Chapter 12: Principles of Neutralization Titrations By: Andie Aquilato

Solutions and Indicators Used Slides 2-6

Standard Solutions: strong acids or strong bases because they will react completely Acids: hydrochloric (HCl), perchloric (HClO 4 ), and sulfuric (H 2 SO 4 ) Bases: sodium hydroxide (NaOH), potassium hydroxide (KOH) Variables: temperature, ionic strength of medium and presence of organic solvents or colloidal particles

Indicators Acid/Base Indicators: a weak organic acid or weak organic base whose undissociated form differs in color from its conjugate form (In would be indicator) HIn + H 2 O  In - + H 3 O or In + H 2 O  HIn + + OH - (acid color) (base color) (base color) (acid color) Ka = [H3O + ][In - ] [HIn]  [H 3 O + ] = Ka[HIn] [In - ]

Indicators (cont’d) HIn pure acid color: [HIn]/[In - ] ≥ 10 HIn pure base color: [HIn]/[In - ] ≤ 0.1 ~The ratios change from indicator to indicator~ Substitute the ratios into the rearranged Ka: [H3O + ] = 10Ka (acid color) [H3O + ] = 0.1Ka (base color) pH range for indicator = pKa ± 1 acid color pH = -log(10Ka) = pKa + 1 base color pH = -log(0.1Ka) = pKa – 1

Indicators (cont’d) IndicatorpH RangeAcidBase Thymol Blue redyellow Thymol blue yellowblue Methyl yellow redyellow Methyl orange redorange Bromcresol green yellowblue Methyl red redyellow Bromcresol purple yellowpurple Bromothymol Blue yellowblue Phenol red yellowred Cresol purple yellowpurple Phenolphthalein colorlessred Thymolphthalein colorlessblue Alizarin yellow GG colorlessyellow Commonly Used Indicators

Calculating pH in Titrations of Strong Acids and Strong Bases Slides 7-11

Titrating a Strong Acid with a Strong Base – calculating pH Preequivalence: calculate the concentration of the acid from is starting concentration and the amount of base that has been added, the concentration of the acid is equal to the concentration of the hydroxide ion and you can calculate pH from the concentration Equivalence: the hydronium and hydroxide ions are present in equal concentrations Postequivalence: the concentration of the excess base is calculated and the hydroxide ion concentration is assumed to be equal to or a multiple of the analytical concentration, the pH can be calculated from the pOH

Do the calculations needed to generate the hypothetical titration curve for the titration of mL of M HCl with M NaOH Initial Point: the solution is M in H 3 O +, so pH = -log(.0500) = 1.30 Preequivalence Point (after addition of 10 mL reagent) c HCl = mmol remaining (original mmol HCl – mmol NaOH added) total volume (mL) = (50.0 mL x M) – (10.00 mL x M) 50.0 mL mL = x M pH = -log(2.500 x 10-2) = Equivalence Point [OH - ] = [H 3 O + ], pH = 7 Postequivalence Point (after addition of mL reagent) c HCl = mmol NaOH added – original mmol HCl total volume solution = (21.10 mL x M) – (50.00 mL x M) 50.0 mL mL = 1.33 x 10-4 M pOH = -log(1.33 x 10-4) = 3.88 pH = 14 – pOH = Calculating pH (cont’d) – Ex.

Other Things to Consider Concentrations: with a higher concentration titrant (0.1 M NaOH versus M NaOH), the change in pH equivalence-point region is large Choosing an indicator: you need to choose an indicator that has a color change in the same range as your equivalence point

Titrating a Strong Base with a Strong Acid – calculating pH Preequivalence: calculate the concentration of the base from is starting concentration and the amount of acid that has been added, the concentration of the base is equal to the concentration of the hydronium ion and you can calculate pOH from the concentration, and then the pH Equivalence: the hydronium and hydroxide ions are present in equal concentrations, so the pH is 7 Postequivalence: the concentration of the excess acid is calculated and the hydronium ion concentration is the same as the concentration of the acid, and the pH can be calculated

Buffer Solutions Slides 12-19

Calculating pH of Buffer Solutions A buffer is a mixture of a weak acid and its conjugate base or a weak base and its conjugate acid that resists change in pH HA + H 2 O  H 3 O + + A - Ka = [H 3 O + ][A - ] [HA] A - + H 2 O  OH - + HA Kb = [OH - ][HA] [A - ] Mass-Balance Equation for [HA]: [HA]=c HA – [H 3 O + ] + [OH - ] Mass-Balance Equation for [A-]: [A-] = c NaA + [H 3 O + ] – [OH - ]

Calculating pH of Buffer Solutions (cont’d) [HA] ≈ c HA [A - ] ≈ c NaA We can eliminate the rest of the mass-balance equations because of the inverse relationship between the hydronium and the hydroxide ion, as well as because the difference in concentration is so small relative to the concentrations of the acid and conjugate base If we substitute the concentration equations for [HA] & [A - ] into the dissociation constant expression, we get [H 3 O + ] = Ka c HA c NaA The hydronium ion concentration is dependent only on the ratio of the molar concentrations of the weak acid and its conjugate base, and is independent of dilution because the molar concentrations change proportionately

Buffer Formed From a Weak Acid and its Conjugate Base What is the pH of a solution that is M in formic acid and 1.00 M in sodium formate? HCOOH + H 2 O  H 3 O+ + HCOO - Ka = 1.80 x HCOO - + H 2 O  HCOOH + OH - Kb = Kw/Ka = 5.56 x  [HCOO - ] ≈ c HCOO- = 1.00 M [HCOOH] ≈ c HCOOH = M [H 3 O + ] = (1.80 x ) (0.400) = 7.20 x (1.00) pH = -log(7.20 x ) = 4.14

Buffer Formed From a Weak Base and its Conjugate Acid Calculate the pH of a solution that is M in NH 3 and M in NH 4 Cl. NH H 2 O  NH 3 + H 3 O + Ka = 5.70 x NH 3 + H 2 O  NH OH - Kb = Kw/Ka = 1.75 x  [NH 4 + ] ≈ c NH4Cl = M [NH 3 ] ≈ c NH3 = M [H 3 O + ] = (5.70 x ) (0.300) = 8.55 x (0.200) pH = -log(8.55 x ) = 9.07

Properties of Buffers Dilution: the pH of a buffer solution is essentially independent of dilution until the concentrations of the species are decreased to the point so that we cannot assume that the differences between the hydronium and hydroxide ion concentrations is negligible when calculating the concentration of the species Added Acids and Bases: buffers are resistant to pH change after addition of small amounts of strong acids or bases

Buffer Capacity (the number of moles of strong acid or strong base that causes one liter of the buffer to change pH by one unit) Calculate the pH change that takes place when a 100 mL portion of M NaOH is added to a 400 mL buffer consisting of 0.2 M NH 3 and 0.3 M NH 4 Cl (see example for “Buffers Formed from a Weak Base and its Conjugate Acid”) An addition of a base converts NH 4 + to NH 3 : NH OH -  NH 3 + H 2 O The concentration of the NH 3 and NH 4 Cl change: c NH3 = original mol base + mol base added total volume c NH3 = (400 x 0.200) + (100 x 0.300) = M 500 c NH4Cl = original mol acid – mol base added total volume c NH3 = (400 x 0.300) + (100 x 0.300) = M 500 [H 3 O + ] = (5.70 x ) (0.230) =7.71 x (0.170) pH = -log(7.71 x ) = 9.11 ∆pH = 9.11 – 9.07 = 0.04

Preparing Buffers In principle the calculations work, but there are uncertainties in numerical values of dissociation constants & simplifications used in calculations How to Prepare/Get: Making up a solution of approximately the desired pH and then adjust by adding acid or conjugate base until the required pH is indicated by a pH meter Empirically derived recipes are available in chemical handbooks and reference works Biological supply houses

Calculating pH in Weak Acid (or Base) Titrations Slides 20-24

Steps 1.At the beginning: pH is calculated from the concentration of that solute and its dissociation constant 2.After various increments of titrant has been added: pH is calculated by the analytical concentrations of the conjugate base or acid and the residual concentrations of the weak acid or base 3.At the equivalence point: the pH is calculated from the concentration of the conjugate of the weak acid or base ~ a salt 4.Beyond the equivalence point: pH is determined by the concentration of the excess titrant

Example Calculation Determine the pH for the titration of mL of M acetic acid after adding 0.00, 5.00, 50.00, and mL of M sodium hydroxide HOAc + H 2 O  H 3 O + + OAc - OAc - + H 2 O  HOAc + OH - Ka = 1.75 x Starting Point: [H 3 O + ] = 1.32 x pH = -log(1.32 x ) = 2.88 After Titrant Has Been Added (5.00 mL NaOH): *the buffer solution now has NaOAc & HOAc* c HOAc = mol original acid – mol base added total volume c HOAc = (50.00 x 0.100) – (5.00 x 0.100) = c NaOAc = mol base added total volume c NaOAc = (5.00 x 0.100) = *we can then substitute these concentrations into the dissociation-constant expression for acetic acid* Ka = [H 3 O + ][OAc - ] [HOAc] Ka = 1.75 x = [H 3 O + ][ ] [0.075] [H 3 O + ] = 1.58 x pH = -log(1.58 x ) = 3.80

Example Calculation (cont’d) Equivalence Point (50.00 mL NaOH): *all the acetic acid has been converted to sodium acetate* [NaOAc]= M *we can substitute this in to the base-dissociation constant for OAc - * Kb = [OH - ][HOAc] = Kw [OAc - ] Ka [HOAc] = [OH - ]  [OH - ] 2 = 1.00 x x [OH - ] = 5.34 x pH = – (-log(5.34 x )) pH = 8.73 Beyond the Equivalence Point (50.01 mL NaOH): *the excess base and acetate ion are sources of the hydroxide ion, but the acetate ion concentration is so small it is negligible* [OH - ] = c NaOH = mol base added – original mol acid total volume [OH - ] = (50.01 x 0.100) – (50.00 x 0.100) [OH - ] = 1.00 x pH = – (-log(1.00 x )) pH = 9.00

The Effect of Variables The Effect of Concentration: the change in pH in the equivalence-point region becomes smaller with lower analyte and reagent concentrations The Effect of Reaction Completeness: pH change in the equivalence-point region becomes smaller as the acid become weaker (the reaction between the acid and the base becomes less complete) Choosing an Indicator: the color change must occur in the equivalence-point region

How do Buffer Solutions Change as a Function of pH? Slides 25-27

Alpha Values Def.: the relative equilibrium concentration of the weak acid/base and its conjugate base/acid (titrating with HOAc with NaOH): *at any point in a titration, c T = c HOAc + c NaOAc * α 0 = [HOAc] c T α 1 = [OAc - ] c T *alpha values are unitless and are equal to one* α 0 + α 1 = 1

Derivation of Alpha Values *alpha values depend only on [H 3 O + ] and Ka, not c T * *mass-balance requires that c T = [HOAc] + [OAc - ]* For α 0, we rearrange the dissociation-constant expression to: [OAc - ] = Ka[HOAc] [H 3 O + ] *substitute mass-balance into the dissociation-constant expression* α 0 = [HOAc] = [ H 3 O + ] c T [H 3 O + ] + Ka For α 1, we rearrange the dissociation-constant expression to: [HOAc] = [H 3 O + ] [OAc - ] Ka *substitute mass-balance into the dissociation-constant expression* α 1 = [OAc - ] = _____Ka________ c T [H 3 O + ] + Ka