H + ] =

Homework, 7.27 7.27 Following the example of Mg(OH) 2 in section 7.5, write the equation needed to find the solubility of Ca(OH) 2. Include activity coefficient when appropriate. K sp Ca(OH) 2 (s)!

28 Homework, Following the example of Mg(OH) 2 in section 7.5, write the equation needed to find the solubility of Ca(OH) 2. Include activity coefficient when appropriate. K sp Ca(OH) 2 (s)! Ca OH - K sp = [Ca 2+ ]γ Ca 2+ [OH - ] 2 2 γ OH - = K 1 Ca 2+ + OH -! Ca(OH) + K 1 = [CaOH + ] γ CaOH + [Ca 2+ ] γ Ca 2+ [OH - ] γ OH - = Kw H 2 O! H + + OH - = = [H + ] γ H + + [OH - ] γ OH - = Charge balance : 2[Ca 2+ ] + [CaOH + ] + [H + ] = [OH - ] Mass balance [OH - ] Species containing OH- = [Ca 2+ ] Species containing Ca2+ { } + [H + ] [OH - ] + [CaOH + ] = 2 [Ca 2+ ] + [CaOH + ] Four equations and four unknowns For the approximation method, activity, α is ignored. After substituting, the solubility equation reduces to- K sp = [Ca 2+ ] [OH - ] 2 = [OH - ] K 1 [OH - ] Excel s Goal Seeker: Ca(OH)2 Solubility Ksp = [OH-] guess = [OH-]^3/(2+K1[OH-]) = 6.50E E E-06 K1 = 2.00E+01 [Ca+2] = H+ = Kw/[OH-]= 1.13E E-03 D24=C24^3/(2+A26*C24) C27 = A24/C24^2 D27=A26*C27*C24 28

Ch.7 ACTIVITY & SYSTEMATIC TREATMENT OF EQ

Ch.7 ACTIVITY & SYSTEMATIC TREATMENT OF EQ 7. Activity Effective concentration under ionic surroundings 7-. The effect of Ionic Strength on Soluility of salts. * Consider a saturated sol. Hg (IO 3 ), concentration