Chemical Equilibria. General representation

Transcription

1 Chemical Equilibria General representation a A + b B c C + d D Where uppercase letters are chemical species and lowercase letters are coefficients (i.e. # of atoms or moles) 1

2 Equilibrium Constant [C] c [D] d K = [A] a [B] b where [ ] = concentration, usually molar 2

3 True Thermodynamic Equilibrium Constant (a o C ) c (a D ) d K = (a A ) a (a B ) b For a A + b B c C + d D K o Defined for standard conditions of 25 o C, 1 atm pressure and I = 0 (infinite dilution) 3

4 Equilibrium Constant [C] c [D] d K = [A] a [B] b where [ ] = concentration, usually molar 4

5 Many types of K s (equilibrium constants) K a for acid dissociation K b for base hydrolysis K w for water auto ionization K sp for solubility product K f for a formation constant K 1, K 2, K 3, etc. for stepwise formation constants β 1, β 2, β 3, etc. for overall formation constants 5

6 Solubility Equilibria Ba 2+ (aq) + SO 4 (aq) BaSO 4(s) or by convention BaSO 4(s) Ba 2+ (aq) + SO 4 (aq) We can write an equilibrium constant for rxn 6

7 Solubility Product (equilibrium constant) [Ba 2+ ] [SO 4 ] K sp = = [Ba 2+ ] [SO 4 ] 1 a Ba a SO4 K sp = = a Ba a SO4 1 activity of solid is defined as = 1 7

8 Solubility Calculated Solubility (S) is the concentration of individual ions generated from an insoluble compound BaSO 4(s) Ba 2+ (aq) + SO 4 (aq) S = [Ba 2+ ] = [SO 4 ] 8

9 Solubility Calculation (continued) Given K SP = [Ba 2+ ][SO 4 ] = 2.0 x Then S = K SP = 2.0 x = 1.4 x 10-5 M So S = [Ba 2+ ] = [SO 4 ] = 1.4 x 10-5 M 9

10 Activity Correction a Ba a SO4 K SP = = a Ba a SO4 Since 1 a Ba = γ Ba [Ba 2+ ] & a SO4 = γ SO4 [SO 4 ] Substituting K SP = a Ba a SO4 = γ Ba [Ba 2+ ]γ SO4 [SO 4 ] 10

11 Solubility Calculation Since (completed) K SP = γ Ba [Ba 2+ ]γ SO4 [SO 4 ] & γ Ba = γ SO4 Then K SP S = γ2 To determine solubility of BaSO 4 in a solution containing other ions (as in SW), you must calculate the activity coefficient (γ) 11

12 Two ways to correct for activity 1) Correct each ion as discussed K SP = a Ba a SO4 = γ Ba [Ba 2+ ]γ SO4 [SO 4 ] 2) Correct the equilibrium constant K K SP K = = [Ba 2+ ] [SO 4 ] γ 2 12

13 Common Ion Effect In seawater the total concentration of sulfate is 2.86 x 10-2 moles/kg must use here K SP = a Ba a SO4 = γ Ba [Ba 2+ ]γ SO4 [SO 4 ] K SP K = = [Ba 2+ ] [SO 4 ] γ 2 13

14 Water Hydrolysis (very important) H 2 O H + + OH - Applying same rules for K expressions a H+ a OH- K w = = a H+ a OH- Where H 2 O (the solvent) is assigned activity = 1 14

15 Remember ph ph is defined as the negative logarithm of the hydrogen ion activity ph = -log a H+ Given the numerical value K w = 1 x & K w = a H+ a OH- then we can always calculate OH - from the ph 15

16 ph Examples At neutral ph a H+ = a OH- and a H+ = K w = 1 x 10-7 = ph 7.00 At seawater ph (e.g., 8.2) a H+ = 1 x = 6.31 x 10-9 M K w 1 x a OH- = = = 1.58 x 10-6 M a H x

17 Hydronium Ion Water actually hydrolyses to form a hydronium ion (H 3 O + ) rather than the lone proton (H + ) (Once again an ion-water interaction akin to those discussed previously) For the sake of simplicity, we will refer to this species as H + which is common practice 17

18 A Note on Strong & Weak Electrolytes Salts, Acids & Bases are all ionic compounds that dissociate (i.e., form ions) in water either partially or completely Complete dissociation = a strong electrolyte NaCl H 2 O Na + + Cl - no equilibrium Partial dissociation = a weak electrolyte H 2 CO 3 H + + HCO 3 - K a1 HCO 3 - H + + CO 3 K a2 Two step equilibrium = forward & back reactions 18

19 Acid-Base Equilibria Fictitious Weak Acid (HA) HA H + + A - [H + ] [A-] a H + a A - K a = or [HA] a HA The smaller the K a the weaker the acid Strong acids have no K a it approaches infinity 19

20 Acid-Base Equilibria Fictitious Weak Base (B) Capable of accepting a proton (H + ) B + H 2 O BH + + OH - [BH + ] [OH - ] a BH + a OH - K b = or [B] a B The smaller the K b the weaker the base Strong bases have no K b it approaches infinity 20

21 Ion Pair or Complex Formation Equilibria Dozens of Ion Pairs form in SW & even more complexes deal with them the same way Mg 2+ (aq) + SO 4 (aq) MgSO 4(aq) a MgSO4 Kf = a Mg a SO4 Larger K f = stronger formation reaction 21

22 Typical Problem in SW Find Various Forms or Species Given total concentration data for certain constituents in SW, find % of species Example:If total Mg is C Mg = 5.28 x 10-2 mol/kg and total SO 4 is C SO4 = 2.82 x 10-2 mol/kg knowing that Mg 2+ (aq) + SO 4 (aq) MgSO 4(aq) and the value of the K f or K MgSO4 = 2.29 x

23 Steps in the Manual Solution of Simple Equilibrium Problems 1) Start with a recipe: C Mg = 5.28 x 10-2 mol/kg C SO4 = 2.82 x 10-2 mol/kg 2) List the species: Mg 2+, SO 4, MgSO 4 3) List reaction(s): Mg 2+ + SO 4 MgSO 4 4) Write Mass Balance equations: C Mg = [Mg 2+ ] + [MgSO 4 ] = 5.28 x 10-2 mol/kg C SO4 = [SO 4 ] + [MgSO 4 ] = 2.82 x 10-2 mol/kg 23

24 Steps in the Manual Solution of Simple Equilibrium Problems 5) Write a Charge Balance equation: Σ Z i+ [i + ] = Σ Z i- [i - ] 6) Write equilibrium constant expression(s): a MgSO4 [MgSO 4 ] K f = or a Mg a SO4 [Mg 2+ ] [SO 4 ] There are 3 species or 3 unknown concentrations There are also 3 equations (actually 4) to solve 24

25 We can solve the 3 equations simultaneously to get an answer Solve for free Mg concentration first = [Mg 2+ ] Rearrange the mass balance equations: C Mg = [Mg 2+ ] + [MgSO 4 ] rearranges to give [MgSO 4 ] = C Mg - [Mg 2+ ] C SO4 = [SO 4 ] + [MgSO 4 ] rearranges giving [SO 4 ] = C SO4 - [MgSO 4 ] We must also substitute the 1st into the 2nd 25

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27 C Mg = [Mg 2+ ] + [MgSO 4 ] rearranges to give [MgSO 4 ] = C Mg - [Mg 2+ ] C SO4 = [SO 4 ] + [MgSO 4 ] rearranges giving [SO 4 ] = C SO4 - [MgSO 4 ] Substituting the 1 st into the 2 nd for [MgSO 4 ] Gives [SO 4 ] = C SO4 - (C Mg - [Mg 2+ ]) Now we can [MgSO 4 ] K f = Substitute into K [Mg 2+ ] [SO 4 ] 27

28 Our resulting equation looks like C Mg - [Mg 2+ ] K MgSO4 = [Mg 2+ ] (C SO4 - (C Mg - [Mg 2+ ])) Be careful of signs in denomenator C Mg - [Mg 2+ ] K MgSO4 = [Mg 2+ ] (C SO4 - C Mg + [Mg 2+ ]) Cast in the form of a quadratic K[Mg 2+ ]C SO4 - K[Mg 2+ ]C Mg + K[Mg 2+ ] 2 = C Mg - [Mg 2+ ] Set equal to zero and solve with the quadratic formula 28

29 Equation from previous slide K[Mg 2+ ]C SO4 - K[Mg 2+ ]C Mg + K[Mg 2+ ] 2 = C Mg - [Mg 2+ ] Set equal to 0 & rearrange in form for quadratic formula K[Mg 2+ ] 2 + K[Mg 2+ ]C SO4 - K[Mg 2+ ]C Mg + [Mg 2+ ] - C Mg = 0 Gather terms K[Mg 2+ ] 2 + (KC SO4 - KC Mg + 1)[Mg 2+ ] - C Mg = 0 Remember the quadratic formula? 29

30 Equation from previous slide K[Mg 2+ ] 2 + (KC SO4 - KC Mg + 1)[Mg 2+ ] - C Mg = 0 Quadratic formula -b + b 2-4 a c x = a Solve for x which for us is [Mg 2+ ] where a = K b = (KC SO4 - KC Mg + 1) c = - C Mg 30

31 Solving this problem with the quadratic formula And substituting in the known values for: K f which equals K γ 2 f Where K f = K MgSO4 = 2.29 x 10 2 and γ = 0.23 C Mg = 5.28 x 10-2 mol/kg C SO4 = 2.82 x 10-2 mol/kg The answer is: x = [Mg 2+ ] = 4.35 x 10-2 mol/kg Since C Mg = 5.28 x 10-2 mol/kg then [Mg 2+ ] = 82 % 31

32 Activity Coefficient At typical ionic strengths for SW I = 0.5 to 0.7 From Davies Equation Mg 2+ activity coefficient ln γ = - A Z 2 [I 0.5 /(1 + I 0.5 ) 0.2 I] If Z = 2 & A = 1.17 then ln γ = & γ =

33 Calculate All Species Given C Mg = 5.28 x 10-2 mol/kg and C SO4 = 2.82 x 10-2 mol/kg We calculated [Mg 2+ ] = 4.35 x 10-2 mol/kg or 82 % By difference [MgSO 4 ] = 9.30 x 10-3 mol/kg or 18 % We can likewise calculate [SO 4 ] concentration & % C SO4 - [MgSO 4 ] = [SO 4 ] = 1.89 x 10-2 mol/kg 33

34 Problems Went through a moderately difficult calculation & only calculated species for 1 reaction in SW If considered more complicated equilibria where several reactions were going on, the math would quickly get out of hand Didn t consider any other reactions involving Mg or SO 4 that might influence our results 34

35 Other Problems Equilibrium constants can vary as much as 5% depending on the source Concentration data vary as well Activity corrections can also vary depending on the method used We only considered activity corrections for charged species, while neutral species may also have γ s that are non unity (e.g., MgSO 4 ) 35

36 Problems Mentioned Only 1 reaction, 1 set of species, simple equilibrium Didn t consider any other reactions involving Mg or SO 4 that might influence our results CO 3 Ca 2+ Na + Mg 2+ (aq) + SO 4 (aq) MgSO 4(aq) F - K + Zn 2+ Other reactions influence amount of MgSO 4 produced 36

37 Must Consider Other Reactions Beside Mg 2+ (aq) + SO 4 (aq) MgSO 4(aq) There are also Ca 2+ (aq) + SO 4 (aq) CaSO 4(aq) Na + (aq) + SO 4 (aq) NaSO 4 -(aq) K + (aq) + SO 4 (aq) KSO 4 -(aq) and Mg 2+ (aq) + CO 3 (aq) MgCO 3(aq) as well as others 37

38 Several Questions to Ponder 1) Based on the knowledge that there are other competing reactions in SW, is our calculation accurate? (82 % free Mg 2+?) 2) How do we know what other reactions are going on in SW that we should consider? 3) How do we include all the other equilibrium reactions that we might consider important? 4) How do we deal with the increased complexity of the mathematics? 5) Why did I take this course? 38

39 Answers to Question #1 Based on the knowledge that there are other competing reactions in SW, is our calculation accurate? (82 % free Mg 2+?) The calculation is only an estimate because we did not consider the formation of other species CO 3 Ca 2+ Na + Mg 2+ (aq) + SO 4 (aq) MgSO 4(aq) F - K + Zn 2+ their influence on the amount of MgSO 4 produced 39

40 Answers to Question #2 How do we know what other reactions are going on in SW that we should consider? We will largely rely on the literature for known reactions in SW (e.g., books & papers such as the handout provided this week). Chemical Intuition also helps 40

41 Answers to Question #3 How do we include all the other equilibrium reactions that we might consider important? Using our manual approach to solving equilibrium problems, we would have to expand our mass balance equations as well as add additional equilibrium constant expressions The number of species for which the concentration is unknown & the number of equations grows rapidly as we add equilibria. 41

42 Answers to Question #4 How do we deal with the increased complexity of the mathematics? Solving 5 or more equations simultaneously can only be handled in two ways: 1) Assumptions or approximations can be made to simplify the equations to something more manageable 2) Computer programs designed to solve ionic equilibrium problems can be used 42

43 Computer programs MINEQL+ we will use this exclusively MINTEQA2 EPA DOS version of MINEQL GEOCHEM-EZ geochemical modeling software Visual MINTEQ user friendly MINTEQA2 PHREEQC USGS modeling software wwwbrr.cr.usgs.gov/projects/gwc_coupled/phreeqc/ WinHumicV Includes humic binding model 43

44 Handout 44

45 Motekaitis & Martell (1987) Table I Note molar concentration units (M) & multiplier 45

46 Motekaitis & Martell (1987) Table I, Major Ions Note: These numbers have been multiplied by 10 3 or 1000 so they are millimolar (mm) or 479 is really M or 4.79 x 10-2 M 46

47 Motekaitis & Martell (1987) Table I, Trace Ions Note: These numbers have been multiplied by 10 9 and are nanomolar (nm) or 4 is really 4 nm or 4 x 10-9 M 47

48 Motekaitis & Martell give a long list of species with equilibrium constants Note: Values are Log β not K, also I (μ) =

49 The difference between β & K K is a stepwise formation constant [CdCl + ] Cd 2+ + Cl - CdCl + K 1 = = 1.0 x 10 [Cd 2+ ][Cl - ] 2 [CdCl 2 ] CdCl + + Cl - CdCl 2 K 2 = = 4.0 x 10 [CdCl + ][Cl - ] 0 β is an overall formation constant [CdCl + ] Cd 2+ + Cl - CdCl + β 1 = = 1.0 x 10 [Cd 2+ ][Cl - ] 2 [CdCl 2 ] Cd Cl - CdCl 2 β 2 = = 4.0 x 10 2 [Cd 2+ ][Cl - ] 2 49

50 Notes on β & K K 1 = β 1 (from previous slide) K 2 is not equal to β 2 (note denominators of each expression) β 2 = K 1 x K 2 [CdCl 2 ] [CdCl + ] [CdCl 2 ] β 2 = = x [Cd 2+ ][Cl - ] 2 [Cd 2+ ][Cl - ] [CdCl + ][Cl - ] β 3 = K 1 x K 2 x K 3 (etc.) 50

51 Motekaitis & Martell (1987) (values expressed as %) 51

52 Computer programs MINEQL+ we will use this exclusively Program still requires setting up the Equilibrium Problem Must list species of interest Must have total concentration data for each constituent Other needed information may include Ionic strength ph CO 2 /Carbonate 52

53 Steps in the MINEQL+ Solution of Simple Equilibrium Problems 1) Start with a recipe: C Mg = 5.28 x 10-2 mol/kg C SO4 = 2.82 x 10-2 mol/kg 2) List the species: Mg 2+, SO 4, MgSO 4 3) Run the program 4) Interpret the results 53

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