D. Harvey - Modern Analytical Chemistry (794078), страница 43
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At the same time, the species involved in thereaction undergo a change in their concentrations. The Gibb’s free energy, therefore, must be a function of the concentrations of reactants and products.As shown in equation 6.3, the Gibb’s free energy can be divided into two terms.∆G = ∆G° + RT ln Q[C]c [D]d[A]a [B]benthalpyA change in enthalpy indicates the heatabsorbed or released during a chemicalreaction at constant pressure.entropyA measure of disorder.6.3The first term, ∆G°, is the change in Gibb’s free energy under standard-state conditions; defined as a temperature of 298 K, all gases with partial pressures of 1 atm, allsolids and liquids pure, and all solutes present with 1 M concentrations.
The secondterm, which includes the reaction quotient, Q, accounts for nonstandard-state pressures or concentrations. For reaction 6.1 the reaction quotient isQ =Gibb’s free energyA thermodynamic function for systemsat constant temperature and pressurethat indicates whether or not a reactionis favorable (∆G < 0), unfavorable(∆G > 0), or at equilibrium (∆G = 0).6.4where the terms in brackets are the molar concentrations of the solutes. Note thatthe reaction quotient is defined such that the concentrations of products are placedstandard stateCondition in which solids and liquids arein pure form, gases have partial pressuresof 1 atm, solutes have concentrations of1 M, and the temperature is 298 K.1400-CH06 9/9/99 7:40 AM Page 138138Modern Analytical Chemistryin the numerator, and the concentrations of reactants are placed in the denominator.In addition, each concentration term is raised to a power equal to its stoichiometriccoefficient in the balanced chemical reaction.
Partial pressures are substituted forconcentrations when the reactant or product is a gas. The concentrations of puresolids and pure liquids do not change during a chemical reaction and are excludedfrom the reaction quotient.At equilibrium the Gibb’s free energy is zero, and equation 6.3 simplifies to∆G° = –RT ln Kequilibrium constantFor a reaction at equilibrium, theequilibrium constant determines therelative concentrations of products andreactants.where K is an equilibrium constant that defines the reaction’s equilibrium position.
The equilibrium constant is just the numerical value obtained when substituting the concentrations of reactants and products at equilibrium into equation 6.4;thus,[C]ceq[D]deqK =6.5[A]aeq[B]beqwhere the subscript “eq” indicates a concentration at equilibrium. Although thesubscript “eq” is usually omitted, it is important to remember that the value of K isdetermined by the concentrations of solutes at equilibrium.As written, equation 6.5 is a limiting law that applies only to infinitely dilutesolutions, in which the chemical behavior of any species in the system is unaffectedby all other species.
Corrections to equation 6.5 are possible and are discussed inmore detail at the end of the chapter.6C Manipulating Equilibrium ConstantsWe will use two useful relationships when working with equilibrium constants.First, if we reverse a reaction’s direction, the equilibrium constant for the new reaction is simply the inverse of that for the original reaction. For example, the equilibrium constant for the reactionA + 2Bt AB2K1 =[AB2 ][A][B]2is the inverse of that for the reactionAB2t A + 2BK2 =1[A][B]2=K1[AB2 ]Second, if we add together two reactions to obtain a new reaction, the equilibriumconstant for the new reaction is the product of the equilibrium constants for theoriginal reactions.A+CAC + CA + 2Ct AC2t ACK1 =t AC2K2 =K3 = K1K2 =[AC][A][C][AC 2 ][AC][C][AC][AC 2 ][AC 2 ]×=[A][C] [AC][C] [A][C]21400-CH06 9/9/99 7:40 AM Page 139Chapter 6 Equilibrium Chemistry139EXAMPLE 6.1Calculate the equilibrium constant for the reaction2A + Bgiven the following informationRxn 1: A + BRxn 2: A + ERxn 3: C + ERxn 4: F + Ct C + 3DtDtC+D+FtBtD+BK1 = 0.40K2 = 0.10K3 = 2.0K4 = 5.0SOLUTIONThe overall reaction is given asRxn 1 + Rxn 2 – Rxn 3 + Rxn 4If Rxn 3 is reversed, givingRxn 5: BtC+EK5 =11== 0.50K32.0then the overall reaction isRxn 1 + Rxn 2 + Rxn 5 + Rxn 4and the overall equilibrium constant isKoverall = K1 × K2 × K5 × K4 = 0.40 × 0.10 × 0.50 × 5.0 = 0.106D Equilibrium Constants for Chemical ReactionsSeveral types of reactions are commonly used in analytical procedures, either inpreparing samples for analysis or during the analysis itself.
The most important ofthese are precipitation reactions, acid–base reactions, complexation reactions, andoxidation–reduction reactions. In this section we review these reactions and theirequilibrium constant expressions.6D.1 Precipitation ReactionsA precipitation reaction occurs when two or more soluble species combine to forman insoluble product that we call a precipitate. The most common precipitation reaction is a metathesis reaction, in which two soluble ionic compounds exchangeparts. When a solution of lead nitrate is added to a solution of potassium chloride,for example, a precipitate of lead chloride forms.
We usually write the balanced reaction as a net ionic equation, in which only the precipitate and those ions involvedin the reaction are included. Thus, the precipitation of PbCl2 is written asPb2+(aq) + 2Cl–(aq)t PbCl2(s)In the equilibrium treatment of precipitation, however, the reverse reaction describing the dissolution of the precipitate is more frequently encountered.PbCl2(s)t Pb2+(aq) + 2Cl–(aq)precipitateAn insoluble solid that forms when twoor more soluble reagents are combined.1400-CH06 9/9/99 7:40 AM Page 140140Modern Analytical Chemistrysolubility productThe equilibrium constant for a reactionin which a solid dissociates into its ions(Ksp).The equilibrium constant for this reaction is called the solubility product, Ksp, andis given asKsp = [Pb2+][Cl–]2 = 1.7 × 10–56.6Note that the precipitate, which is a solid, does not appear in the Ksp expression.
Itis important to remember, however, that equation 6.6 is valid only if PbCl2(s) ispresent and in equilibrium with the dissolved Pb2+ and Cl–. Values for selected solubility products can be found in Appendix 3A.6D.2 Acid–Base ReactionsacidA proton donor.baseA proton acceptor.A useful definition of acids and bases is that independently introduced by Johannes Brønsted (1879–1947) and Thomas Lowry (1874–1936) in 1923. In theBrønsted-Lowry definition, acids are proton donors, and bases are proton acceptors. Note that these definitions are interrelated.
Defining a base as a proton acceptor means an acid must be available to provide the proton. For example, in reaction 6.7 acetic acid, CH3COOH, donates a proton to ammonia, NH3, which servesas the base.CH3COOH(aq) + NH3(aq)t CH3COO–(aq) + NH4+(aq)6.7When an acid and a base react, the products are a new acid and base. For example, the acetate ion, CH3COO–, in reaction 6.7 is a base that reacts with the acidicammonium ion, NH4+, to produce acetic acid and ammonia. We call the acetate ionthe conjugate base of acetic acid, and the ammonium ion is the conjugate acid ofammonia.Strong and Weak Acids The reaction of an acid with its solvent (typically water) iscalled an acid dissociation reaction.
Acids are divided into two categories based onthe ease with which they can donate protons to the solvent. Strong acids, such asHCl, almost completely transfer their protons to the solvent molecules.HCl(aq) + H2O(l) → H3O+(aq) + Cl–(aq)In this reaction H2O serves as the base. The hydronium ion, H3O+, is the conjugate acid of H2O, and the chloride ion is the conjugate base of HCl. It is the hydronium ion that is the acidic species in solution, and its concentration determines the acidity of the resulting solution.
We have chosen to use a single arrow(→) in place of the double arrows ( ) to indicate that we treat HCl as if it werecompletely dissociated in aqueous solutions. A solution of 0.10 M HCl is effectively 0.10 M in H3O+ and 0.10 M in Cl–. In aqueous solutions, the commonstrong acids are hydrochloric acid (HCl), hydroiodic acid (HI), hydrobromic acid(HBr), nitric acid (HNO3), perchloric acid (HClO4), and the first proton of sulfuric acid (H2SO4).Weak acids, of which aqueous acetic acid is one example, cannot completelydonate their acidic protons to the solvent. Instead, most of the acid remains undissociated, with only a small fraction present as the conjugate base.tacid dissociation constantThe equilibrium constant for a reactionin which an acid donates a proton to thesolvent (Ka).CH3COOH(aq) + H2O(l)t H3O+(aq) + CH3COO–(aq)The equilibrium constant for this reaction is called an acid dissociation constant,Ka, and is written asKa =[H 3O + ][CH3COO – ]= 1.75 × 10 –5[CH 3COOH]1400-CH06 9/9/99 7:40 AM Page 141Chapter 6 Equilibrium Chemistry141Note that the concentration of H2O is omitted from the Ka expression because itsvalue is so large that it is unaffected by the dissociation reaction.* The magnitudeof Ka provides information about the relative strength of a weak acid, with asmaller Ka corresponding to a weaker acid.
The ammonium ion, for example,with a Ka of 5.70 × 10–10, is a weaker acid than acetic acid.Monoprotic weak acids, such as acetic acid, have only a single acidic protonand a single acid dissociation constant. Some acids, such as phosphoric acid, candonate more than one proton and are called polyprotic weak acids.
Polyprotic acidsare described by a series of acid dissociation steps, each characterized by it own aciddissociation constant. Phosphoric acid, for example, has three acid dissociation reactions and acid dissociation constants.H 3 PO4 (aq) + H 2 O(l)Ka1 =[H 2 PO4– ][H 3O + ]= 7.11 × 10 –3[H 3 PO4 ]H 2 PO4– (aq) + H 2 O(l)Ka2 =t H3O + (aq) + H2 PO4– (aq)t H3O + (aq) + HPO42 − (aq)[HPO4 2 − ][H 3O + ]= 6.32 × 10 –8[H 2 PO4– ]t H3O + (aq) + PO43 − (aq)HPO4 2 − (aq) + H 2 O(l)Ka3 =[PO4 3 − ][H 3O + ]= 4.5 × 10 –13[HPO4 2 − ]The decrease in the acid dissociation constant from Ka1 to Ka3 tells us that each successive proton is harder to remove.
Consequently, H3PO4 is a stronger acid thanH2PO4–, and H2PO4– is a stronger acid than HPO42–.Strong and Weak Bases Just as the acidity of an aqueous solution is a measure ofthe concentration of the hydronium ion, H3O+, the basicity of an aqueous solutionis a measure of the concentration of the hydroxide ion, OH–. The most commonexample of a strong base is an alkali metal hydroxide, such as sodium hydroxide,which completely dissociates to produce the hydroxide ion.NaOH(aq) → Na+(aq) + OH–(aq)Weak bases only partially accept protons from the solvent and are characterized bya base dissociation constant, Kb. For example, the base dissociation reaction andbase dissociation constant for the acetate ion areCH 3COO – (aq) + H 2 O(l)Kb =t OH – (aq) + CH3COOH(aq)[CH 3COOH][OH − ]= 5.71 × 10 –10[CH 3COO – ]Polyprotic bases, like polyprotic acids, also have more than one base dissociation reaction and base dissociation constant.Amphiprotic Species Some species can behave as either an acid or a base.
For example, the following two reactions show the chemical reactivity of the bicarbonateion, HCO3–, in water.*The concentration of pure water is approximately 55.5 Mbase dissociation constantThe equilibrium constant for a reactionin which a base accepts a proton fromthe solvent (Kb).1400-CH06 9/9/99 7:40 AM Page 142142Modern Analytical Chemistryt H3O+(aq) + CO32–(aq)HCO3–(aq) + H2O(l) t OH–(aq) + H2CO3(aq)HCO3–(aq) + H2O(l)amphiproticA species capable of acting as both anacid and a base.6.86.9A species that can serve as both a proton donor and a proton acceptor is called amphiprotic.
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