Chapter 12 Principles of Neutralization Titrations By Andie

Chapter 12: Principles of Neutralization Titrations By: Andie Aquilato

Solutions and Indicators Used Slides 2 -6

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

Indicators l 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 (acid color) (base color) or In + H 2 O HIn+ + OH(base color) (acid color) Ka = [H 3 O+][In-] [HIn] [H 3 O+] = Ka[HIn] [In-]

Indicators (cont’d) l l HIn pure acid color: [HIn]/[In-] ≥ 10 HIn pure base color: [HIn]/[In-] ≤ 0. 1 ~The ratios change from indicator to indicator~ l Substitute the ratios into the rearranged Ka: [H 3 O+] = 10 Ka (acid color) [H 3 O+] = 0. 1 Ka (base color) l p. H range for indicator = p. Ka ± 1 l acid color p. H = -log(10 Ka) = p. Ka + 1 l base color p. H = -log(0. 1 Ka) = p. Ka – 1

Indicators (cont’d) Commonly Used Indicators Indicator p. H Range Acid Base Thymol Blue 1. 2 -2. 8 red yellow Thymol blue 8. 0 -9. 6 yellow blue Methyl yellow 2. 9 -4. 0 red yellow Methyl orange 3. 1 -4. 4 red orange Bromcresol green 4. 0 -5. 6 yellow blue Methyl red 4. 4 -6. 2 red yellow Bromcresol purple 5. 2 -6. 8 yellow purple Bromothymol Blue 6. 2 -7. 8 yellow blue Phenol red 6. 4 -8. 0 yellow red Cresol purple 7. 6 -9. 2 yellow purple Phenolphthalein 8. 0 -10. 0 colorless red Thymolphthalein 9. 4 -10. 6 colorless blue Alizarin yellow GG 10. 0 -12. 0 colorless yellow

Calculating p. H in Titrations of Strong Acids and Strong Bases Slides 7 -11

Titrating a Strong Acid with a Strong Base – calculating p. H l l l 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 p. H 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 p. H can be calculated from the p. OH

Calculating p. H (cont’d) – Ex. Do the calculations needed to generate the hypothetical titration curve for the titration of 50. 00 m. L of 0. 0500 M HCl with 0. 1000 M Na. OH l Initial Point: the solution is 0. 0500 M in H 3 O+, so p. H = -log(. 0500) = 1. 30 l Preequivalence Point (after addition of 10 m. L reagent) c. HCl = mmol remaining (original mmol HCl – mmol Na. OH added) total volume (m. L) = (50. 0 m. L x 0. 0500 M) – (10. 00 m. L x 0. 1000 M) 50. 0 m. L + 10. 00 m. L = 2. 500 x 10 -2 M p. H = -log(2. 500 x 10 -2) = 1. 602 l Equivalence Point [OH-] = [H 3 O+], p. H = 7 l Postequivalence Point (after addition of 25. 10 m. L reagent) c. HCl = mmol Na. OH added – original mmol HCl total volume solution = (21. 10 m. L x 0. 1000 M) – (50. 00 m. L x 0. 0500 M) 50. 0 m. L + 25. 10 m. L = 1. 33 x 10 -4 M p. OH = -log(1. 33 x 10 -4) = 3. 88 p. H = 14 – p. OH = 10. 12

Other Things to Consider l Concentrations: with a higher concentration titrant (0. 1 M Na. OH versus 0. 001 M Na. OH), the change in p. H equivalence-point region is large l 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 p. H l l l 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 p. OH from the concentration, and then the p. H Equivalence: the hydronium and hydroxide ions are present in equal concentrations, so the p. H 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 p. H can be calculated

Buffer Solutions Slides 12 -19

Calculating p. H of Buffer Solutions l 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 p. H HA + H 2 O H 3 O+ + AKa = [H 3 O+][A-] [HA] A- + H 2 O OH- + HA Kb = [OH-][HA] [A-] l Mass-Balance Equation for [HA]: [HA]=c. HA – [H 3 O+] + [OH-] l Mass-Balance Equation for [A-]: [A-] = c. Na. A + [H 3 O+] – [OH-]

Calculating p. H of Buffer Solutions (cont’d) [HA] ≈ c. HA [A-] ≈ c. Na. A l 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 l If we substitute the concentration equations for [HA] & [A-] into the dissociation constant expression, we get [H 3 O+] = Ka c. HA c. Na. A l 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 p. H of a solution that is 0. 400 M in formic acid and 1. 00 M in sodium formate? HCOOH + H 2 O H 3 O+ + HCOOKa = 1. 80 x 10 -4 HCOO- + H 2 O HCOOH + OHKb = Kw/Ka = 5. 56 x 10 -11 [HCOO-] ≈ c. HCOO- = 1. 00 M [HCOOH] ≈ c. HCOOH = 0. 400 M [H 3 O+] = (1. 80 x 10 -4) (0. 400) = 7. 20 x 10 -5 (1. 00) p. H = -log(7. 20 x 10 -5) = 4. 14

Buffer Formed From a Weak Base and its Conjugate Acid Calculate the p. H of a solution that is 0. 200 M in NH 3 and 0. 300 M in NH 4 Cl. NH 4+ + H 2 O NH 3 + H 3 O+ Ka = 5. 70 x 10 -10 NH 3 + H 2 O NH 4+ + OHKb = Kw/Ka = 1. 75 x 10 -5 [NH 4+] ≈ c. NH 4 Cl = 0. 300 M [NH 3] ≈ c. NH 3 = 0. 200 M [H 3 O+] = (5. 70 x 10 -10) (0. 300) = 8. 55 x 10 -10 (0. 200) p. H = -log(8. 55 x 10 -10) = 9. 07

Properties of Buffers l l Dilution: the p. H 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 p. H 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 p. H by one unit) Calculate the p. H change that takes place when a 100 m. L portion of 0. 0500 M Na. OH is added to a 400 m. L 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 4+ + OH- NH 3 + H 2 O The concentration of the NH 3 and NH 4 Cl change: c. NH 3 = original mol base + mol base added total volume c. NH 3 = (400 x 0. 200) + (100 x 0. 300) = 0. 170 M 500 c. NH 4 Cl = original mol acid – mol base added total volume c. NH 3 = (400 x 0. 300) + (100 x 0. 300) = 0. 230 M 500 [H 3 O+] = (5. 70 x 10 -10) (0. 230) =7. 71 x 10 -10 (0. 170) p. H = -log(7. 71 x 10 -10) = 9. 11 ∆p. H = 9. 11 – 9. 07 = 0. 04

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

Calculating p. H in Weak Acid (or Base) Titrations Slides 20 -24

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

Example Calculation Determine the p. H for the titration of 50. 00 m. L of 0. 1000 M acetic acid after adding 0. 00, 50. 00, and 50. 01 m. L of 0. 100 M sodium hydroxide HOAc + H 2 O H 3 O+ + OAc- + H 2 O HOAc + OHKa = 1. 75 x 10 -5 l Starting Point: [H 3 O+] = 1. 32 x 10 -3 p. H = -log(1. 32 x 10 -3) = 2. 88 l After Titrant Has Been Added (5. 00 m. L Na. OH): *the buffer solution now has Na. OAc & HOAc* c. HOAc = mol original acid – mol base added total volume c. HOAc = (50. 00 x 0. 100) – (5. 00 x 0. 100) = 0. 075 60. 0 c. Na. OAc = mol base added total volume c. Na. OAc = (5. 00 x 0. 100) = 0. 008333 60. 0 *we can then substitute these concentrations into the dissociation-constant expression for acetic acid* Ka = [H 3 O+][OAc-] [HOAc] Ka = 1. 75 x 10 -5 = [H 3 O+][0. 008333] [0. 075] [H 3 O+] = 1. 58 x 10 -4 p. H = -log(1. 58 x 10 -4) = 3. 80

Example Calculation (cont’d) l l Equivalence Point (50. 00 m. L Na. OH): *all the acetic acid has been converted to sodium acetate* [Na. OAc]= 0. 0500 M *we can substitute this in to the base-dissociation constant for OAc-* Kb = [OH-][HOAc] = Kw [OAc-] Ka [HOAc] = [OH-]2 = 1. 00 x 10 -14 0. 0500 1. 75 x 10 -5 [OH-] = 5. 34 x 10 -6 p. H = 14. 00 – (-log(5. 34 x 10 -6)) p. H = 8. 73 Beyond the Equivalence Point (50. 01 m. L Na. OH): *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. Na. OH = mol base added – original mol acid total volume [OH-] = (50. 01 x 0. 100) – (50. 00 x 0. 100) 100. 01 [OH-] = 1. 00 x 10 -5 p. H = 14. 00 – (-log(1. 00 x 10 -5)) p. H = 9. 00

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

How do Buffer Solutions Change as a Function of p. H? Slides 25 -27

Alpha Values l Def. : the relative equilibrium concentration of the weak acid/base and its conjugate base/acid (titrating with HOAc with Na. OH): *at any point in a titration, c. T = c. HOAc + c. Na. OAc* 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