7-4 Systematic Treatment of Equilibrium

Transcription

1 7-4 Systematic Treatment of Equilibrium The equilibrium problem can be solved by working n equations and n unknowns. n equations n-2 : chemical equilibrium conditions 2 : Charge balance Mass balance 1) Charge Balance Solutions must have zero total charge positive charges = negative charges n 1 [C 1 ] + n 2 [C 2 ] + = m 1 [A 1 ] + m 2 [A 2 ] + [C i ], n i = concentration, charge of i th cation [A i ], m i = concentration, charge of i th anion * Activity coefficients do not appear in the charge balance.

2 7-4 Systematic Treatment of Equilibrium 2) Mass Balance Example) When MA and HA both are added to a solution C T,A = [HA] + [A - ] C T,A : sum of the moles of HA and MA added per liter of solution * Activity coefficients do not appear in the mass balance * It really refers to conservation of atoms, not to mass

3 Example) Analysis of Vittel (France) : Mineral Water mg/l M.W. mm Na K Mg Ca mm Error = (0.05/13.27) 100 = 0.38% HCO Cl NO SO mm * Water Analysis : ± 2% Wastewater or Sea water Analysis : ± 5% generally acceptable

4 Systematic Treatment of Equilibrium General Prescription Step 1 : Write all the pertinent chemical equations. * Because of not knowing all the chemical reactions, we undoubtedly oversimplify many equilibrium problems. Step 2 : Write all the charge balance Write the mass balance Step 3 : Write the equilibrium constant for each chemical reaction. - This is one step in which activity coefficients appear. - Although it is proper to write all equilibrium constants in terms of activities, the algebraic complexity of manipulating the activity coefficients often obscures the chemistry of a problem

5 7-5 Applying the systematic Treatment of Equilibrium The Dependence of Solubility on ph. (Coupled Equilibria ; the product of one reaction is a reactant in the next reaction) i) CaF 2 (s) Ca F - K sp = [Ca 2+ ][F - ] 2 = ii) F - + H 2 O HF(aq) + OH - iii) H 2 O H + + OH - K w = [H + ][OH - ] = iv) Mass Balance ; [F - ] + [HF] = 2[Ca 2+ ] v) Charge Balance ; [H + ] + 2[Ca 2+ ] = [OH - ] + [F - ] - [HF][OH ] K b = = [F] Five unknowns [H + ], [OH - ], [Ca 2+ ], [F - ], [HF] * To solve the problems is no simple matter for these 5 equations. - If we fix the ph, it becomes simpler. In this case, we don t use charge balance equations (v) because we don t know charge balance exactly due to the addition of chemicals to get, for example, ph=3. -11

6 Fig. ph dependence of the concentration of Ca 2+, F - and HF in a saturated solution of CaF 2 -

7 What is the ph of saturated solution with CaF 2? From Charge Balance, [H + ] + 2[Ca 2+ ] = [OH - ] + [F - ] Assumption ; 2[Ca 2+ ] >> [H + ] [F - ] >> [OH - ] 2[Ca 2+ ] = [F - ] log2 + log[ca 2+ ] = log[f - ] log[f - ] = log[ca 2+ ] Find a point on pc-ph diagram (Fig. 9-3) where log[ca 2+ ] is located 0.3 log unit vertically below log[f - ]. ph 7.11, Check the assumptions.

8 What is the ph of distilled water in equilibrium with the air? [CO 2 (aq)] [H 2 CO 3 *] = K H P CO2 = = M [H ][HCO ] K a1 = = * [H 2CO3 ] [H ][CO ] K a2 = = [HCO3 ] K w = [H ][OH ] = charge balance [H + ] = [HCO 3- ] + 2[CO 3 2- ] + [OH - ] unknowns : [H + ], [HCO 3- ], [CO 3 2- ], [OH - ] equation : 4 Solving the equations,,,, then ph = 5.7

9 * What is the ph of distilled water in equilibrium with the air? * About 100 years ago, Friedrich Kohlrausch measured conductivity of water. In removing ionic impurities from water, they found it necessary to distill the water 42 consecutive times under vacuum to reduce conductivity to a limiting value.

10 Box 6-4. Carbonic Acid (p. 134) K a1 for H 2 CO 3 is about 10 2 to 10 4 times smaller than K a for other carboxylic acid.

11 Erosion of Marble by Acidic Rain

12 Erosion of Marble by Acidic Rain

13 Erosion of Marble by Acidic Rain Water Stabilization CaCO 3 (s) Ca 2+ + CO 2-3 K sp H + + CO 2-3 HCO - 3 1/Ka2 CaCO 3 (s) + H + Ca 2+ + HCO - 3 K= Ksp /Ka [Ca ][HCO3] [Ca ][HCO3] [H ] eq = = K Ksp / Ka2 [H + ] eq: [H + ] in water if it were in equilibrium with CaCO 3 (s) at the existing solution concentration of HCO - 3 and Ca 2 [H + ] : actual [H + ] in water [H + ] > [H + ] eq undersaturated (dissolution) [H + ] < [H + ] eq supersaturated (precipitation)

14 Erosion of Marble by Acidic Rain Langelier Saturation Index, I = ph - ph eq I < 0 I > 0 undersaturated (dissolution) supersaturated (precipitation) ph eq = log K sp /Ka2 Log [Ca 2+ ] log [HCO 3- ] ph eq : ph of water if it were in equilibrium with CaCO 3 (s) at the existing solution concentration of HCO 3 - and Ca 2+, ph eq = -log [H + ] eq. ph : actual ph of water, ph = -log [H + ] Ksp, Ka 2 must be adjusted for temperature and ionic strength to Ksp and Ka 2 1 µ = CiZ 2 i C = molar conc. of ion i 2 i Z = valence of ion i ph eq = (pk a2 pk sp ) + pca 2+ + phco 3 -

15 Harris: Quantitative Chemical Analysis, Eight Edition CHAPTER 09: POLYPROTIC ACID-BASE EQUILIBRIA

16 9-5 Fractional Composition Equations Monoprotic Systems

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18 CHAPTER 09: Equation 9.18 Diprotic Systems

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20 9-4 Which is the principal Species? Figure 9-2 The predominant molecular form of a triprotic system (H 3 A) in the various ph intervals. - Figure 9-2 summarizes how we think of a triprotic system. - We determine the principal species by comparing the ph of the solution with the pka values.

21 9-6 Isoelectric and Isoionic ph Isoionic Point (= Isoionic ph) : the ph obtained when the pure, neutral, polyprotic acid HA (the neutral zwitterion) is dissolved in water. Isoelectric point ( = Ioselectric ph) : the ph at which the average charge of the polyprotic acid is 0. [ H 2 A + ] = [A - ] Isoelectric ph = 1/2 (pk1 + pk2 )

22 9-1 Diprotic Acids and Bases The intermediate Form, HL

23 9-1 Diprotic Acids and Bases The intermediate Form, HL

24 9-1 Diprotic Acids and Bases The intermediate Form, HL

25 9-1 Diprotic Acids and Bases The intermediate Form, HL Intermediate form of diprotic acid ( HA) : (Textbook p.192)

26 9-1 Diprotic Acids and Bases The intermediate Form, HL Problem Derivation of Equation 9-10 for ph of the intermediate from of Na + HA -. Answer Mass balance: [Na + ] = [H 2 A] + [HA - ] + [A 2- ] Charge balance: [Na + ] + [H + ] = [HA - ] + 2[A 2- ] +[OH - ] Equlibria : K 1 = [H + ][HA - ] / [H 2 A] K 2 = [H + ][A 2- ]/ [HA - ] K w = [H + ][OH - ] [H 2 A] + [HA - ] + [A 2- ] +[H + ] = [HA - ] + 2[A 2- ] +[OH - ] [ H 2 A] + [H + ] = [A 2- ] +[OH - ]

27 9-1 Diprotic Acids and Bases [ H 2 A] + [H + ] = [A 2- ] +[OH - ] [H + ][HA - ]/ K 1 + [H + ] = K 2 [HA - ]/[H + ] + K w /[H + ] [H + ] 2 [HA - ]/ K 1 + [H + ] 2 = K 2 [HA - ] + K w [H + ] 2 {[HA - ]/ K } = K 2 [HA - ] + K w [H + ] 2 = { K 2 [HA - ] + K w }/ {[HA - ]/ K } = { K 1 K 2 [HA - ] + K 1 K w }/ {[HA - ] + K 1 } (Equation 9-10 on page 190) = { K 1 K 2 F + K 1 K w }/ {F + K 1 } (Equation 9-11 on page 190)

28 9-1 Diprotic Acids and Bases - A summary of results for leucine. Solution ph [H + ] (M) [H₂L + ] (M) [HL] (M) [L - ] (M) M H₂A X X X X M HA X X X X M HA X X X X Notice the relative concentrations of H2L+, HL, L - in each solution. - Notice the ph of each solution.

29 Harris: Quantitative Chemical Analysis, Eight Edition CHAPTER 10: ACID-BASE TITRATIONS

30 10-9 The leveling Effect Dissociation reactions and relative strengths of some common acids and their conjugated bases -The acids become progressively weaker from top to bottom. i) Perchloric acid (HClO 4 ) and HCl are completely dissociated. ii) 1 % of acetic acid (CH 3 CO 2 H ) is dissociated. iii) 0.01 % of ammonium ion (NH 4 + ) is dissociated into hydronium ion (H 3 O + ) & NH 3

31 10-9 The leveling Effect Figure. Dissociation reactions and relative strengths of some common acids and their conjugated bases. - The weaker acid forms the stronger conjugate base. i) Perchlorate (ClO 4- ) and chloride (Cl - ) ions have no affinity for protons. ii) Ammonia (NH 3 ) has a much stronger affinity for protons than any other base above.

32 10-9 The leveling Effect The tendency of a solvent to accept or to donate protons determines the strength of a solute (acid or base) dissolved in it. 1) Perchloric and hydrochloric acids are strong acids in water. They dissociate completely in water. HClO 4 + H 2 O H 3 O + + ClO 4 HCl + H 2 O H 3 O + + Cl i) Water is a leveling solvent for perchloric, hydrochloric, and nitric acids because all three are completely ionized in this solvent and shows no differences in strength. ii) In a leveling solvent, several acids are completely dissociated and show the same strength.

33 10-9 The leveling Effect The tendency of a solvent to accept or to donate protons determines the strength of a solute (acid or base) dissolved in it. 2) If anhydrous acetic acid (a weaker proton acceptor than water) is substituted as solvent, neither of these acids undergo complete dissociation; instead, equilibria are established as following: HClO 4 + CH 3 CO 2 H CH 3 CO 2 H ClO 4 K = 1.3 x 10-5 HCl + CH 3 CO 2 H CH 3 CO 2 H Cl K = 2.8 x 10-9 i) HClO 4 is stronger than HCl by about 5,000 times. ii) Acetic acid (CH 3 CO 2 H ) thus acts as a differenciating solvent toward the two acids by revealing the inherent differences in their acidities. iii) In a differenciating solvent, various acids dissociate to different degrees and have different strengths.

34 10-9 The leveling Effect ph Solvent ; isobutyl ketone Titrant ; tetrabutylammonium hydroxide Analytes ; mixture of acids

35 10-9 The leveling Effect The strongest acid that can exist in water is H 3 O + and the strongest base is OH -. If an acid stronger than H 3 O + (or an acid whose conjugate base is less stronger than H 2 O )is dissolved in water, it protonates H 2 O to make H 3 O +. If a base stronger than OH - is dissolved in water, it deprotonates H 2 O to make OH -. HClO 4 + H 2 O H 3 O + + ClO 4 HCl + H 2 O H 3 O + + Cl - In acetic acid solvent, which is less basic than H 2 O, HClO 4 and HCl are not leveled to the same strength: HClO 4 + CH 3 CO 2 H CH 3 CO 2 H ClO 4 K = 1.3 x 10-5 HCl + CH 3 CO 2 H CH 3 CO 2 H Cl K = 2.8 x 10-9