D. Harvey - Modern Analytical Chemistry (794078), страница 50
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Furthermore, our third assumption that the [Ag(NH3)2+] issignificantly less than the [NH3] also is reasonable. Our second assumption was[NH4+] << [NH3] + [Ag(NH3)2+]To verify this assumption, we solve the Kb equation for [NH4+][NH 4+ ][OH – ] [NH 4+ ]2== 1.75 × 10 –5[NH3 ]0.10 Mgiving[NH4+] = 1.3 × 10–3 MAlthough the [NH4+] is not significantly smaller than the combined concentrationsof NH3 and Ag(NH3)2+, the error is only about 1%.
Since this is not an excessivelylarge error, we will accept this approximation as reasonable.Since one mole of AgI produces one mole of I–, the solubility of AgI is the sameas the concentration of iodide, or 3.7 × 10–6 mol/L.6H Buffer SolutionsAdding as little as 0.1 mL of concentrated HCl to a liter of H2O shifts the pH from7.0 to 3.0.
The same addition of HCl to a liter solution that is 0.1 M in both a weakacid and its conjugate weak base, however, results in only a negligible change in pH.Such solutions are called buffers, and their buffering action is a consequence of therelationship between pH and the relative concentrations of the conjugate weakacid/weak base pair.bufferA solution containing a conjugate weakacid/weak base pair that is resistant to achange in pH when a strong acid orstrong base is added.1400-CH06 9/9/99 7:41 AM Page 168168Modern Analytical ChemistryA mixture of acetic acid and sodium acetate is one example of an acid/basebuffer. The equilibrium position of the buffer is governed by the reactionCH3COOH(aq) + H2O(l)t H3O+(aq) + CH3COO–(aq)and its acid dissociation constantKa =[H 3O + ][CH3COO – ]= 1.75 × 10 –5[CH 3COOH]6.43The relationship between the pH of an acid–base buffer and the relative amounts ofCH3COOH and CH3COO– is derived by taking the negative log of both sides ofequation 6.43 and solving for the pHpH = pKa + log[CH 3COO – ][CH 3COO – ]= 4.76 + log[CH 3COOH][CH 3COOH]6.44Buffering occurs because of the logarithmic relationship between pH and the ratioof the weak base and weak acid concentrations.
For example, if the equilibriumconcentrations of CH3COOH and CH3COO– are equal, the pH of the buffer is 4.76.If sufficient strong acid is added such that 10% of the acetate ion is converted toacetic acid, the concentration ratio [CH3COO–]/[CH3COOH] changes to 0.818,and the pH decreases to 4.67.6H.1 Systematic Solution to Buffer ProblemsEquation 6.44 is written in terms of the concentrations of CH 3 COOH andCH3COO– at equilibrium. A more useful relationship relates the buffer’s pH to theinitial concentrations of weak acid and weak base. A general buffer equation can bederived by considering the following reactions for a weak acid, HA, and the salt ofits conjugate weak base, NaA.NaA(aq) → Na+(aq) + A–(aq)HA(aq) + H2O(l)2H2O(l)t H3O+(aq) + A–(aq)t H3O+(aq) + OH–(aq)Since the concentrations of Na+, A–, HA, H3O+, and OH– are unknown, five equations are needed to uniquely define the solution’s composition.
Two of these equations are given by the equilibrium constant expressionsKa =[H 3O + ][A – ][HA]K w = [H 3O + ][OH – ]The remaining three equations are given by mass balance equations on HA andNa+CHA + CNaA = [HA] + [A–]6.45CNaA = [Na+]6.46and a charge balance equation[H3O+] + [Na+] = [OH–] + [A–]1400-CH06 9/9/99 7:41 AM Page 169Chapter 6 Equilibrium Chemistry169Substituting equation 6.46 into the charge balance equation and solving for [A–]gives[A–] = CNaA – [OH–] + [H3O+]6.47which is substituted into equation 6.45 to give the concentration of HA[HA] = CHA + [OH–] – [H3O+]6.48Finally, substituting equations 6.47 and 6.48 into the Ka equation for HA and solving for pH gives the general buffer equationpH = pKa + logCNaA – [OH – ] + [H 3O + ]CHA + [OH – ] – [H 3O + ]If the initial concentrations of weak acid and weak base are greater than [H3O+] and[OH–], the general equation simplifies to the Henderson–Hasselbalch equation.pH = pK a + logCNaACHA6.49As in Example 6.13, the Henderson–Hasselbalch equation provides a simple way tocalculate the pH of a buffer and to determine the change in pH upon adding astrong acid or strong base.EXAMPLE 6.13Calculate the pH of a buffer that is 0.020 M in NH3 and 0.030 M in NH4Cl.What is the pH after adding 1.00 mL of 0.10 M NaOH to 0.10 L of this buffer?SOLUTIONThe acid dissociation constant for NH4+ is 5.70 × 10–10; thus the initial pH ofthe buffer isC0.020pH = 9.24 + log NH 3 = 9.24 + log= 9.06CNH +0.0304Adding NaOH converts a portion of the NH4+ to NH3 due to the followingreactionNH4+(aq) + OH–(aq) NH3(aq) + H2O(l)tSince the equilibrium constant for this reaction is large, we may treat thereaction as if it went to completion.
The new concentrations of NH4+ and NH3are thereforeCNH 4+ ==CNH 3 ==moles NH 4+ – moles OH –Vtot(0.030 M)(0.10 L) – (0.10 M)(1.00 × 10 –3 L)= 0.029 M0.101 Lmoles NH3 + moles OH –Vtot(0.020 M)(0.10 L) + (0.10 M)(1.00 × 10 –3 L)= 0.021 M0.101 LHenderson–Hasselbalch equationEquation showing the relationshipbetween a buffer’s pH and the relativeamounts of the buffer’s conjugate weakacid and weak base.1400-CH06 9/9/99 7:41 AM Page 170170Modern Analytical ChemistrySubstituting the new concentrations into the Henderson–Hasselbalch equationgives a pH ofpH = 9.24 + log0.021= 9.100.029Multiprotic weak acids can be used to prepare buffers at as many different pH’sas there are acidic protons. For example, a diprotic weak acid can be used to preparebuffers at two pH’s and a triprotic weak acid can be used to prepare three differentbuffers.
The Henderson–Hasselbalch equation applies in each case. Thus, buffers ofmalonic acid (pKa1 = 2.85 and pKa2 = 5.70) can be prepared for whichpH = 2.85 + logCHM –CH 2 MpH = 5.70 + logCM 2 –CHM –where H2M, HM–, and M2– are the different forms of malonic acid.The capacity of a buffer to resist a change in pH is a function of the absoluteconcentration of the weak acid and the weak base, as well as their relative proportions. The importance of the weak acid’s concentration and the weak base’s concentration is obvious.
The more moles of weak acid and weak base that a bufferhas, the more strong base or strong acid it can neutralize without significantlychanging the buffer’s pH. The relative proportions of weak acid and weak base affect the magnitude of the change in pH when adding a strong acid or strong base.Buffers that are equimolar in weak acid and weak base require a greater amount ofstrong acid or strong base to effect a change in pH of one unit.2 Consequently,buffers are most effective to the addition of either acid or base at pH values nearthe pKa of the weak acid.Buffer solutions are often prepared using standard “recipes” found in thechemical literature.3 In addition, computer programs have been developed to aid inthe preparation of other buffers.4 Perhaps the simplest means of preparing a buffer,however, is to prepare a solution containing an appropriate conjugate weak acidand weak base and measure its pH. The pH is easily adjusted to the desired pH byadding small portions of either a strong acid or a strong base.Although this treatment of buffers was based on acid–base chemistry, the ideaof a buffer is general and can be extended to equilibria involving complexation orredox reactions.
For example, the Nernst equation for a solution containing Fe2+and Fe3+ is similar in form to the Henderson–Hasselbalch equation.° 3+E = EFe/ Fe 2 + − 0.05916 log[Fe2 + ][Fe3 + ]Consequently, solutions of Fe2+ and Fe3+ are buffered to a potential near thestandard-state reduction potential for Fe3+.6H.2 Representing Buffer Solutions with Ladder DiagramsLadder diagrams provide a simple graphical description of a solution’s predominatespecies as a function of solution conditions. They also provide a convenient way toshow the range of solution conditions over which a buffer is most effective. For ex-1400-CH06 9/9/99 7:41 AM Page 171Chapter 6 Equilibrium ChemistrypHp EDTAF–4.17ECa2+Sn4+11.690.184pKa,HF = 3.172.17E°Sn4+/Sn2+ = 0.154log Kf,Ca(EDTA)2– = 10.699.690.124Ca(EDTA)2–HF171Sn2+Figure 6.13(a)(b)(c)ample, an acid–base buffer can only exist when the relative abundance of the weakacid and its conjugate weak base are similar.
For convenience, we will assume thatan acid–base buffer exists when the concentration ratio of weak base to weak acid isbetween 0.1 and 10. Applying the Henderson–Hasselbalch equationpH = pKa + log1= pKa – 110pH = pKa + log10= pKa + 11shows that acid–base buffer exists within the range of pH = pKa ± 1. In the samemanner, it is easy to show that a complexation buffer for the metal–ligand complexML n exists when pL = log K f ± 1, and that a redox buffer exists forE = E° ± (0.05916/n). Ladder diagrams showing buffer regions for several equilibriaare shown in Figure 6.13.6I Activity EffectsSuppose you need to prepare a buffer with a pH of 9.36. Using the Henderson–Hasselbalch equation, you calculate the amounts of acetic acid and sodium acetateneeded and prepare the buffer.
When you measure the pH, however, you find thatit is 9.25. If you have been careful in your calculations and measurements, what canaccount for the difference between the obtained and expected pHs? In this section,we will examine an important limitation to our use of equilibrium constants andlearn how this limitation can be corrected.Careful measurements of the solubility of AgIO3 show that it increases in thepresence of KNO3, even though neither K+ or NO3– participates in the solubility reaction.5 Clearly the equilibrium position for the reactionAgIO3(s)t Ag+(aq) + IO3–(aq)Ladder diagrams showing buffer regions for(a) HF/F– acid–base buffer;(b) Ca2+/Ca(EDTA)2– metal–ligandcomplexation buffer; and (c) SN4+/Sn2+oxidation–reduction buffer.1400-CH06 9/9/99 7:41 AM Page 172172Modern Analytical Chemistrydepends on the composition of the solution.
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