Chemical Oceanography Dr. David K. Ryan Department of Chemistry University of Massachusetts Lowell & Intercampus Marine Sciences Graduate Program University of Massachusetts http://faculty.uml.edu/David_Ryan/84.653 1
Ion-Ion Interactions Many types – non-specific, bonding, contact, solvent shared, solvent separated Non-specific i.e., long range interactions and the concepts of ionic strength, activity & activity coefficient Specific interactions e.g. complexation, ion pairing (strong or weak) Millero cartoons http://fig.cox.miami.edu/~lfarmer/MSC215/MSC215.HTM 2
Long Range (Non-Specific) Attraction δ – Oriented Non-specific Outward Interactions - electrostatic in nature & limit Long Range effectiveness of (Non-Specific) the ion Repulsion δ + Oriented 3 Outward
Non-specific Interaction Electrostatic in nature Limits effectiveness of ion in solution Use concept of activity to quantify effect (accounting for non-ideal behavior in solution) a i = [i] F γ F (i) where a i = activity of ion i [i] F = free ion conc. (m or M) γ F (i) = activity coefficient a = [i] γ In short of ion I ( < 1) 4
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) 5
Equilibrium Constant [C] c [D] d K = --------------- [A] a [B] b where [ ] = concentration, usually molar 6
True Thermodynamic Equilibrium Constant o (a 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) 7
Equilibrium Constant [C] c [D] d K = --------------- [A] a [B] b where [ ] = concentration, usually molar 8
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 9
Solubility Equilibria Ba 2+ 2- (aq) + SO 4 BaSO 4(s) (aq) or by convention BaSO 4(s) Ba 2+ (aq) + SO 4 2- (aq) We can write an equilibrium constant for rxn 10
Solubility Product (equilibrium constant) [Ba 2+ ] [SO 42- ] K sp = ------------------ = [Ba 2+ ] [SO 42- ] 1 a Ba a SO4 sp = ------------------ = a Ba a K SO4 1 activity of solid is defined as = 1 11
Solubility Calculated Solubility (S) is the concentration of individual ions generated from an insoluble compound Ba 2+ 2- BaSO 4(s) (aq) + SO 4 (aq) S = [Ba 2+ ] = [SO 4 2- ] 12
Solubility Calculation (continued) K SP = [Ba 2+ ][SO 4 2- ] = 2.0 x 10 -10 Given Then S = √ K SP = √ 2.0 x 10 -10 = 1.4 x 10 -5 M S = [Ba 2+ ] = [SO 4 2- ] = 1.4 x 10 -5 M So 13
Activity Correction a Ba a SO4 K SP = --------------- = a Ba a SO4 1 Since a Ba = γ Ba [Ba 2+ ] & a SO4 = γ SO4 [SO 4 2- ] Substituting K SP = a Ba a SO4 = γ Ba [Ba 2+ ] γ SO4 [SO 4 2- ] 14
Solubility Calculation (completed) Since K SP = γ Ba [Ba 2+ ] γ SO4 [SO 4 2- ] & γ Ba = γ SO4 K SP Then S = ------- √ γ 2 To determine solubility of BaSO 4 in a solution containing other ions (as in SW), you must calculate the activity coefficient ( γ ) 15
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- ] 2) Correct the equilibrium constant K K SP K ´ = --------- = [Ba 2+ ] [SO 4 2- ] γ 2 16
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 2- ] K SP K ´ = --------- = [Ba 2+ ] [SO 4 2- ] γ 2 17
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- 1 Where H 2 O (the solvent) is assigned activity = 1 18
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 10 -14 & K w = a H+ a OH- then we can always calculate OH - from the pH 19
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 10 -8.2 = 6.31 x 10 -9 M K w 1 x 10 -14 a OH- = -------- = ------------ = 1.58 x 10 -6 M a H+ 6.31 x 10 -9 20
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 21
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 + + HCO 3 H 2 CO 3 - K a1 H + + CO 3 HCO 3 - 2- K a2 Two step equilibrium = forward & back reactions 22
Acid-Base Equilibria Fictitious Weak Acid (HA) H + + A - HA [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 23
Acid-Base Equilibria Fictitious Weak Base (B) Capable of accepting a proton (H + ) BH + + OH - B + H 2 O [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 24
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 2- MgSO 4(aq) (aq) a MgSO 4 K f = ---------------- a Mg a SO4 Larger K f = stronger formation – reaction 25
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 2- MgSO 4(aq) (aq) and the value of the K f or K MgSO4 = 2.29 x 10 2 26
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 2- , MgSO 4 3) List reaction(s): Mg 2+ + SO 4 2- MgSO 4 4) Write Mass Balance equations: C Mg = [Mg 2+ ] + [MgSO 4 ] = 5.28 x 10 -2 mol/kg 2- ] + [MgSO 4 ] = 2.82 x 10 -2 mol/kg C SO4 = [SO 4 27
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 MgSO 4 [MgSO 4 ] K f = ----------------- or ------------------- a Mg a SO 4 [Mg 2+ ] [SO 4 2- ] There are 3 species or 3 unknown concentrations There are also 3 equations (actually 4) to solve 28
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 2- ] + [MgSO 4 ] rearranges giving [SO 4 2- ] = C SO4 - [MgSO 4 ] We must also substitute the 1st into the 2nd 29
30
C Mg = [Mg 2+ ] + [MgSO 4 ] rearranges to give [MgSO 4 ] = C Mg - [Mg 2+ ] C SO4 = [SO 4 2- ] + [MgSO 4 ] rearranges giving [SO 4 2- ] = C SO4 - [MgSO 4 ] Substituting the 1 st into the 2 nd for [MgSO 4 ] Gives [SO 4 2- ] = C SO4 - (C Mg - [Mg 2+ ]) Now we can [MgSO 4 ] K f = ------------------- 2- ] Substitute into K [Mg 2+ ] [SO 4 31
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 32
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 ? 33
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 = ------------------------ 2 a Solve for x which for us is [Mg 2+ ] where a = K b = (KC SO4 - KC Mg + 1) c = - C Mg 34
Solving this problem with the quadratic formula And substituting in the known values for: K f ´ which equals K f γ 2 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 % 35
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 = - 1.47 & γ = 0.23 36
Recommend
More recommend
