D. Harvey - Modern Analytical Chemistry (794078), страница 49
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Thus, the [H3O+] is 2.6 × 10–2 M, andthe pH to two significant figures is still 1.59.This same approach can be extended to find the pH of a monoprotic weakbase, replacing Ka with Kb, CHF with the weak base’s concentration, and solving forthe [OH–] in place of [H3O+].EXAMPLE 6.11Calculate the pH of 0.050 M NH3. State any assumptions made in simplifyingthe calculation, and verify that the error is less than 5%.SOLUTIONSince NH3 is a weak base (Kb = 1.75 × 10–5), we assume that[OH–] >> [H3O+]andCNH3 = 0.050 MWith these assumptions, we find (be sure to check the derivation)[OH – ] =K b CNH 3 =(1.75 × 10 –5 )(0.050) = 9.35 × 10 –4 M1400-CH06 9/9/99 7:41 AM Page 163163Chapter 6 Equilibrium ChemistryBoth assumptions are acceptable (again, verify that this is true).
Theconcentration of H3O+ is calculated using Kw[H 3 O + ] =Kw1.00 × 10 –14= 1.07 × 10 –11=9.35 × 10 –4[OH – ]giving a pH of 10.97.CH3+H3NCHCH3pKa1 = 2.348+HCOOHCH3NH2L+CH3pKa2 = 9.867COO –H2NHLCHCOO –Figure 6.11Acid–base equilibria for theamino acid alanine.L–6G.5 pH of a Polyprotic Acid or BaseA more challenging problem is to find the pH of a solution prepared from apolyprotic acid or one of its conjugate species. As an example, we will use theamino acid alanine whose structure and acid dissociation constants are shown inFigure 6.11.CH3H2L+pH of 0.10 MAlanine hydrochloride is a salt consisting of the diprotic weakacid H2L+ and Cl–.
Because H2L+ has two acid dissociation reactions, a completesystematic solution to this problem will be more complicated than that for a monoprotic weak acid. Using a ladder diagram (Figure 6.12) can help us simplify theproblem. Since the areas of predominance for H2L+ and L– are widely separated, wecan assume that any solution containing an appreciable quantity of H2L+ will contain essentially no L–. In this case, HL is such a weak acid that H2L+ behaves as if itwere a monoprotic weak acid.To find the pH of 0.10 M H2L+, we assume thatH2NCOO –L–9.867CH3pH+H3N[H3O+] >> [OH–]CHCOO –HLBecause H2L+ is a relatively strong weak acid, we cannot simplify the problem further, leaving us withKa =CH2.348CH3O + ]2[H 3CH 2L+ – [H 3O + ]Solving the resulting quadratic equation gives the [H3O+] as 1.91 × 10–2 M or apH of 1.72.
Our assumption that [H3O+] is significantly greater than [OH–] isacceptable.pH of 0.10 M L– The alaninate ion is a diprotic weak base, but using the ladder diagram as a guide shows us that we can treat it as if it were a monoprotic weak base.Following the steps in Example 6.11 (which is left as an exercise), we find that thepH of 0.10 M alaninate is 11.42.+H3NCHCOOHH2L+Figure 6.12Ladder diagram for the amino acid alanine.1400-CH06 9/9/99 7:41 AM Page 164164Modern Analytical ChemistrypH of 0.1 M HL Finding the pH of a solution of alanine is more complicated thanthat for H2L+ or L– because we must consider two equilibrium reactions involvingHL.
Alanine is an amphiprotic species, behaving as an acidHL(aq) + H2O(l)and a baseHL(aq) + H2O(l)t H3O+(aq) + L–(aq)t OH–(aq) + H2L+(aq)As always, we must also consider the dissociation of water2H2O(l)t H3O+(aq) + OH–(aq)This leaves us with five unknowns ([H2L+], [HL], [L–], [H3O+], and [OH–]), forwhich we need five equations. These equations are Ka2 and Kb2 for HL,Ka2 =K b2 =[H 3O + ][L– ][HL][OH – ][H 2 L+ ]Kw=[HL]Ka1the Kw equation,Kw = [H3O+][OH–]a mass balance equation on HL,CHL = [H2L+] + [HL] + [L–]and a charge balance equation[H2L+] + [H3O+] = [OH–] + [L–]From the ladder diagram it appears that we may safely assume that the concentrations of H2L+ and L– are significantly smaller than that for HL, allowing us to simplify the mass balance equation toCHL = [HL]L+]Next we solve Kb2 for [H2[H 2 L+ ] =K w[HL][HL][H 3O + ] CHL[H 3O + ]==Ka1Ka1Ka1[OH – ]and Ka2 for [L–][L– ] =Ka2 [HL]K C= a2 HL[H 3 O + ][H 3 O + ]Substituting these equations, along with the equation for Kw, into the charge balance equation gives usCHL[H 3O + ]KwK C+ [H 3 O + ] =+ a2 HLKa1[H 3 O + ] [H 3 O + ]which simplifies toC1[H 3O + ] HL + 1 =(K w + Ka2 CHL ) Ka1 [H 3 O + ][H 3O + ]2 =[H 3 O + ] =Ka2 CHL + K w(CHL / Ka1 ) + 1Ka1Ka2 CHL + Ka1K wCHL + Ka16.411400-CH06 9/9/99 7:41 AM Page 165Chapter 6 Equilibrium ChemistryWe can simplify this equation further if Ka1Kw << Ka1Ka2CHL, and if Ka1 << CHL,giving[H 3 O + ] =Ka1Ka2O+]For a solution of 0.10 M alanine, the [H3[H 3 O + ] =6.42is(4.487 × 10 –3 )(1.358 × 10 –10 ) = 7.807 × 10 –7 Mor a pH of 6.11.
Verifying that the assumptions are acceptable is left as an exercise.Triprotic Acids and Bases, and Beyond The treatment of a diprotic acid or base iseasily extended to acids and bases having three or more acid–base sites. For a triprotic weak acid such as H3PO4, for example, we can treat H3PO4 as if it was a monoprotic weak acid, H2PO4– and HPO42– as if they were intermediate forms of diproticweak acids, and PO43– as if it was a monoprotic weak base.EXAMPLE 6.12Calculate the pH of 0.10 M Na2HPO4.SOLUTIONWe treat HPO42– as the intermediate form of a diprotic weak acidH2PO4–(aq)t HPO42–(aq) t PO43–(aq)where the equilibrium constants are Ka2 = 6.32 × 10–8 and Ka3 = 4.5 × 10–13.Since the value of Ka3 is so small, we use equation 6.41 instead of equation 6.42.[H 3 O + ] =(6.32 × 10 –8 )(4.5 × 10 –13 )(0.10) + (6.32 × 10 –8 )(1.00 × 10 –14 )0.10 + 6.32 × 10 –8= 1.86 × 10 –10or a pH of 9.73.6G.6 Effect of Complexation on SolubilityThe solubility of a precipitate can be improved by adding a ligand capable offorming a soluble complex with one of the precipitate’s ions.
For example, thesolubility of AgI increases in the presence of NH3 due to the formation of thesoluble Ag(NH3)2+ complex. As a final illustration of the systematic approachto solving equilibrium problems, let us find the solubility of AgI in 0.10 MNH3.We begin by writing the equilibria that we need to considerAgI(s)t Ag+(aq) + I–(aq)Ag+(aq) + 2NH3(aq)NH3(aq) + H2O(l)2H2O(l)t Ag(NH3)2+(aq)t OH–(aq) + NH4+(aq)t H3O+(aq) + OH–(aq)1651400-CH06 9/9/99 7:41 AM Page 166166Modern Analytical ChemistryCounting unknowns, we find that there are seven—[Ag+], [I–], [Ag(NH 3)2+],[NH3], [NH4+], [OH–], and [H3O+]. Four of the equations needed to solve thisproblem are given by the equilibrium constant expressionsKsp = [Ag + ][I – ] = 8.3 × 10 –17β2 =Kb =[Ag(NH 3 )2+ ]= 1.7 × 107[Ag + ][NH 3 ]2[NH 4+ ][OH – ]= 1.75 × 10 –5[NH 3 ]K w = [H 3O + ][OH – ] = 1.00 × 10 –14Three additional equations are needed. The first of these equations is a mass balancefor NH3.CNH3 = [NH3] + [NH4+] + 2 × [Ag(NH3)2+]Note that in writing this mass balance equation, the concentration of Ag(NH3)2+must be multiplied by 2 since two moles of NH3 occurs per mole of Ag(NH3)2+.
Thesecond additional equation is a mass balance on iodide and silver. Since AgI is theonly source of I– and Ag+, every iodide in solution must have an associated silverion; thus[I–] = [Ag+] + [Ag(NH3)2+]Finally, the last equation is a charge balance equation[Ag+] + [Ag(NH3)2+] + [NH4+] + [H3O+] = [I–] + [OH–]Our problem looks challenging, but several assumptions greatly simplify the algebra. First, since the formation of the Ag(NH3)2+ complex is favorable, we will assume that[Ag+] << [Ag(NH3)2+]Second, since NH3 is a base, we will assume that[H3O+] << [OH–][NH4+] << [NH3] + [Ag(NH3)2+]Finally, since Ksp is significantly smaller than β2, it seems likely that the solubility ofAgI is small and[Ag(NH3)2+] << [NH3]Using these assumptions allows us to simplify several equations.
The mass balance for NH3 is nowCNH3 = [NH3]and the mass balance forI–is[I–] = [Ag(NH3)2+]Simplifying the charge balance expression by dropping [H3O+] and [Ag+], and replacing [Ag(NH3)2+] with [I–] gives1400-CH06 9/9/99 7:41 AM Page 167Chapter 6 Equilibrium Chemistry167[NH4+] = [OH–]We continue by multiplying together the equations for Ksp and β2, givingKspβ2 =[Ag(NH 3 )2+ ][I – ]= 1.4 × 10 –9[NH 3 ]2Substituting in the new mass balance equations for NH3 and I–[I – ]2= 1.4 × 10 –9(CNH 3 )2and solving for the [I–] gives[I – ]2= 1.4 × 10 –9(0.10 M)2[I – ] = 3.7 × 10 –6 MBefore accepting this answer, we first check our assumptions. Using the Kspequation we calculate the [Ag+] to be[Ag + ] =Ksp[I – ]=8.3 × 10 –17= 2.2 × 10 –11 M3.7 × 10 –6From the simplified mass balance equation for I–, we have[Ag(NH3)2+] = [I–] = 3.7 × 10–6 MOur first assumption that the [Ag+] is significantly smaller than the [Ag(NH3)2+],therefore, is reasonable.
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