Chapter 12 principles of neutralization titrations
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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 )

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Chapter 12: Principles of Neutralization Titrations

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Chapter 12 principles of neutralization titrations

Chapter 12: Principles of Neutralization Titrations

By: Andie Aquilato


Solutions and indicators used

Solutions and Indicators Used

Slides 2-6


Chapter 12 principles of neutralization titrations

  • Standard Solutions: strong acids or strong bases because they will react completely

    • Acids: hydrochloric (HCl), perchloric (HClO4), and sulfuric (H2SO4)

    • Bases: sodium hydroxide (NaOH), potassium hydroxide (KOH)

  • Variables: temperature, ionic strength of medium and presence of organic solvents or colloidal particles


Indicators

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 + H2O In- + H3O or In + H2O HIn+ + OH-

    (acid color) (base color) (base color) (acid color)

    Ka = [H3O+][In-]

    [HIn]

    [H3O+] = Ka[HIn]

    [In-]


Indicators cont d

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 d1

Indicators (cont’d)

Commonly Used Indicators


Calculating ph in titrations of strong acids and strong bases

Calculating pH in Titrations of Strong Acids and Strong Bases

Slides 7-11


Titrating a strong acid with a strong base calculating ph

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


Calculating ph cont d ex

Calculating pH (cont’d) – Ex.

Do the calculations needed to generate the hypothetical titration curve for the titration

of 50.00 mL of 0.0500 M HCl with 0.1000 M NaOH

  • Initial Point: the solution is 0.0500 M in H3O+, so pH = -log(.0500) = 1.30

  • Preequivalence Point (after addition of 10 mL reagent)

    cHCl = mmol remaining (original mmol HCl – mmol NaOH added)

    total volume (mL)

    = (50.0 mL x 0.0500 M) – (10.00 mL x 0.1000 M)

    50.0 mL + 10.00 mL

    = 2.500 x 10 -2 M

    pH = -log(2.500 x 10-2) = 1.602

  • Equivalence Point

    [OH-] = [H3O+], pH = 7

  • Postequivalence Point (after addition of 25.10 mL reagent)

    cHCl = mmol NaOH added – original mmol HCl

    total volume solution

    = (21.10 mL x 0.1000 M) – (50.00 mL x 0.0500 M)

    50.0 mL + 25.10 mL

    = 1.33 x 10-4 M

    pOH = -log(1.33 x 10-4) = 3.88

    pH = 14 – pOH = 10.12


Other things to consider

Other Things to Consider

  • Concentrations: with a higher concentration titrant (0.1 M NaOH versus 0.001 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

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

Buffer Solutions

Slides 12-19


Calculating ph of buffer solutions

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 + H2O  H3O+ + A-

    Ka = [H3O+][A-]

    [HA]

    A- + H2O  OH- + HA

    Kb = [OH-][HA]

    [A-]

    • Mass-Balance Equation for [HA]: [HA]=cHA – [H3O+] + [OH-]

    • Mass-Balance Equation for [A-]: [A-] = cNaA + [H3O+] – [OH-]


Calculating ph of buffer solutions cont d

Calculating pH of Buffer Solutions (cont’d)

[HA] ≈ cHA

[A-] ≈ cNaA

  • 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

    [H3O+] = Ka cHA

    cNaA

    • 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

Buffer Formed From a Weak Acid and its Conjugate Base

What is the pH of a solution that is 0.400 M in formic acid and 1.00

M in sodium formate?

HCOOH + H2O  H3O+ + HCOO-Ka = 1.80 x 10-4

HCOO- + H2O  HCOOH + OH-Kb = Kw/Ka = 5.56 x 10-11

[HCOO-] ≈ cHCOO- = 1.00 M

[HCOOH] ≈ cHCOOH = 0.400 M

[H3O+] = (1.80 x 10-4) (0.400) = 7.20 x 10-5

(1.00)

pH = -log(7.20 x 10-5) = 4.14


Buffer formed from a weak base and its conjugate acid

Buffer Formed From a Weak Base and its Conjugate Acid

Calculate the pH of a solution that is 0.200 M in NH3 and 0.300 M in

NH4Cl.

NH4+ + H2O  NH3 + H3O+Ka = 5.70 x 10-10

NH3 + H2O  NH4+ + OH-Kb = Kw/Ka = 1.75 x 10-5

[NH4+] ≈ cNH4Cl = 0.300 M

[NH3] ≈ cNH3 = 0.200 M

[H3O+] = (5.70 x 10-10) (0.300) = 8.55 x 10-10

(0.200)

pH = -log(8.55 x 10-10) = 9.07


Properties of buffers

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


Chapter 12 principles of neutralization titrations

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 0.0500 M NaOH is

added to a 400 mL buffer consisting of 0.2 M NH3 and 0.3 M NH4Cl (see example for “Buffers

Formed from a Weak Base and its Conjugate Acid”)

An addition of a base converts NH4+ to NH3: NH4+ + OH- NH3 + H2O

The concentration of the NH3 and NH4Cl change:

cNH3 = original mol base + mol base added

total volume

cNH3 = (400 x 0.200) + (100 x 0.300) = 0.170 M

500

cNH4Cl = original mol acid – mol base added

total volume

cNH3 = (400 x 0.300) + (100 x 0.300) = 0.230 M

500

[H3O+] = (5.70 x 10-10) (0.230) =7.71 x 10-10

(0.170)

pH = -log(7.71 x 10-10) = 9.11

∆pH = 9.11 – 9.07 = 0.04


Preparing buffers

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

Calculating pH in Weak Acid (or Base) Titrations

Slides 20-24


Steps

Steps

  • At the beginning: pH is calculated from the concentration of that solute and its dissociation constant

  • 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

  • At the equivalence point: the pH is calculated from the concentration of the conjugate of the weak acid or base ~ a salt

  • Beyond the equivalence point: pH is determined by the concentration of the excess titrant


Example calculation

Example Calculation

Determine the pH for the titration of 50.00 mL of 0.1000 M acetic acid after adding

0.00, 5.00, 50.00, and 50.01 mL of 0.100 M sodium hydroxide

HOAc + H2O  H3O+ + OAc-

OAc- + H2O  HOAc + OH-

Ka = 1.75 x 10 -5

  • Starting Point:

    [H3O+] = 1.32 x 10-3

    pH = -log(1.32 x 10-3) = 2.88

  • After Titrant Has Been Added (5.00 mL NaOH):

    *the buffer solution now has NaOAc & HOAc*

    cHOAc = mol original acid – mol base added

    total volume

    cHOAc = (50.00 x 0.100) – (5.00 x 0.100) = 0.075

    60.0

    cNaOAc = mol base added

    total volume

    cNaOAc = (5.00 x 0.100) = 0.008333

    60.0

    *we can then substitute these concentrations into the dissociation-constant expression for acetic acid*

    Ka = [H3O+][OAc-]

    [HOAc]

    Ka = 1.75 x 10-5 = [H3O+][0.008333]

    [0.075]

    [H3O+] = 1.58 x 10-4

    pH = -log(1.58 x 10-4) = 3.80


Example calculation cont d

Example Calculation (cont’d)

  • Equivalence Point (50.00 mL NaOH):

    *all the acetic acid has been converted to sodium acetate*

    [NaOAc]= 0.0500 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 10-14

    0.0500 1.75 x 10-5

    [OH-] = 5.34 x 10-6

    pH = 14.00 – (-log(5.34 x 10-6))

    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-] = cNaOH = 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

    pH = 14.00 – (-log(1.00 x 10-5))

    pH = 9.00


The effect of variables

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

How do Buffer Solutions Change as a Function of pH?

Slides 25-27


Alpha values

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, cT = cHOAc + cNaOAc*

    α0 = [HOAc]

    cT

    α1 = [OAc-]

    cT

    *alpha values are unitless and are equal to one*

    α0 + α1 = 1


Derivation of alpha values

Derivation of Alpha Values

*alpha values depend only on [H3O+] and Ka, not cT*

*mass-balance requires that cT = [HOAc] + [OAc-]*

For α0, we rearrange the dissociation-constant expression to:

[OAc-] = Ka[HOAc]

[H3O+]

*substitute mass-balance into the dissociation-constant expression*

α0 = [HOAc] = [ H3O+ ]

cT [H3O+] + Ka

For α1, we rearrange the dissociation-constant expression to:

[HOAc] = [H3O+] [OAc-]

Ka

*substitute mass-balance into the dissociation-constant expression*

α1 = [OAc-] = _____Ka________

cT [H3O+] + Ka


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