3 STRONG ACIDS AND BASESStrong acids and bases react nearly “completely” to produce H+ and OH- equilibrium constants are largee.g.: HCl H+ + Cl-Complete dissociation:largesmallCommon strong acids:HCl, HBr, HI, H2SO4, HNO3, HClO4(Why is HF not a strong acid?)Common strong bases:LiOH, NaOH, KOH, RbOH, CsOH, R4NOH
4 LiOH Li+ + OH- Example: Calculate the pH of 0.1M LiOH. 0.1M 0.1M Strong base Dissociates completelyLiOH Li OH-0.1MStart:0.1M0.1MComplete rxn:pOH = -log [OH-]= -log (0.1) = 1 pH = = 13
5 IMPOSSIBLE!!!!!! Problem: What is the pH of 1x10-8 M KOH? As before: pOH = -log (1x10-8) = 8pH = 14 – 8 = 6BUTpH 6 acidic conditions and KOH is a strong baseIMPOSSIBLE!!!!!!
6 We do this by systematic treatment of equilibrium. Since the concentration of KOH is so low (1x10-8 M), we need to take the ionisation of water into account.In pure water [OH-] = 1x10-7 M, which is greater than the concentration of OH- from KOH.We do this by systematic treatment of equilibrium.Charge balance: [K+] + [H+] = [OH-]Mass balance: [K+] = 1x10-8 MEquilibria: [H+][OH-] = Kw = 1x10-14 M3 equations + 3 unknowns solve simultaneouslyFind: pH = 7.02Hint: You end up with a quadratic equation which you solve using the formula.
7 Also note that:Only pure water produces 1x10-7 M H+ and OH-.If there is say 1x10-4 M HBr in solution,pH = 4 and [OH-] = 1x10-10 MBut the only source of OH- is from the dissociation of water. if water produces 1x10-10 M OH- it can only produce 1x10-10 M H+ due to the dissociation of water. pH in this case is due mainly to the dissociation of HBr and not the dissociation of water.It is thus important to look at the concentrationof acid and bases present.
8 Some guidelines regarding the concentrations of acids and bases: When conc > 1x10-6 M calculate pH as usualWhen conc < 1x10-8 M pH = 7(there is not enough acid or base to affect the pH of water)When conc 1x x10-6 M Effect of water ionisation and added acid and bases are comparable, thus:use the systematic treatment of equilibrium approach.
9 WEAK ACIDS AND BASESWeak acids and bases react only “partially” to produce H+ and OH- equilibrium constants are smallHA H+ + A-Partial dissociationsmalllargeAcid dissociation constantCommon weak acids:carboxylic acids(e.g. acetic acid = CH3COOH)ammonium ions(e.g. RNH3+, R2NH2+, R3NH+)Common weak bases:carboxylate anions(e.g. acetate = CH3COO-)amines(e.g. RNH2, R2NH, R3N)
10 base hydrolysis constant/ base “dissociation” constant B + H2O BH+ + OH-Weak basepartial dissociationKb smallbase hydrolysis constant/ base “dissociation” constantNOTE:pKa = -log KapKb = -log KbAs K increases, its p-function decreases and vice versa.
11 Problem:Find the pH of a solution of formic acid given that the formal concentration is 2 M and Ka = 1.80x10-4.HCOOH H+ + HCOO-H2O H+ + OH-Systematic treatment of equilibria:Charge balance: [H+] = [HCOO-] + [OH-]Mass balance: 2 M = [HCOOH] + [HCOO-]Equilibria:4 equations 4 unknowns difficult to solve
12 [HCOO-] >> [OH-] Make an assumption:[H+] due to acid dissociation [H+] due to water dissociationProduces HCOO-Produces OH-[HCOO-] large[OH-] small [HCOO-] >> [OH-] Charge balance: [H+] [HCOO-]
13 Charge balance: [H+] [HCOO-] Mass balance: 2 M = [HCOOH] + [H+]Equilibria:Let [H+] = [HCOO-] = xOr x = No negative conc’s[H+] = [HCOO-] = M pH = 1.7
14 OR since [HCOOH] > 1x10-6, we can calculate pH as usual Weak acidequilibrium conditionsHCOOH H HCOO-2MStart:2-xxxEquilibrium:Solve as before
15 FRACTION OF DISSOCIATION, Fraction of acid in the form A-For the above problem: =[HCOO-]F=0.019 M2 M= Acid is 0.95% dissociated at 2 M formal concentrationWeak electrolytes dissociate more as they are diluted.
16 FRACTION OF ASSOCIATION WEAK BASE EQUILIBRIAB + H2O BH+ + OH-Charge balance: [BH+] = [OH-]Mass balance: F = [B] + [BH+]Equilibria:Let [BH+] = [OH-] = xFRACTION OF ASSOCIATION
17 CONJUGATE ACIDS AND BASES Relationship between Ka and Kb for a conjugate acid- base pair:Ka.Kb = Kw = 1x at 25oCIf Ka is very large (strong acid)Then Kb must be very small (weak conjugate base)And vice versaBase so weak it is nota base at all in waterIf Ka is very small, say 1x10-6 (weak acid)Then Kb must be small, 1x10-8 (weak conjugate base)Greater acid strength, weaker conjugate base strength, and vice versa.
18 Calculate the pH of 0.1 M NH3, given that pKa = 9.244 for ammonia. Problem:Calculate the pH of 0.1 M NH3, given that pKa = for ammonia.KbNH3 + H2O NH4+ + OH-baseacidpKa = -log KaKa = 5.70x10-10Kb =KwKa=1x10-145 .70x10-10= x10-5
19 Solve for x using the quadratic equation Problem:Calculate the pH of 0.1 M NH3, given that pKa = for ammonia.Kb =x2F - xWhere x = [OH-] = [NH+]=x20.1 - x= x10-5Solve for x using the quadratic equationNegative value discardedFind x = 1.31x10-3 M = [OH-]pOH = -log [OH-]= 2.88pH = 14 – 2.88 = 11.12
20 BUFFERS Mixture of an acid and its conjugate base. Buffer solution resists change in pH when acids or bases are added or when dilution occurs.Mix:A moles of weak acid + B moles of conjugate baseFind:moles of acid remains close to A, andmoles of base remains close to B Very little reactionHA H+ + A-Le Chatelier’s principle
21 HENDERSON-HASSELBALCH EQUATION For acids:When [A-] = [HA], pH = pKaFor bases:pKa applies to this acidKbB + H2O BH+ + OH-Kabaseacidacidbase
23 The acid or base is consumed by A- or HA respectively Why does a buffer resist change in pH when small amounts of strong acid or bases is added??The acid or base is consumed by A- or HA respectivelyA buffer has a maximum capacity to resist change to pH.Buffer capacity, : Measure of how well solution resists change in pH when strong acid/base is added.Larger more resistance to pH change
24 A buffer is most effective in resisting changes in pH when: pH = pKai.e.: [HA] = [A-] Choose buffer whose pKa is as close as possible to the desired pH.pKa 1 pH unit
25 Problem:Calculate the pH of a solution containing M NH3 and M NH4Cl given that the acid dissociation constant for NH4+ is 5.7x10-10.NH3 + H2O NH4+ + OH-acidbaseKapKa = 9.244pKa applies to this acidpH = log(0.200)(0.300)pH = 9.07
26 POLYPROTIC ACIDS AND BASES Can donate or accept more than one proton.In general:Diprotic acid:H2L HL- + H+ Ka1 K1HL L2- + H+ Ka2 K2Diprotic base:L2- + H2O HL- + OH- Kb1HL- + H2O H2L + OH- Kb2Relationships between Ka’s and Kb’s:Ka1. Kb2 = KwKa2. Kb1 = Kw
27 Using pKa values and mass balance equations, the fraction of each species can be determined at a given pH.
29 ACID-BASE TITRATIONSWe will construct graphs to see how pH changes as titrant is added.Start by:writing chemical reaction between titrant and analyteusing the reaction to calculate the composition and pH after each addition of titrant
30 TITRATION OF STRONG BASE WITH STRONG ACID Example:Titrate ml of M KOH with M HBr.HBr + KOH KBr + H2OWhat is of interest to us in an acid-base titration:H+ + OH- H2OMix strong acid and strong base reaction goes to completionH+ + OH- H2O
31 * Calculate volume of HBr needed to reach the equivalence point, Veq: Example:Titrate ml of M KOH with M HBr.* Calculate volume of HBr needed to reach the equivalence point, Veq:C1V1C2V2n1n2=But n1 = n2 = 1 CHBrVeq = CKOHVKOH( M)Veq = ( M)(50.00 ml)Veq = ml
32 There are 3 parts to the titration curve: 1Before reaching the equivalence point excess OH- presentAt the equivalence point [H+] = [OH-]23After reaching the equivalence point excess H+ present
33 Before reaching the equivalence point excess OH- present HBr + KOH KBr + H2OSay 2.00 ml HBr has been added.Starting nOH-= (0.02 M)(0.050 L)= 1x10-3 molCOH- =nunreactedVtotalnH+ added= (0.1 M)(0.002 L)= 2x10-4 molCOH- = MVtotal = mL= 52 mL= LnOH- unreacted= 8x10-4 molKw = [H+][OH-]1x10-14 = [H+]( M)[H+] = 6.500x10-13 M pH = 12.19
34 At the equivalence point nH+ = nOH-pH is determined by dissociation of H2O:H2O H+ + OH-xxKw = [H+][OH-]1x10-14 = x2x = 1x10-7 M [H+] = 1x10-7 M pH = 7pH = 7 at the equivalence point ONLY for strong acid – strong base titrations!!
35 After reaching the equivalence point excess H+ present HBr + KOH KBr + H2OSay ml HBr has been added.Starting nOH-= 1x10-3 molCH+ =nexcessVtotalnH+ added= (0.1 M)( L)= 1.010x10-3 molCH+ = 1.664x10-4 MpH = 3.78nH+ excess= 1x10-5 molVtotal = mL= 60.1 mL= L
36 Note:A rapid change in pH near the equivalence point occurs.Equivalence point where:slope is greatestsecond derivative iszero (point of inflection)
37 Calculate titration curve by calculating pH values after a number of additions of HBr.
38 TITRATION OF WEAK ACID WITH STRONG BASE Example:Titrate ml of M formic acid with M NaOH.HCO2H + NaOH HCO2Na + H2OOR HCO2H + OH- HCO2- + H2OHAA-pKa = 3.745Equilibrium constant so large reaction “goes to completion” after each addition of OH-Ka = 1.80x10-4Kb = 5.56x10-11Strong and weak react completely
39 * Calculate volume of NaOH needed to reach the equivalence point, Veq: Example:Titrate ml of M formic acid with M NaOH.HCO2H + OH- HCO2- + H2O* Calculate volume of NaOH needed to reach the equivalence point, Veq:C1V1C2V2n1n2=But n1 = n2 = 1 CNaOHVeq = CFAVFA( M)Veq = ( M)(50.00 ml)Veq = ml
40 There are 4 parts to the titration curve: Before base is added HA and H2O present. HA weak acid, pH determined by equilibrium:HA H+ + A-KaFrom first addition of NaOH to immediately before equivalence point mixture of unreacted HA and A-HA + OH- A- + H2OBUFFER!! use Henderson-Hasselbalch eqn for pH21
41 At the equivalence point all HA converted to A-. A- is a weak base whose pH is determined by reaction:A- + H2O HA + OH-Kb434) Beyond the equivalence point excess OH- added to A-.Good approx: pH determined by strong base (neglect small effect from A-)
42 Before base is added HA and H2O present. HA = weak acid.Ka = 1.80x10-4HA H+ + A-F- xx1.80x10-4 =x2xx x10-4x – 3.60x10-6 = 0x = 1.81x10-3 [H+] = 1.81x10-3 pH = 2.47
43 Say 2.00 ml NaOH has been added. From first addition of NaOH to immediately before equivalence point mixture of unreacted HA and A- . BUFFER!!HCO2H + OH- HCO2- + H2OSay 2.00 ml NaOH has been added.Starting nHA= (0.02 M)(0.05 L)= 1x10-3 molnOH- added= (0.1 M)(0.002 L)= 2x10-4 molHA + OH A- + H2O1x10-32x10-48x10-4Start-EndpH = log2x10-48x10-4 pH = 3.14
45 Special condition:When volume of titrant = ½ VeqpH = pKaSince:nHA = nA-
46 At the equivalence point all HA converted to A-. A- = weak base (nHA = nNaOH)Starting nHA= 1x10-3 molHA + OH A- + H2O1x10-3Start-End nOH-= 1x10-3 molSolution contains just A- a solution of weak base
47 A- + H2O HA + OH- Kb = 5.56x10-11 F- x x x FA- = nA- V = 1x10-3 mol Vtotal = mL= 60 mL = L= M5.56x10-11 =x2x[OH-] = 9.63x10-7 Mx x10-11x – 9.27x10-13 = 0pOH = 6.02x = 9.63x10-7pH = 7.98pH is slightly basic at equivalence point for strong base-weak acid titrations
49 Titration curve depends on Ka of HA. As HA becomes a weaker acid the inflection near the equivalence point decreases until the equivalence point becomes too shallow to detect not practical to titrate an acid or base that is too weak.
50 Titration curve depends on extent of dilution of HA. As HA becomes a more dilute the inflection near the equivalence point decreases until the equivalence point becomes too shallow to detect not practical to titrate a very dilute acid or base.
51 TITRATION OF WEAK BASE WITH STRONG ACID This is the reverse of the titration of weak base with strong acid.The titration reaction is:B + H+ BH+Recall:Strong and weak react completely
52 There are 4 parts to the titration curve: Before acid is added B and H2O present.B weak base pH determined by equilibrium:B + H2O BH+ + OH-KbF-xx
53 From first addition of acid to immediately before equivalence point mixture of unreacted B and BH+B + H BH+BUFFER!! use Henderson-Hasselbalch equation for pHpKa applies to this acid
54 At the equivalence point all B converted to BH+. BH+ is a weak acid determined pH by reaction:BH B + H+KaF-xxFBH+ =nVtotalTake dilution into accountpH is slightly acidic (pH below 7) for strong acid-weak base titrations
55 4) Beyond the equivalence point excess H+ added to BH+.Good approx: pH determined by strong acid (neglect small effect from BH+)Example:50.00 ml of 0.05 M NaCN is titrated with 0.1 M HCl. Ka for NaCN = 6.20x1010Draw the titration curve by calculating pH at various volumes of HCl.
59 FINDING END POINTS WITH A pH ELECTRODE After each small addition of titrant the pH is recorded and a titration curve is plotted.2 ways of determining end points from this:using derivativesusing a Gran plot
61 But there are autotitrators! Titrando from Metrohm
62 USING DERIVATIVES End point is taken where the slope is greatest Or where the 2nd derivative is zero
63 USING A GRAN PLOT A problem with using derivatives titration data is most to obtain near the end pointExample – titration of a weak acid, HAKa =[H+]H+[A-] A-[HA] HAHA H+ + A-NOTE:pH electrode responds to hydrogen ion ACTIVITY, not concentration
64 Say we titrated Va ml of HA (formal conc = Fa) with Vb ml of NaOH (formal conc = Fb) : HA + OH- A- + H2O[A-] =nOH(titrated)VTotal=VbFbVa + Vb[HA] =nHA(initial) – nOH(titrated)VTotal=VaFa- VbFbVa + VbSubstitute into the equilibrium constant:Ka =[H+]H+[A-] A-[HA] HA
66 Gran plot equation:Gran plot Graph of Vb10-pH vs VbIf is constant, then:A-HASlope = -KaA-HAandx-intercept = VeUse data taken before end point to find end pointCan determine Ka from slope
67 Use only linear portion of graph Extrapolate graph to get Ve
68 FINDING END POINTS WITH INDICATORS Acid-base indicator acid or base itselfVarious protonated species have different coloursHIn H+ + In-
69 Choose indicator whose colour change is as close as possible to the pH of the end point Indicators transition range overlaps the steepest part of the titration curve
70 Indicator error: difference between the observed end point (colour change) and the true equivalence point.Systematic errorRandom errorVisual uncertainty associated with distinguishing the colour of the indicator reproduciblyWhy do we only add a few drops of indicator?Indicator is an acid/base itself will react with analyte/titrantFew drops neglible relative to amount of analyte