Principles of Neutralization Titration Version 2012 Updated on 042312 Copyright © All rights reserved Dong-Sun Lee, Prof., Ph.D. Chemistry, Seoul Women’s University Chapter 12 Principles of Neutralization Titration

Standard solutions for Acid-Base Titrations Analyte Titrant vs Standard solution N V = N’V’ The standard reagents used in acid-base titrations are always strong acids or strong bases, most commonly HCl, HClO4, H2SO4, NaOH, KOH. Weak acids and bases are never used as standard reagents because they react incompletely with analyte. N= unknown V= measure N’= known V’= known

Primary standards for standardizing Acids Potassium acid phthalate Sulfamic acid ( H2NSO3H) HCl Potassium hydrogen iodate Bases TRIS(hydroxymethylaminomethane) Sodium carbonate Borax (= sodium tetraborate ) HgO 

General goal and procedure of acid-base titration  1) Standardization 2) Determination 3) Titration curve 4) Interpret titration curve , understand what is happening during titration Primary standard solution known normality(N) known volume(V) Acid or base standard solution unknown normality(N’) unknown volume(V’) known normality(N’) Analyte Sample solution unknown Ns known Vs Plots of pH versus volume of titrant

Finding the end point 1) Indicator 2) Potentiometry : Titration curve ( pH or mV vs Va ) 1st derivative titration curve 2nd derivative titration curve Gran plot 3) Conductometry 4) Spectrometry

Indicator Ex. Thymol blue H2In  HIn–  In2 – Transition range An acid-base indicator is itself an acid or base whose different protonated species have different colors. Ex. Thymol blue H2In  HIn–  In2 – Red Yellow Blue H2In  H+ + HIn– pKa1 = 1.7 pH = pKa1 + log [HIn– ]/[H2In] if [HIn– ]/[H2In]  1:10 pH = 0.70 Red color [HIn– ]/[H2In] = 1:1 pH = pKa1 = 1.7 Orange color [HIn– ]/[H2In]  10:1 pH = 2.70 Yellow color HIn–  H+ + In2 – pKa2 = 8.9 Transition range pKa11 The approximate pH transition range of most acid-type indicator is roughly pKa 1

Choosing an indicator PP 8.0~9.0 MR 4.8~6.0 MO 3.1~4.4 Phenolphthalein Volume of 0.1000M HCl (ml) 5 10 15 20 25 30 35 0.00 2.00 4.00 6.00 8.00 10.00 12.00 14.00 PP 8.0~9.0 MR 4.8~6.0 MO 3.1~4.4 pH Phenolphthalein acid form : colorless transition range(pH): 8.0~9.0 Phenolphthalein base form : pink The calculated pH titration curve for the titration of 10.00ml of 0.1000M Na2CO3 with 0.1000M HCl.

Calculated titration curve for the reaction of 100 mL of 0 Calculated titration curve for the reaction of 100 mL of 0.0100M base (pKb = 5.00) with 0.0500 M HCl.

Indicator color as a function of pH (pKa=5.0)

Experiments. Standardization of 0.1000N NaOH Primary standard KHP 204.22g/1000ml=1.0000N 0.35g/25ml = xN x = 0.06860N Titration of 25.00ml of KHP with NaOH End point=18.50ml 0.06860N×25.00ml= x N×18.50ml x=0.09270N

Calculation of concentration (Finding of NaOH concentration) HOOCC6H4COOK (mw 204.44)  NaOH (mw 40.01)  1 Eq wt. 204.22 g  1 N × 1000 mL 20.422 mg  0.1 N × 1 mL a g  x N’ × Ve mL  x N’ = (a g × 1000 mL) / (204.22 g × Ve mL)

Gran plot HA = H+ + A– Ka = [H+] fH+ [A–]fA– / [HA]fHA HA vs NaOH [A–] = (moles of OH– delivered) / (total volume) = VbFb / (Vb + Va) [HA] = (original moles of HA – moles of OH –) / (total volume) = (VaFa – VbFb ) / (Vb + Va) Ka = [H+] f H+ VbFb fA– / (VaFa – VbFb ) fHA [H+] fH+ Vb = (fHA / fA– ) Ka{(VaFa – VbFb )/ Fb} 10–pH Vb = (fHA / fA– ) Ka(Ve – Vb) = –(fHA / fA– ) Ka Vb + (fHA / fA– ) KaVe Slope =–(fHA / fA– ) Ka 10–pH Vb Ve Vb(l)

Gran plot for the first equivalence point Gran plot for the first equivalence point. This plot gives an estimate of Ve that differs from that in Figure 12.7 by 0.2μL (88.4 versus 88.2μL). The last 10-20% of volume prior to Ve is normally used form a Gran plot.

Conductometric end-point detection

Titration of strong acid with strong base V a (mL) 2 4 6 8 10 12 14 16 18 20 B C D pH Neutralization : HCl + NaOH  NaCl + H2O Ve mL×0.1000M = 50.00 mL × 0.02000 M Ve = 10.00 mL 0.02000M NaOH 50.00 mL 0.1000M HCl Calculated titration curve for the reaction of 50.00 mL of 0.02000 M NaOH with 0.1000 M HCl.

A. Before the beginning of titration : Va = 0.00 mL. NaOH  Na+ + OH–, 0.02000 M [OH–] = 0.02000 M [H+] = 1.00×10–14/ 2.000×10–2 pH = 12.30 B. Before the equivalence point : 0 < Va < Ve remaining NaOH Initial NaOH amount =F × Vi added HCl amount = used NaOH amount = F× Va Remaining NaOH amount = [OH–] = {(Ve – Va )/ Ve}FNaOH {Vi /(Vi + Va)} V a Ve Ex. Va = 3.00 mL [OH–] = {(10.00 – 3.00)/10.00}(0.02000){50.00 /(50.00 +3.00)} = 0.0132 M [H+] = 1.00×10–14/ 0.0132 = 7.58×10–13 pH = 12.12

C. At the equivalence point : Va = Ve H2O = H+ + OH–, [H+] = 1.00×10–7 pH = 7.00 D. After the equivalence point : Va > Ve excess HCl added HCl amount = FHCl× Va initial NaOH amount = used HCl = F × Vi Excess HCl amount = FHCl(Va–Vi) = FHCl (Va–Ve) [H+] = FHCl {(Va– Ve) /(Vi + Va)} NaOH Vi V a Ve Ex. Va = 10.50 mL [H+] = (0.1000) {10.50 –10.00) /(50.00 + 110.50)} = 8.26×10–4 pH = 3.08

Titration curves for NaOH with HCl. 50.0 mL of 0.0500 M NaOH with 0.1000 M HCl. 50.00 mL of 0.00500 M NaOH with 0.0100 M HCl.

Titration curves for HCl with NaOH. A. 50.0 mL of 0.0500 M HCl with 0.1000 M NaOH. 50.00 mL of 0.000500 M HCl with 0.00100 M NaOH.

The effects of titrant and analyte concentrations on neutralization titration curves. For the titration with diluted concentrations, the change in pH in equivalence point region is markedly less than concentrated solution V a of NaOH (mL) 5 10 15 20 25 30 35 0.00 1.00 2.00 3.00 4.00 5.00 6.00 7.00 8.00 9.00 10.00 11.00 12.00 13.00 14.00 A B pH Changes in pH during the totration of strong acid with strong base. A: 50 mL of 0.0500 M HCl vs 0.1000 M NaOH B: 50 mL of 0.000500 M HCl vs 0.001000 M NaOH

Weak acid titrated with strong base    Weak acid titrated with strong base 1) Titration reaction : CH3COOH + NaOH  CH3COO– Na+ + H2O strong + weak  complete reaction K = 1/Kb = 1.76 ×109 2) Calculation of equivalence point : [CH3COOH] Vi mL = [NaOH] Ve mL 0.1000 M×5.00 mL = 0.1000 M × Ve mL Ve= 5.00 mL 3) Titration curve : V a of NaOH(mL) 1 2 3 4 5 6 7 8 0.00 2.00 4.00 6.00 8.00 10.00 12.00 14.00 pH A B Changes in pH during the titration of a weak acid with a strong base. A: 5.00 mL of 0.1000M HOAC vs 0.1000 M NaOH B: 5.00 ml of 0.001000M HOAC vs 0.001000M NaOH

Titration curves for the titration of acetic acid with NaOH. A. 0.1000 M acetic acid with 0.1000M NaOH. B. 0.001000 M acetic acid with 0.00100 M NaOH.

A. Before adding any titrant : Va = 0.00    HAC = H+ + AC– [H+] = KaF =  1.75 ×10–5 ×0.1000 = 1.323 × 10–3 pH= 2.88    V a of 0.1000 M NaOH (mL) 1 2 3 4 5 6 7 pH 0.00 2.00 4.00 6.00 8.00 10.00 12.00 14.00 A 0.1000M NaOH 0.1000M HAC 25.00ml Titration curve. 0.1000 M HAC vs 0.1000 M NaOH.

B. Before equivalence point : 0 < Va < Ve    HAC + NaOH  H2O + Na+AC– [H+] = Ka[HAC] / [AC–] initial HAC amount = F×Ve – added NaOH amount = used amount HAC= F×Va Remaining HAC amount =F(Vi–Va) [HAC] = {(F×Ve) – (F×Va)}/(Vi +Va) [AC–] = (F×Va) / (Vi +Va) pH = pKa + log [AC–] /[HAC] at Va = Ve/2 pH= pKa = 4.76    V a of 0.1000 M NaOH (mL) 1 2 3 4 5 6 7 pH 0.00 2.00 4.00 6.00 8.00 10.00 12.00 14.00 B Ve/2 Vi Titration curve. 0.1000 M HAC vs 0.1000 M NaOH. Va Ve

C. Equivalence point Va = Ve    C. Equivalence point Va = Ve HAC + NaOH  H2O + Na+AC– initial 1 1 final 0 0 1 AC– = HAC + OH– [OH–] = KbF’ =  KwF’ /Ka F’=( F×Vi) / (Vi +Va) V a of 0.1000 M NaOH (mL) 1 2 3 4 5 6 7 pH 0.00 2.00 4.00 6.00 8.00 10.00 12.00 14.00 C Titration curve. 0.1000 M HAC vs 0.1000 M NaOH.

D. After equivalence point Va> Ve    D. After equivalence point Va> Ve added NaOH amount = F×Va initial HAC amount = used NaOH amount = F×Ve Excess NaOH amount = F×(Va –Ve) [OH–] = FNaOH ×(Va– Vi) / (Vi +Va) V a of 0.1000 M NaOH (mL) 1 2 3 4 5 6 7 pH 0.00 2.00 4.00 6.00 8.00 10.00 12.00 14.00 D Vi Titration curve. 0.1000 M HAC vs 0.1000 M NaOH. Va Ve

The effect of acid strength (dissociation constant) on titration curves. Each curve represents the titration of 50.00 mL of 0.1000 M acid with 0.1000 M base.

Weak base titrated with strong acid 1) Titration reaction : NH3 + HCl  NH4+Cl– 2) Equivalence point : NH3 HCl 0.1100 M×20.00 mL = 0.1000 M×Ve mL Ve = 22.00 mL 3) Titration curve : A. Initial Va = 0 NH3 + H2O = NH4+ + OH– Kb = Kw / Ka = [NH4+][OH– ]/[NH3] 1.79×10–5 = [OH– ]2 / 0.1100–[OH– ] [OH– ]=  KbF =1.4×10–3 pH = 11.15 0.1000M HCl volume (mL) 10 20 30 40 50 pH 0.00 2.00 4.00 6.00 8.00 10.00 12.00 14.00 A The calculated titration curve for the titration of 20.00 mL of 0.1100 M ammonia with 0.1000 M HCl.

B Ve/2 B. Before the equivalence point 0<Va< Ve Major constituents: NH3 + NH4+Cl– [OH]= Kb [NH3]/ [NH4+] initial NH3 amount = F×Ve – added HCl amount = used amount NH3 = F×Va Remaining NH3 amount =F(Vi–Va) [NH3] = {(F×Ve) – (F×Va)}/(Vi +Va) [NH4+] = (F×Va) / (Vi +Va) pH = pKb + log [NH4+] / [NH3] at Va = Ve/2 pOH= pKb = 4.74 pH = 9.26 B Ve/2 Vi Va Ve

C. Equivalence point Va = Ve NH3 + HCl  NH4+Cl– initial 1 1 final 0 0 1 NH4+ = NH3 + H+ [H+] = KaF’ F’=( F×Vi) / (Vi +Va) C

D. After equivalence point Va> Ve added HCl amount = F×Va initial NH3 amount = used HCl amount = F×Ve Excess HCl amount = F×(Va –Ve) [H+] = FHCl ×(Va– Ve) / (Vi +Va) D Vi Ve Va

The effect of base strength (dissociation constant) on titration curves. Each curve represents the titration of 50.00 ml of 0.1000 M base with 0.1000 M HCl.

Comparison of Weak Acid/ Base with Strong Base/Acid Titration of Weak Base with Strong Acid Comparison of Weak Acid/ Base with Strong Base/Acid [H+] = KaF’ F’=( F×Vi) / (Vi +Va) [OH–] = KbF’ =  KwF’ /Ka Equivalence point After equivalence point (Va>Ve) pH = pKb + log[NH4+]/[NH3] pH = pKa + log [A–] /[HA] Before the equivalence point (0<V a<V e) [OH-] =  KbF =1.4×10–3 [H+] = KaF Initial B + H2O → BH+ + OH- HA + OH- → H2O + A- Titration reaction Weak Base with Strong Base Weak Acid with Strong Base (Vi + Va) } (Va– Ve) { FHCl = [H+] FNaOH [OH-]

Determining the pK values for amino acids Amino acids contain both an acidic and a basic group. alanine All naturally occurring amino acids are left-handed (L) form. The amine group behaves as a base, while the carboxyl group acts as an acid. Aspartic acid

Curves for the titration of 20. 00ml of 0. 1000M alanine with 0 Curves for the titration of 20.00ml of 0.1000M alanine with 0.1000 M NaOH and 0.1000M HCl. Note that the zwitterion is present before any acid or base has been added. Adding acid protonates the carboxylate group with a pKa of 2.35. Adding base reacts with the protonated amine group with a pKa of 9.89.

Plots of relative amounts of acetic acid and acetate ion during a titration.

Acid base titration in non-aqueous media HOOCCHNH2 + HClO4  HOOCCHNH3+ClO4– R CH3COOH R (solvent) HClO4 + CH3COOK  CH3COOH + K+ClO4– CH3COOH(solvent) Amino acid in acetic acid + 0.020M HClO4 in acetic acid 0.010M CH3COOK

Titrants used in non-aqueous titrimetry Acidic titrants Perchloric acid p- Toluenesulfonic acid 2,4-Dinitrobenzenesulfonic acid Basic titrants Tetrabutylammonium hydroxide Sodium acetate Potassium methoxide Sodium aminoethoxide

Selected solvents for non-aqueous titration Solvent Autoprotolysis constant Dielectric (pKHS) constant Amphiprotic Glacial acetic acid 14.45 6.1 Ethylenediamine 15.3 12.9 Methanol 16.7 32.6 Aprotic or basic Dimethylformamide 36.7 Benzene 2.3 Methyl isobutylketone 13.1 Pyridine 12.3 Dioxane 2.2 n-Hexane 1.9

neither acid is completely dissociated. Acid and base strengths that are not distinguished in aqueous solution may be distinguishable in non-aqueous solvents. Ex. Perchloric acid is a stronger acid than hydrochloric acid in acetic acid solvent, neither acid is completely dissociated. HClO4 + CH3COOH = ClO4– + CH3COOH2+ K = 1.3×10–5 strong acid strong base weak base weak acid HCl + CH3COOH = Cl– + CH3COOH2+ K = 5.8×10–8 Differentiate acidity or basicity of different acids or bases differentiating solvent for acids …… acetic acid, isobutyl ketone differentiating sovent for bases …… ammonia, pyridine

Hammett acidity function, Ho, for aqueous solutions of acids. Acid Name Ho H2SO4 (100%) Sulfuric acid –11.93 H2SO4 SO3 Fuming sulfuric acid –14.14 H SO3F Fluorosulfuric acid –15.07 H SO3F+10%SbF5 Super acid –18.94 H SO3F+7%SbF53SO3 –19.35 Hammett acidity function, Ho, for aqueous solutions of acids.

Titration of a mixture of acids with tetrabutylammonium hydroxide in methyl isobutyl ketone solvent shows that the order of acid strength is HClO4 > HCl > 2-hydroxybenzoic acid > acetic acid > hydroxybenzene. Measurements were made with a glass electrode and a platimum reference electrode. The ordinate is proportional to pH, with increasing pH as the potential becomes more positive.

Summary Acid-Base titration Titration curve Indicator Detection of end-point Potentiometry Gran plot Conductometry Non-aqueous titration Hammett acidity function, Ho