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These lecture notes cover the topics of acid-base reactions, complexation reactions, and oxidation-reduction reactions in analytical chemistry. The notes provide definitions of acids and bases, as well as information on strong and weak acids. The notes also cover the dissociation of water and the development of the pH scale. Additionally, the notes discuss the relationship between the strength of a base and the strength of its conjugate acid. These notes are from a course taught by Assist. Prof. Dr. Hamad K. Abdulkadir at the University of Anbar's College of Engineering.
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Department of Chem. & Petrochemical Engineering 2 nd^. Stage / Analytical Chemistry
2.4.2 Acid–Base Reactions
2.4.4 Oxidation–Reduction (Redox) Reactions
Department of Chem. & Petrochemical Engineering 2 nd^. Stage / Analytical Chemistry 2.4.2 Acid–Base Reactions A useful definition of acids and bases is that independently introduced in 1923 by Johannes Brønsted and Thomas Lowry. In the Brønsted-Lowry definition, an acid is a proton donor and a base is a proton acceptor. Note : the connection in these definitions—defining a base as a proton acceptor implies that there is an acid available to donate the proton. For example, in reaction 2.4.4 acetic acid, CH 3 COOH, donates a proton to ammonia, NH 3 , which serves as the base. CH3COOH (aq) + NH 3 (aq) ⇌ NH 4 +^ (aq) + CH3COO−(aq) (2.4.4) When an acid and a base react, the products are a new acid and a new base. For example, the acetate ion, CH 3 COO–, in reaction 2.4.4 is a base that can accept a proton from the acidic ammonium ion, NH 4 +, forming acetic acid and ammonia. We call the acetate ion the conjugate base of acetic acid, and the ammonium ion is the conjugate acid of ammonia. Strong and Weak Acids The reaction of an acid with its solvent (typically water) is an acid dissociation reaction. We divide acids into two categories—strong and weak—based on their ability to donate a proton to the solvent. A strong acid, such as HCl, almost completely transfers its proton to the solvent, which acts as the base. HCl (aq) + H 2 O (l) → H 3 O+(aq) +Cl−(aq) (2.4.5) We use a single arrow (→) in place of the equilibrium arrow (⇋) because we treat HCl as if it completely dissociates in aqueous solutions. In water, the common strong acids are hydrochloric acid (HCl), hydroiodic acid (HI), hydrobromic acid (HBr), nitric acid (HNO 3 ), perchloric acid (HClO 4 ), and the first proton of sulfuric acid (H 2 SO 4 ). Note : In a different solvent, HCl may not be a strong acid. For example, HCl does not act as a strong acid in methanol. In this case we use the equilibrium arrow when writing the acid–base reaction. HCl (aq) + CH 3 OH (l) ⇌ CH3OH 2 +^ (aq) + Cl−(aq) (2.4.6) A weak acid, of which aqueous acetic acid is one example, does not completely donate its acidic proton to the solvent. Instead, most of the acid remains undissociated, with only a small fraction present as the conjugate base. CH 3 COOH(aq) + H 2 O (l) ⇌ H 3 O+(aq) + CH 3 COO−^ (aq) (2.4.7) The equilibrium constant for this reaction is an acid dissociation constant , K a, which we write as
Department of Chem. & Petrochemical Engineering 2 nd^. Stage / Analytical Chemistry The decrease in the acid dissociation constants from K a1 to K a3 tells us that each successive proton is harder to remove. Consequently, H 3 PO 4 is a stronger acid than H 2 PO 4 – , and H 2 PO 4 –^ is a stronger acid than HPO 42 –. Strong and Weak Bases The most common example of a strong base is an alkali metal hydroxide, such as sodium hydroxide, NaOH, which completely dissociates to produce hydroxide ion. NaOH (s) → Na+^ (aq) + OH−^ (aq) (2.4.16) A weak base, such as the acetate ion, CH 3 COO–, only partially accepts a proton from the solvent, and is characterized by a base dissociation constant , K b. For example, the base dissociation reaction and the base dissociation constant for the acetate ion are CH 3 COO−^ (aq) + H 2 O (l) ⇌ OH−(aq) + CH3COOH (aq) (2.4.17) (2.4.18) A polyprotic weak base, like a polyprotic acid, has more than one base dissociation reaction and more than one base dissociation constant. Amphiprotic Species Some species can behave as either a weak acid or as a weak base. For example, the following two reactions show the chemical reactivity of the bicarbonate ion, HCO 3 – , in water. HCO 3 −^ (aq) + H 2 O(l) ⇌ H 3 O+(aq) + CO 3 2−^ (aq) (2.4.19) HCO 3 −^ (aq) + H 2 O (l) ⇌ OH−(aq) + H 2 CO 3 (aq) (2.4.20) A species that is both a proton donor and a proton acceptor is called amphiprotic. Whether an amphiprotic species behaves as an acid or as a base depends on the equilibrium constants for the competing reactions. For bicarbonate, the acid dissociation constant for reaction 2.4. (2.4.21) is smaller than the base dissociation constant for reaction 2.4.20. (2.4.22)
Department of Chem. & Petrochemical Engineering 2 nd^. Stage / Analytical Chemistry Because bicarbonate is a stronger base than it is an acid, we expect an aqueous solution of HCO 3 –^ to be basic. Dissociation of Water Water is an amphiprotic solvent because it can serve as an acid or as a base. An interesting feature of an amphiprotic solvent is that it is capable of reacting with itself in an acid–base reaction. 2H 2 O (l) ⇌ H 3 O+(aq) + OH−(aq) (2.4.23) We identify the equilibrium constant for this reaction as water’s dissociation constant, Kw, Kw= [H 3 O+] [OH−] = 1.00×10−14^ (2.4.24) which has a value of 1.0000 × 10–^14 at a temperature of 24oC. The value of K w varies substantially with temperature. For example, at 20oC K w is 6.809 × 10–^15 , while at 30 oC K w is 1.469 × 10–^14. At 25oC, K w is 1.008 × 10–^14 , which is sufficiently close to 1.00 × 10–^14 that we can use the latter value with negligible error. An important consequence of equation 6.4.24 is that the concentration of H 3 O+^ and the concentration of OH−^ are related. If we know [H 3 O+] for a solution, then we can calculate [OH−] using Equation 6.4.. Example 2. What is the [OH–] if the [H 3 O+] is 6.12 × 10-^5 M? Solution (2.4.25) The pH Scale Equation 6.4.24 allows us to develop a pH scale that indicates a solution’s acidity (pH = – log[H 3 O+]). When the concentrations of H 3 O+and OH–^ are equal a solution is neither acidic nor basic; that is, the solution is neutral. Letting [H3O+] = [OH−] (2.4.26) substituting into equation 2.4. Kw =[H 3 O+]^2 = 1.00×10−14^ (2.4.27)
Department of Chem. & Petrochemical Engineering 2 nd^. Stage / Analytical Chemistry 2H 2 O (l) ⇌ H 3 O+(aq) + OH−(aq) (2.4.32) for which the equilibrium constant is K w. Because adding together two reactions is equivalent to multiplying their respective equilibrium constants, we may express K w as the product of K a for CH 3 COOH and Kb for CH 3 COO–. Kw = Ka,CH 3 COOH × Kb,CH 3 COO−^ (2.4.33) For any weak acid, HA, and its conjugate weak base, A–, we can generalize this to the following equation. Kw = Ka, HA × Kb, A−^ (2.4.34) The relationship between Ka and Kb for a conjugate acid–base pair simplifies our tabulation of acid and base dissociation constants. Appendix 11 includes acid dissociation constants for a variety of weak acids. To find the value of Kb for a weak base, use equation 6.4.34and the Ka value for its corresponding weak acid. Note : A common mistake when using equation 2.4.34 is to forget that it applies only to a conjugate acid–base pair. Example 2. Using Appendix 11 , calculate values for the following equilibrium constants. (a) Kb for pyridine, C 5 H 5 N (b) Kb for dihydrogen phosphate, H 2 PO− Solution When finding the K b value for polyprotic weak base, you must be careful to choose the correct K a value. Remember that equation 2.4.34 applies only to a conjugate acid–base pair. The conjugate acid of H 2 PO 4 –^ is H 3 PO 4 , not HPO 42 –. Practice Exercise 2.2 Using Appendix 11 , calculate the K b values for hydrogen oxalate, HC 2 O 4 –^ , and oxalate C 2 O 42 – .The K b for hydrogen oxalate is
Department of Chem. & Petrochemical Engineering 2 nd^. Stage / Analytical Chemistry and the K b for oxalate is As we expect, the K b value for C 2 O 42 –^ is larger than that for HC 2 O 4 −. 2H 2 O (l) ⇌ H 3 O+(aq) + OH−(aq) ( 9 - 9) (9-10) K (55.6)^2 = Kw= [H 3 O+] [OH−] = 1.00×10−14^ at 25 C (9-11)
Department of Chem. & Petrochemical Engineering 2 nd^. Stage / Analytical Chemistry Example 2.
A more general definition of acids and bases was proposed in1923 by G. N. Lewis. The Brønsted-Lowry definition of acids and bases focuses on an acid’s proton-donating ability and a base’s proton-accepting ability. Lewis theory, on the other hand, uses the breaking and forming of covalent bonds to describe acid–base characteristics. In this treatment, an acid is an electron pair acceptor and a base in an electron pair donor. Although we can apply Lewis theory to the treatment of acid–base reactions, it is more useful for treating complexation reactions between metal ions and ligands. The following reaction between the metal ion Cd2+^ and the ligand NH 3 is typical of a complexation reaction. Cd2+^ (aq) + 4:NH 3 (aq) ⇌ Cd(:NH 3 ) 4 2+^ (aq) (2.4.35) The product of this reaction is a metal–ligand complex. In writing this reaction we show ammonia as :NH 3 , using a pair of dots to emphasize the pair of electrons it donates to Cd2+. In subsequent reactions we will omit this notation.
Department of Chem. & Petrochemical Engineering 2 nd^. Stage / Analytical Chemistry Metal-Ligand Formation Constants We characterize the formation of a metal–ligand complex by a formation constant , K f. The complexation reaction between Cd2+^ and NH 3 , for example, has the following equilibrium constant. (2.4.36) The reverse of reaction 2.4.35 is a dissociation reaction, which we characterize by a dissociation constant , K d, that is the reciprocal of K f. Many complexation reactions occur in a stepwise fashion. For example, the reaction between Cd2+^ and NH 3 involves four successive reactions. Cd2+^ (aq) + NH 3 (aq) ⇌ Cd(NH 3 )2+(aq) (2.4.37) Cd(NH 3 )2+(aq) + NH 3 (aq) ⇌ Cd(NH3) 2 2+^ (aq) (2.4.38) Cd(NH 3 ) 2 2+^ (aq) +NH 3 (aq) ⇌ Cd(NH 3 ) 3 2+^ (aq) (2.4.39) Cd(NH3) 3 2+^ (aq) + NH 3 (aq) ⇌ Cd(NH3) 4 2+^ (aq) (2.4.40) To avoid ambiguity, we divide formation constants into two categories. Stepwise formation constants , which we designate as Ki for the i th step, describe the successive addition of one ligand to the metal–ligand complex from the previous step. Thus, the equilibrium constants for reactions 2.4.37–2.4.40 are, respectively, K 1 , K 2 , K 3 , and K 4. Overall, or cumulative formation constants , which we designate as β i , describe the addition of i ligands to the free metal ion. The equilibrium constant in equation 2.4.36 is correctly identified as β 4 , where β 4 = K 1 × K 2 × K 3 × K 4 (2.4.41) In general Βi = K 1 × K 2 × ... × Ki (2.4.42) Stepwise and overall formation constants for selected metal–ligand complexes are in Appendix 12.
Department of Chem. & Petrochemical Engineering 2 nd^. Stage / Analytical Chemistry Practice Exercise 2. What is the equilibrium constant for the following reaction? You will find appropriate equilibrium constants in Appendix 10 andAppendix 11. AgBr (s) + 2S 2 O 3 2−^ (aq) ⇌ Ag(S 2 O 3 ) 2 3−(aq) + Br−(aq) (2.4.52) We can write the reaction as a sum of three other reactions. The first reaction is the solubility of AgBr(s), which we characterize by its K sp. AgBr (s) ⇌ Ag+^ (aq) + Br−^ (aq) The remaining two reactions are the stepwise formation of Ag(S 2 O 3 ) 23 – , which we characterize by K 1 and K 2. Ag+^ (aq) + S 2 O 3 2−^ (aq) ⇌ Ag(S 2 O 3 )−^ (aq) Ag(S 2 O 3 )−^ (aq) + S 2 O 3 2−^ (aq) ⇌ Ag(S 2 O 3 ) 2 3−^ (aq) Using values for K sp, K 1 , and K 2 from Appendix 10 and Appendix 11 , we find that the equilibrium constant for our reaction is K = Ksp × K 1 × K 2 = (5.0×10−13)(6.6×10^8 )(7.1×10^4 ) = 23 2.4.4 Oxidation–Reduction (Redox) Reactions An oxidation–reduction reaction occurs when electrons move from one reactant to another reactant. As a result of this electron transfer, these reactants undergo a change in oxidation state. Those reactants that experience an increase in oxidation state undergo oxidation , and those experiencing a decrease in oxidation state undergo reduction. For example, in the following redox reaction between Fe3+^ and oxalic acid, H 2 C 2 O 4 , iron is reduced because its oxidation state changes from +3 to +2. 2Fe3+^ (aq) + H 2 C 2 O 4 (aq) + 2H 2 O(l) ⇌ 2Fe2+^ (aq) + 2CO 2 (g) + 2H 3 O+^ (aq) (2.4.53) Oxalic acid, on the other hand, undergoes oxidation because the oxidation state for carbon increases from +3 in H 2 C 2 O 4 to +4 in CO 2. We can divide a redox reaction, such as reaction 2.4.53, into separate half-reactions that show the oxidation and the reduction processes. H 2 C 2 O 4 (aq) + 2H 2 O (l) ⇌ 2CO 2 (g) + 2H 3 O+(aq) + 2 e − (2.4.54) Fe3+^ (aq) + e − ⇌ Fe2+^ (aq) (2.4.55) It is important to remember, however, that an oxidation reaction and a reduction reaction occur as a pair. We formalize this relationship by identifying as a reducing agent the reactant undergoing oxidation, because it provides the electrons for the reduction half- reaction. Conversely, the reactant undergoing reduction is an oxidizing agent. In reaction 2.4.53, Fe3+^ is the oxidizing agent and H 2 C 2 O 4 is the reducing agent.
Department of Chem. & Petrochemical Engineering 2 nd^. Stage / Analytical Chemistry The products of a redox reaction also have redox properties. For example, the Fe2+^ in reaction 2.4.53 can be oxidized to Fe3+, while CO 2 can be reduced to H 2 C 2 O 4. Borrowing some terminology from acid–base chemistry, Fe2+^ is the conjugate reducing agent of the oxidizing agent Fe3+, and CO 2 is the conjugate oxidizing agent of the reducing agent H 2 C 2 O 4. Thermodynamics of Redox Reactions Unlike precipitation reactions, acid–base reactions, and complexation reactions, we rarely express the equilibrium position of a redox reaction using an equilibrium constant. Because a redox reaction involves a transfer of electrons from a reducing agent to an oxidizing agent, it is convenient to consider the reaction’s thermodynamics in terms of the electron. For a reaction in which one mole of a reactant undergoes oxidation or reduction, the net transfer of charge, Q , in coulombs is Q = nF (2.4.56) where n is the moles of electrons per mole of reactant, and F is Faraday’s constant (96,485 C/mol). The free energy, Δ G , to move this charge, Q , over a change in potential , E , is ΔG = EQ (2.4.57) The change in free energy (in kJ/mole) for a redox reaction, therefore, is ΔG = − nFE (2.4.58) where Δ G has units of kJ/mol. The minus sign in equation 6.4.58 is the result of a difference in the conventions for assigning a reaction’s favorable direction. In thermodynamics, a reaction is favored when Δ G is negative, but a redox reaction is favored when E is positive. Substituting equation 2.4.58 into equation 2.2. − nFE = − nFE∘^ + RTln Q (2.4.59) and dividing by - nF , leads to the well-known Nernst equation (2.4.60) where E o^ is the potential under standard-state conditions. Substituting appropriate values for R and F , assuming a temperature of 25oC (298 K), and switching from ln to log (ln( x ) = 2.303log( x )) gives the potential in volts as (2.4.61)
Department of Chem. & Petrochemical Engineering 2 nd^. Stage / Analytical Chemistry E∘^ = E∘Ag+/Ag − E∘Cd2+/Cd = 0.7996−(−0.4030) =1.2026 V (2.4.66) (b) To calculate the equilibrium constant we substitute appropriate values into equation 2.4.62. (2.4.67) Solving for K gives the equilibrium constant as Log K = 40.6558 (2.4.68) K = 4.527 × 1040 (2.4.69) (c) To calculate the potential when [Ag+] is 0.020 M and [Cd2+] is 0.050 M, we use the appropriate relationship for the reaction quotient, Q , in equation 2.4.61. (2.4.70) (2.4.71) E = 1.14V (2.4.72) Practice Exercise 2. For the following reaction at 25oC 5Fe2+(aq) + MnO 4 −^ (aq) + 8H+(aq) ⇌ 5Fe3+(aq) + Mn2+(aq) + 4H 2 O(l) (6.4.73) calculate (a) the standard potential, (b) the equilibrium constant, and (c) the potential under these conditions: [Fe2+] = 0.50 M, [Fe3+] = 0.10 M, [MnO 4 – ] = 0.025 M, [Mn2+] = 0.015 M, and a pH of 7.00. See Appendix 13 for standard state reduction potentials. The two half-reactions are the oxidation of Fe2+^ and the reduction of MnO 4 –. Fe2+(aq) ⇌ Fe3+(aq) + e− MnO 4 −(aq) + 8H+(aq) + 5e− ⇌ Mn2+(aq) + 4H 2 O(l) From Appendix 13 , the standard state reduction potentials for these half-reactions are E∘Fe3+/Fe2+ = 0.771 V E∘^ MnO4−/Mn2+ = 1.51 V (a) The standard state potential for the reaction is E∘^ = E∘MnO4−^ / Mn2+^ − E∘^ Fe3+^ / Fe2+^ = 1.51V − 0.771V = 0.74V
Department of Chem. & Petrochemical Engineering 2 nd^. Stage / Analytical Chemistry (b) To calculate the equilibrium constant we substitute appropriate values into equation 6.4.62. Solving for K gives its value as Log K = 62.5 K = 3.2 ×10 62 (c) To calculate the potential under these non-standard state conditions, we make appropriate substitutions into the Nernst equation. Note : When writing precipitation, acid–base, and metal–ligand complexation reaction, we represent acidity as H 3 O+. Redox reactions are more commonly written using H+^ instead of H 3 O+. For the reaction in Practice Exercise 6.4, we could replace H+^ with H 3 O+^ and increase the stoichiometric coefficient for H 2 O from 4 to 12, e.g., 5Fe2+(aq) + MnO 4 −^ (aq) + 8H 3 O+(aq) ⇌ 5Fe3+(aq) + Mn2+(aq) + 12H 2 O(l) (2.4.74)
Department of Chem. & Petrochemical Engineering 2 nd^. Stage / Analytical Chemistry Example 2.
Department of Chem. & Petrochemical Engineering 2 nd^. Stage / Analytical Chemistry Equilibrium Constants For Chemical Reactions References
