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Polyprotic Acids A monoprotic acid is a molecule that is capable of donating only a single proton to an acceptor molecule.A polyprotic acid is a molecule capable of donating two or more protons to acceptor molecule(s).H2SO4(aq) + H2O() H3O+(aq) + HSO4-(aq) strong acidHSO4-(aq) + H2O() H3O+(aq) + SO42-(aq) weak acidHSO4-(aq) is amphoteric, it can act as a proton donor or acceptor.

Weak Polyprotic Acids H2CO3(aq) + H2O() H3O+(aq) + HCO3-(aq)Ka1 = [H3O+][HCO3-] / [H2CO3]HCO3-(aq) + H2O() H3O+(aq) + CO32-(aq)Ka2 = [H3O+][CO32-] / [HCO3-]The [H3O+] in the two ionization equilibria are one in the same.Successive ionization constants usually decrease by a factor of ~104 to 106. The negative charge left behind by the loss of a proton causes the next proton to be more tightly bound, such that each step in the ionization of a polyprotic acid occurs to a much lesser extent than the previous step.Weak Polyprotic Acids The exact calculation of simultaneous equilibria can be complex.

However, the solution to the simultaneous equilibria simplifies considerably if . . .The original concentration is not too small and Ka1 and Ka2 differ by more than a factor of 100, which is generally the case.Diprotic Acid with K1 >> K2H2A + H2O H3O+ + HA- HA- + H2O H3O+ + A2- (1) K1 > 100K2(2) Solution contains mainly H2A with a little HA-(3) Second equilibrium can be ignored (4) Calculate pH from the expression for K1K1 = [H3O+][HA-] / [H2A]Diprotic Acid with K1 >> K2Calculate the pH of a 0.15 M solution of malonic acid (pK1 = 2.85 and pK2 = 5.66)K1 > 100K2 but K1 is large enough that [H2A] C H2AH2A + H2O H3O+ + HA- At equilibrium (1) [HA-] = [H3O+](2) [H2A] = C H2A - [H3O+]K1 = [H3O+][HA-] / [H2A] = [H3O+]2 / 0.15 [H3O+]Solve the quadratic equation giving [H3O+] = 1.38 x 10-2pH = 1.86 Half Salt of a Diprotic AcidNaHA Na+ + HA-H2A + H2O H3O+ + HA- (1)HA- + H2O H3O+ + A2- (2)Both equlibria effect [HA-](1) HA- is removed in (1) and converted to [H2A](2) HA- is converted in (2) to A2-Mass balance [A2-] = [H3O+] + [H2A] (eliminate HA-)From K2 = [H3O+][A2-] / [HA-] [A2-] = K2[HA-] / [H3O+]Substitute for [A2-] in the mass balanceK2[HA-] / [H3O+] = [H3O+] + [H2A](3) Half Salt of a Diprotic Acid ContdFrom K1 = [H3O+] [HA-] / [H2A][H2A] = [H3O+] [HA-] / K1 Substitute for [H2A] in (3)[H3O+] + ([H3O+] [HA-] / K1) = K2[HA-] / [H3O+]Multiply through by [H3O+]K1 and rearrange[H3O+]2(K1 + [HA-]) = K1 K2[HA-][H3O+]2 = K1 K2[HA-] / (K1 + [HA-])(formal solution)

Notice: for typical situations [HA-] >> K1 K1 + [HA-] [HA-]then [H3O+]2 = K1 K2 pH =(1/2)(pK1 + pK2)

Salt of a Diprotic Acid Na2A 2 Na+ + A2-H2A + H2O H3O+ + HA- (1)HA- + H2O H3O+ + A2- (2)The solution contains HA- and A2- with little H2A Equilibrium (1) is not important and pH can be calculated from (2) K2 = [H3O+][A2-] / [HA-]

Calculate the pH of a solution containing hydrogen malonate [HA-] = 0.15 M and malonate [A2-] = 0.05 M (K2 = 2.2 x 10-6 )2.2 x 10-6 = [H3O+][0.05] / [0.15][H3O+] = 6.6 x 10-6 and pH = 5.18 Diprotic BuffersPrepared from the diprotic acid and the half salt of the same diprotic acidpH = pK1 + log [HA-] / [H2A]Prepared from the half salt and fully substituted salt of the same diprotic acidpH = pK2 + log [A2-] / [HA-] 11Amino Acids The amino acids are the building blocks of proteins and are examples of diprotic acids (or triprotic acids).

The R group is different for all of the amino acids, it can be nonpolar, polar, acidic or basic.

Amino Acids13Diprotic Acids and Bases The zwitterionic form of an amino acid (molecule with both positive and negative charges) is that found in solution at neutral pH because the carboxyl group is more acidic than the ammonium group.

The charge state of both groups depends upon the pH, both can be protonated or deprotonated.

The structures and acid dissociation constants of the common amino acids are summarized in Table 9-1.14Amino Acids Amino acids can be thought of as diprotic acids (and triprotic acids if the side chain is acidic or basic).H2L+ + H2O HL + H3O+Ka1 = K1HL + H2O L- + H3O+ Ka2 = K2L- + H2O HL + OH-Kb1HL + H2O H2L+ + OH-Kb2Recall the relationship between Ka and Kb for polyprotic acids and basesKa1 Kb2 = KwandKa2 Kb1 = KwFor most diprotic acids, Ka1 >> Ka2 and it can be approximated as a monoprotic acid with Ka = Ka1.15Example:Calculate the pH of a 0.100 M H2L+ solution (made by dissolving H2L+Cl- in solution, for L=glycine).

H2L+ + H2O HL + H3O+Ka1 = K1 = 4.47 x 10-3HL + H2O L- + H3O+Ka2 = K2 = 1.67 x 10-10Ka1 =[HL][H3O+]/[H2L+] = x2/(0.100-x) = 4.47 x 10-3x2 + 4.47 x 10-3 x 4.47 x 10-4 = 0x = 1.90 x 10-2 MpH = 1.721[HL] = [H3O+] = 1.90 x 10-2 M[H2L+] = {0.100 1.90 x 10-2} M = 0.081 M[L-] = Ka2[HL]/[H3O+] = 1.67 x 10-1016Example: Calculate the pH of a 0.100 M L- solution (made by dissolving NaL in solution, for L=glycine).

L- + H2O HL + OH-Kb1 = Kw/Ka2 = 5.99 x 10-5HL + H2O H2L+ + OH-Kb2 = Kw/Ka1 = 2.24 x 10-12Kb1 =[HL][OH-]/[L-] = x2/(0.100-x) = 5.99 x 10-5x2 + 5.99 x 10-5 x 5.99 x 10-6 = 0x = 2.42 x 10-3 MpOH = 2.616pH = 11.384[HL] = [OH-] = 2.42 x 10-3 M[L-] = {0.100 2.42 x 10-3} M = 0.098 M[H2L+] = Kb2[HL]/[OH-] = 2.24 x 10-1217Amino AcidsThe situation is not so simple if we start out with the intermediate form of the diprotic acid or base, HL.Both equilibria must be considered simultaneously. The problem simplifies somewhat because HL acts as both a weak acid and a weak base and thus will exist primarily in the HL form and can often be approximated as the initial concentration of HL.HL + H2O L- + H3O+Ka = 1.67 x 10-10HL + H2O H2L+ + OH-Kb = 2.24 x 10-1218Example: Calculate the pH of a 0.100 M HL solution.HL + H2O L- + H3O+Ka = 1.67 x 10-10HL + H2O H2L+ + OH-Kb = 2.24 x 10-12Ka = [L-][H+]/[HL] = 1.67 x 10-10(1)Kb = [H2L+][OH-]/[HL] = 2.24 x 10-12(2)Kw = [H+][OH-] = 1.00 x10-14(3)charge balance[H+] + [H2L+] = [OH-] + [L-](4)mass balance0.100 = [H2L+] + [HL] + [L-](5)19Example: Calculate the pH of a 0.100 M HL solution.

Solving equation (2) for [H2L+] gives . . .[H2L+] = Kb[HL]/[OH-] = Kb[HL][H3O+]/KwSolving equation (1) for [L-] gives . . .[L-] = Ka[HL]/[H3O+]Substituting both of these into equation (4) gives . . .[H3O+] + Kb[HL][H3O+]/Kw = Kw/[H3O+] + Ka[HL]/[H3O+]Solving for [H3O+] gives . . .{1 + Kb[HL]/Kw}[H3O+]2 = Kw + Ka[HL][H3O+] = ({Kw + Ka[HL]}/{1 + Kb[HL]/Kw})20Example: Calculate the pH of a 0.100 M HL solution.

The major species will be HL so this can be approximated by the initial concentration, 0.100 M

[H3O+] ({Kw + 0.100Ka}/{1 + 0.100Kb/Kw}) = ({1.00 x 10-14 +0.100(1.67 x 10-10)} {1 + 0.100(2.24 x 10-12/1.00 x 10-14)}) = 8.45 x 10-7 M

pH = -log[H3O+] = 6.0721Example: Calculate the pH of a 0.100 M HL solution.

[L-] = Ka[HL]/[H3O+] = (1.67 x 10-10)(0.100)/(8.45 x 10-7) = 1.98 x 10-5 M[H2L+] = Kb[HL][H3O+]/Kw = (2.24 x 10-12)(0.100)(8.45 x 10-7)/1.00 x 10-14) = 1.89 x 10-5 MThe approximation that [HL] = 0.100 M was a good one because the concentrations of L- and H2L+ are both pKa the conjugate base will dominateAt pH < pKa, the acid will dominate.pH = pKa + log [A-]/[HA]The analogous argument applies to polyprotic systems, but all pKas must be considered.Species in Solution for a Monoprotic AcidHA + H2O H3O+ + A-Mass Balance CHA = [A-] + [HA] Ka = [H3O+][A-] / [HA]

Ka = [H3O+](CHA - [HA]) / [HA] (substitute for [A-]) [HA] = CHA [H3O+]/{[H3O+] + Ka}(rearrange for [HA])

aHA = fraction of all species containing A that is HA aHA = [HA]/{[HA] + [A-]} = [HA]/CHA = [H3O+]/{[H3O+] + Ka}

aA = fraction of all species containing A that is A- aA- = [A-]/{[HA] + [A-]} = [A-]/CA- = Ka/{[H3O+] + Ka}

Each species given as an expression that contains only pH and an equilibrium constant

Fractional composition diagram for a monoproticacid with pKa = 5Species in Solution for a Diprotic Acid28 Fraction of species containing A present as H2AaH2A = [H2A]/CH2A = [H3O+]2/{[H3O+]2 + K1[H3O+] + K1K2}Fraction of species containing A present as HA-aHA- = [HA-]/C = K1[H3O+]/{[H3O+]2 + K1[H3O+] + K1K2}Fraction of species containing A present as HA-aA2- = [A2-]/C = K1K2/{[H3O+]2 + K1[H3O+] + K1K2}

Fractional compositiondiagram for a diprotic acid

Low pH H2A dominates

Intermediate pH HA- dominates

High pH A2- dominates Titration Reactions

A volumetric analysis is any procedure in which the volume of reagent needed to react with analyte is determined.

In a titration, small increments of a reagent solution (the titrant) are added to an analyte solution until their reaction is complete.

From the quantity of titrant required, the unknown quantity of analyte present can be determined. Titration Reactions

Requirements for a titration reaction . . .The reaction must have a large equilibrium constant.The titration reaction must proceed rapidly.

Common types of titration reactionsacid-baseoxidation-reductioncomplex formationprecipitation32 Acid-Base Titrations

A titration curve is a graph showing how the pH changes as titrant is added.Reaction goes to completion as titrant is added.

It exhibits a rapid change in pH near the equivalence point.At the equivalence pointThe slope dpH/dV The second derivative d2pH/dV2 = 0 (an inflection point).33 Titration of Strong Acid w/ Strong BaseTitration Reaction:H3O+(aq) + OH-(aq) 2 H2O() Region 1: Before the Equivalence PointpH is determined by excess acid in solution not yet neutralized Region 2: At the Equivalence PointpH is determined by the autoionization of H2O pH = 7.00 for titration of strong acid with strong base.Region 3: After the Equivalence PointpH is determined by excess titrant (base) added to the solution 34 Titration of Strong Acid with Strong BaseRegion 1: Before the equivalence pointRegion 2: At the equivalence pointRegion 3: After the equivalence pointInflection pointdpH2/dV2=0VbasepH

Titration of a Strong Base with a Strong AcidRegion 1Before the equivalence pointRegion 2At the equivalence pointRegion 3After the equivalence point36Example: Consider the titration of 100 mL of 0.100 M HCl with 0.500 M NaOH.

a. Determine the volume of NaOH that must be added to reach equivalence.moles acid = moles baseMacidVacid = MbaseVbase

Vbase = MacidVacid / Mbase = (0.100M)(100mL)/(0.500M) = 20.0 mL37Example: Consider the titration of 100 mL of 0.100 M HCl with 0.500 M NaOH.

b. Determine the pH after 5.00 mL of NaOH have been added.pH is determined by the amount of excess acid still present.MnewVnew = M1 V1 x (% remaining) [H+] x 105 = 0.100 x 100 mL x (Ve-Vbase)/Ve

= (0.100M)(100.mL) x (20.0-5.00)mL/20.0mL /105mL= 0.0714 M

pH = -log[H+] = -log(0.0714) = 1.15

38Example: Consider the titration of 100. mL of 0.100 M HCl with 0.500 M NaOH.

d. Determine the pH after 25.0 mL of NaOH have been added.pH is determined by the excess amount of titrant (NaOH) that has been added.Vexcess base = Vbase - Ve = 25.0 20.0 = 5.0 mLMnewVnew = M1V1 (excess base)[OH-] x 125 = 0.500 x 5.0 mL =0.500 x 5.0 / 125 = 0.020 M

pH = -log[H+] = -log{Kw/[OH-]} = -log{1.00x10-14/0.020} = 12.3039Titration of Weak Acid with Strong BaseTitration Reaction:HA (aq) + OH- (aq) A-(aq) + H2O()

Reaction goes to completion as strong base is added.

Region 1:Before Base is AddedpH is determined by the equilibrium for the weak acid.HA + H2O H3O+ + A-Ka = [H3O+][A-]/[HA]40Titration of Weak Acid with Strong BaseRegion 2: Before the Equivalence PointThe Henderson-Hasselbach equation is used to determine the pH because the solution is a buffer (HA and A-) are present.pH = pKa + log{[A-]/[HA]}Region 3: At the Equivalence PointAll of the acid has been converted to its conjugated base.pH is determined by the equilibrium for the conjugate baseA- + H2O HA + OH-Kb = Kw/Ka = [HA][OH-]/[A-]41Titration of Weak Acid with Strong BaseRegion 4: After the Equivalence PointpH is determined by the excess titrant (OH-, base) added to the solution.42

Titration of Weak Acid with Strong BaseRegion 1: Region 4: Region 2: Region 3: CHAPTER 10: Figure 10.2

4344Example: Consider the titration of 50.0 mL of 0.200 M acetic acid (HOAc) with 0.250 M NaOH. HOAcpKa = 4.76

a. Determine the volume of NaOH that must be added to reach equivalence.moles acid = moles baseMacidVacid = MbaseVbaseVbase = MacidVacid / Mbase = (0.200M)(50.0mL)/(0.250M) = 40.0 mL45Example: Consider the titration of 50.0 mL of 0.200 M acetic acid (HOAc) with 0.250 M NaOH. HOAcpKa = 4.76c. Determine the pH after 10.0 mL of titrant (NaOH) have been added.The Henderson-Hasselbach equation is used to determine the pH because the solution is a buffer (HA and A-) are present.pH = pKa + log{[A-]/[HA]}

Titration Reaction:HA + OH- A- + H2ORelative initial quantities1Relative final quantities = 4.76 + log ( / ) = 4.76 + log(0.25/0.75) = 4.28 10 of 40 (equivalence point)46Example: Consider the titration of 50.0 mL of 0.200 M acetic acid (HOAc) with 0.250 M NaOH. HOAcpKa = 4.76d. Determine the pH after 40.0 mL of titrant (NaOH) have been added.This is the equivalence point. All of the acid has been converted to its conjugate base, pH is determined by the equilibrium for the conjugate baseA- + H2O OH- + HAKb = [OH-][HA]/[A-] 0.200(50/90)-y y y

Kb = [HA][OH-]/[A-] = Kw/Ka = {1.00x10-14/1.74x10-5} = 5.75 x10-10= y2/0.111-y y2/0.111 assume y 8.9)Transition range pK 156Choosing an Indicator

An indicator whose transition range overlaps the steepest part of the titration curve as closely as possible will allow the most accurate determination of the end point.

The difference between the observed end point (color change) and the true equivalence point is called the indicator error indicator error.Indicators are also acids (bases) and therefore react with the titrant and influence the indicator error.

Choosing an Indicator

Both indicators could be used for this titration, but bromocresol purple would lead to a slightly smaller indicator error because the color transitions at a pH closer to the equivalence pH58