D. Harvey - Modern Analytical Chemistry (794078), страница 48
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Substituting back gives thecalculated concentration of Pb2+ at equilibrium as[Pb2+] = 1.0 × 10–4 + 2.50 × 10–5 = 1.25 × 10–4 Ma value that differs by 25% from our approximation that the equilibriumconcentration is 1.0 × 10–4 M. This error seems unreasonably large. Rather thanshouting in frustration, we make a new assumption. Our first assumption thatthe concentration of Pb 2+ is 1.0 × 10 –4 M was too small. The calculatedconcentration of 1.25 × 10–4 M, therefore, is probably a little too large. Let usassume that[Pb2+] = 1.0 × 10–4 + x ≈1.2 × 10–4 MSubstituting into the solubility product equation and solving for x gives usx = 2.28 × 10–5or a concentration of Pb2+ at equilibrium of[Pb2+] = 1.0 × 10–4 + (2.28 × 10–5) = 1.23 × 10–4 Mwhich differs from our assumed concentration of 1.2 × 10–4 M by 2.5%.
Thisseems to be a reasonable error since the original concentration of Pb2+ isgiven to only two significant figures. Our final solution, to two significantfigures, is[Pb2+] = 1.2 × 10–4 M[IO3–] = 4.6 × 10–5 Mand the solubility of Pb(IO3)2 is 2.3 × 10–5 mol/L. This iterative approach tosolving an equation is known as the method of successive approximations.1400-CH06 9/9/99 7:40 AM Page 159Chapter 6 Equilibrium Chemistry1596G.3 Systematic Approach to Solving Equilibrium ProblemsCalculating the solubility of Pb(IO3)2 in a solution of Pb(NO3)2 was more complicated than calculating its solubility in distilled water. The necessary calculations, however, were still relatively easy to organize, and the assumption used tosimplify the problem was fairly obvious. This problem was reasonably straightforward because it involved only a single equilibrium reaction, the solubility ofPb(IO3)2.
Calculating the equilibrium composition of a system with multipleequilibrium reactions can become quite complicated. In this section we willlearn how to use a systematic approach to setting up and solving equilibriumproblems.As its name implies, a systematic approach involves a series of steps:1. Write all relevant equilibrium reactions and their equilibrium constantexpressions.2. Count the number of species whose concentrations appear in the equilibriumconstant expressions; these are your unknowns.
If the number of unknownsequals the number of equilibrium constant expressions, then you have enoughinformation to solve the problem. If not, additional equations based on theconservation of mass and charge must be written. Continue to add equationsuntil you have the same number of equations as you have unknowns.3. Decide how accurate your final answer needs to be. This decision will influenceyour evaluation of any assumptions you use to simplify the problem.4.
Combine your equations to solve for one unknown (usually the one you aremost interested in knowing). Whenever possible, simplify the algebra bymaking appropriate assumptions.5. When you obtain your final answer, be sure to check your assumptions. If anyof your assumptions prove invalid, then return to the previous step andcontinue solving.
The problem is complete when you have an answer that doesnot violate any of your assumptions.Besides equilibrium constant equations, two other types of equations are usedin the systematic approach to solving equilibrium problems.
The first of these is amass balance equation, which is simply a statement of the conservation of matter.In a solution of a monoprotic weak acid, for example, the combined concentrationsof the conjugate weak acid, HA, and the conjugate weak base, A–, must equal theweak acid’s initial concentration, CHA.*The second type of equation is a charge balance equation. A charge balanceequation is a statement of solution electroneutrality.Total positive charge from cations = total negative charge from anionsMathematically, the charge balance expression is expressed asnmi =1j =1∑ (z + ) i × [M z + ] i = ∑ (z – ) j × [A z – ]j[Mz+][Az–]wherei andj are, respectively, the concentrations of the ith cationand the jth anion, and (z+)i and (z –)j are the charges of the ith cation and the jthanion.
Note that the concentration terms are multiplied by the absolute valuesof each ion’s charge, since electroneutrality is a conservation of charge, not concentration. Every ion in solution, even those not involved in any equilibrium*You may recall that this is the difference between a formal concentration and a molar concentration.mass balance equationAn equation stating that matter isconserved, and that the total amount of aspecies added to a solution must equalthe sum of the amount of each of itspossible forms present in solution.charge balance equationAn equation stating that the totalconcentration of positive charge in asolution must equal the totalconcentration of negative charge.1400-CH06 9/9/99 7:40 AM Page 160160Modern Analytical Chemistryreactions, must be included in the charge balance equation.
The charge balanceequation for an aqueous solution of Ca(NO3)2 is2 × [Ca2+] + [H3O+] = [OH–] + [NO3–]Note that the concentration of Ca2+ is multiplied by 2, and that the concentrationsof H3O+ and OH– are also included. Charge balance equations must be writtencarefully since every ion in solution must be included. This presents a problemwhen the concentration of one ion in solution is held constant by a reagent of unspecified composition.
For example, in many situations pH is held constant using abuffer. If the composition of the buffer is not specified, then a charge balance equation cannot be written.EXAMPLE 6.10Write a mass balance and charge balance equations for a 0.10 M solution ofNaHCO3.SOLUTIONIt is easier to keep track of what species are in solution if we write down thereactions that control the solution’s composition. These reactions are thedissolution of a soluble saltNaHCO3(s) → Na+(aq) + HCO3–(aq)and the acid–base dissociation reactions of HCO3– and H2Ot H3O+(aq) + CO32–(aq)HCO3–(aq) + H2O(l) t OH–(aq) + H2CO3(aq)2H2O(l) t H3O+(aq) + OH–(aq)HCO3–(aq) + H2O(l)The mass balance equations are0.10 M = [H2CO3] + [HCO3–] + [CO32–]0.10 M = [Na+]The charge balance equation is[Na+] + [H3O+] = [OH–] + [HCO3–] + 2 × [CO32–]6G.4 pH of a Monoprotic Weak AcidTo illustrate the systematic approach, let us calculate the pH of 1.0 M HF.
Twoequilbria affect the pH of this system. The first, and most obvious, is the acid dissociation reaction for HFHF(aq) + H2O(l)t H3O+(aq) + F–(aq)for which the equilibrium constant expression isKa =[H 3O + ][F – ]= 6.8 × 10 –4[HF]6.351400-CH06 9/9/99 7:40 AM Page 161Chapter 6 Equilibrium ChemistryThe second equilibrium reaction is the dissociation of water, which is an obviousyet easily disregarded reaction2H2O(l)t H3O+(aq) + OH–(aq)Kw = [H3O+][OH–] = 1.00 × 10–146.36Counting unknowns, we find four ([HF], [F–], [H3O+], and [OH–]). To solve thisproblem, therefore, we need to write two additional equations involving these unknowns. These equations are a mass balance equationCHF = [HF] + [F–]6.37[H3O+] = [F–] + [OH–]6.38and a charge balance equationWe now have four equations (6.35, 6.36, 6.37, and 6.38) and four unknowns([HF], [F–], [H3O+], and [OH–]) and are ready to solve the problem.
Before doingso, however, we will simplify the algebra by making two reasonable assumptions.First, since HF is a weak acid, we expect the solution to be acidic; thus it is reasonable to assume that[H3O+] >> [OH–]simplifying the charge balance equation (6.38) to[H3O+] = [F–]6.39Second, since HF is a weak acid we expect that very little dissociation occurs, and[HF] >> [F–]Thus, the mass balance equation (6.36) simplifies toCHF = [HF]6.40For this exercise we will accept our assumptions if the error introduced by each assumption is less than ±5%.Substituting equations 6.39 and 6.40 into the equilibrium constant expressionfor the dissociation of HF (equation 6.35) and solving for the concentration ofH3O+ gives usKa =[H 3 O + ] =Ka CHF =[H 3O + ][H 3O + ]CHF(6.8 × 10 –4 )(1.0) = 2.6 × 10 –2 MBefore accepting this answer, we must verify that our assumptions are acceptable.The first assumption was that the [OH–] is significantly smaller than the [H3O+].
Tocalculate the concentration of OH– we use the Kw expression (6.36)[OH – ] =Kw1.00 × 10 –14= 3.8 × 10 –13 M=2.6 × 10 –2[H 3 O + ]Clearly this assumption is reasonable. The second assumption was that the [F–] issignificantly smaller than the [HF]. From equation 6.39 we have[F–] = 2.6 × 10–2 M1611400-CH06 9/9/99 7:41 AM Page 162162Modern Analytical ChemistrySince the [F–] is 2.6% of CHF, this assumption is also within our limit that the errorbe no more than ±5%. Accepting our solution for the concentration of H3O+, wefind that the pH of 1.0 M HF is 1.59.How does the result of this calculation change if we require our assumptions tohave an error of less than ±1%.
In this case we can no longer assume that [HF] >>[F–]. Solving the mass balance equation (6.37) for [HF][HF] = CHF – [F–]and substituting into the Ka expression along with equation 6.39 givesKa =[H 3O + ]2CHF – [H 3O + ]Rearranging leaves us with a quadratic equation[H3O+]2 = KaCHF – Ka[H3O+][H3O+]2 + Ka[H3O+] – KaCHF = 0which we solve using the quadratic formulax =–b ± b 2 – 4ac2awhere a, b, and c are the coefficients in the quadratic equation ax2 + bx + c = 0.Solving the quadratic formula gives two roots, only one of which has any chemicalsignificance. For our problem the quadratic formula gives roots ofx ==–6.8 × 10 –4 ± (6.8 × 10 –4 )2 – (4)(1)(–6.8 × 10 –4 )(1.0)2(1)–6.8 × 10 –4 ± 5.22 × 10 –22= 2.57 × 10 –2 or – 2.63 × 10 –2Only the positive root has any chemical significance since the negative root impliesthat the concentration of H3O+ is negative.
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