Acid-base Titrations 2

7/27/2019 Acid-base Titrations 2

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Chapter 12: Principles of Neutralization Titrations

By: Andie Aquilato


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Standard Solutions: strong acids or

strong bases because they will reactcompletely

Acids: hydrochloric (HCl), perchloric (HClO 4),and sulfuric (H 2SO 4)Bases: sodium hydroxide (NaOH), potassiumhydroxide (KOH)

Variables: temperature, ionic strength of medium and presence of organicsolvents or colloidal particles


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Indicators Acid/Base Indicators: a weak organic acid or weak organic base whose undissociated formdiffers in color from its conjugate form (Inwould be indicator)

HIn + H 2O In-

+ H 3O or In + H 2O HIn+

+ OH-

(acid color) (base color) (base color) (acid color) Ka = [H3O +][In-]

[HIn]

[H3O+] = Ka[HIn][In-]


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Indicators (contd) 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 1acid color pH = -log(10Ka) = pKa + 1base color pH = -log(0.1Ka) = pKa 1


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Calculating pH in

Titrations of StrongAcids and Strong Bases

Slides 7-11


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Titrating a Strong Acid with aStrong Base calculating pH

Preequivalence: calculate the concentration of theacid from is starting concentration and the amountof base that has been added, the concentration of the acid is equal to the concentration of thehydroxide ion and you can calculate pH from theconcentrationEquivalence: the hydronium and hydroxide ionsare present in equal concentrationsPostequivalence: the concentration of the excessbase is calculated and the hydroxide ionconcentration is assumed to be equal to or amultiple of the analytical concentration, the pH

can be calculated from the pOH


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Do the calculations needed to generate the hypothetical titration curve for the titrationof 50.00 mL of 0.0500 M HCl with 0.1000 M NaOH

Initial Point: the solution is 0.0500 M in H 3O+, 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 0.0500 M) (10.00 mL x 0.1000 M)

50.0 mL + 10.00 mL

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

Equivalence Point[OH -] = [H 3O+], pH = 7

Postequivalence Point (after addition of 25.10 mL reagent)c HCl = 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

Calculating pH (contd) Ex.


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Other Things to Consider

Concentrations: with a higher concentrationtitrant (0.1 M NaOH versus 0.001 M NaOH),the change in pH equivalence-point region islarge

Choosing an indicator: you need to choosean indicator that has a color change in thesame range as your equivalence point


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Titrating a Strong Base with aStrong Acid calculating pH

Preequivalence: calculate the concentration of the basefrom is starting concentration and the amount of acid thathas been added, the concentration of the base is equal tothe concentration of the hydronium ion and you cancalculate pOH from the concentration, and then the pHEquivalence: the hydronium and hydroxide ions arepresent in equal concentrations, so the pH is 7Postequivalence: the concentration of the excess acid is

calculated and the hydronium ion concentration is thesame as the concentration of the acid, and the pH can becalculated


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Buffer Solutions

Slides 12-19


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Calculating pH of Buffer Solutions

A buffer is a mixture of a weak acid and its conjugate base or aweak base and its conjugate acid that resists change in pH

HA + H 2O H3O+ + A -

Ka = [H 3O+][A-][HA]

A - + H 2O OH - + HAKb = [OH -][HA]

[A-]

Mass-Balance Equation for [HA]: [HA]= c HA [H3O+] + [OH -]

Mass-Balance Equation for [A-]: [A-] = c NaA + [H 3O+] [OH -]


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Calculating pH of Buffer Solutions (contd)

[HA] c HA[A-] c NaA

We can eliminate the rest of the mass-balance equationsbecause of the inverse relationship between the hydronium andthe hydroxide ion, as well as because the difference inconcentration is so small relative to the concentrations of the acidand conjugate base

If we substitute the concentration equations for [HA] & [A -] intothe dissociation constant expression, we get

[H3O+] = Ka c HAc 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 concentrationschange proportionately


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Buffer Formed From a WeakAcid and its Conjugate BaseWhat is the pH of a solution that is 0.400 M in formic acid and 1.00M in sodium formate?

HCOOH + H 2O H3O+ + HCOO - Ka = 1.80 x 10 -4

HCOO - + H 2O HCOOH + OH - Kb = Kw/Ka = 5.56 x 10 -11

[HCOO -] c HCOO- = 1.00 M[HCOOH] c HCOOH = 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


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Buffer Formed From a WeakBase and its Conjugate AcidCalculate the pH of a solution that is 0.200 M in NH 3 and 0.300 M inNH4Cl.

NH4+ + H 2O NH3 + H 3O + Ka = 5.70 x 10 -10

NH3 + H 2O NH4+ + OH - Kb = Kw/Ka = 1.75 x 10 -5

[NH4+] c NH4Cl = 0.300 M

[NH3] c NH3 = 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


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Buffer Capacity (the number of moles of strongacid or strong base that causes one liter of the buffer tochange pH by one unit)

Calculate the pH change that takes place when a 100 mL portion of 0.0500 M NaOH isadded to a 400 mL buffer consisting of 0.2 M NH 3 and 0.3 M NH 4Cl (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 - NH3 + H 2O

The concentration of the NH 3 and NH 4Cl change:c NH3 = original mol base + mol base added

total volumec NH3 = (400 x 0.200) + (100 x 0.300) = 0.170 M

500c NH4Cl = original mol acid mol base added

total volumec NH3 = (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


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Preparing Buffers

In principle the calculations work, but there areuncertainties in numerical values of dissociationconstants & simplifications used in calculations

How to Prepare/Get:Making up a solution of approximately the desired pH andthen adjust by adding acid or conjugate base until therequired pH is indicated by a pH meter

Empirically derived recipes are available in chemicalhandbooks and reference worksBiological supply houses


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Calculating pH in WeakAcid (or Base) Titrations

Slides 20-24


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Steps

1. At the beginning: pH is calculated from the concentrationof that solute and its dissociation constant

2. After various increments of titrant has been added: pH iscalculated by the analytical concentrations of theconjugate base or acid and the residual concentrations of the weak acid or base

3. At the equivalence point: the pH is calculated from theconcentration of the conjugate of the weak acid or base ~

a salt4. Beyond the equivalence point: pH is determined by the

concentration of the excess titrant


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Example CalculationDetermine the pH for the titration of 50.00 mL of 0.1000 M acetic acid after adding0.00, 5.00, 50.00, and 50.01 mL of 0.100 M sodium hydroxide

HOAc + H 2O H3O+ + OAc -

OAc - + H 2O HOAc + OH -

Ka = 1.75 x 10 -5Starting 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*c HOAc = mol original acid mol base added

total volumec HOAc = (50.00 x 0.100) (5.00 x 0.100) = 0.075

60.0c NaOAc = mol base added

total volumec NaOAc = (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 3O+][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


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Example Calculation (contd) 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 -6pH = 14.00 (-log(5.34 x 10 -6))

pH = 8.73Beyond 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)100.01

[OH -] = 1.00 x 10 -5

pH = 14.00 (-log(1.00 x 10 -5))pH = 9.00


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The Effect of Variables

The Effect of Concentration: the change in pH in theequivalence-point region becomes smaller withlower analyte and reagent concentrations

The Effect of Reaction Completeness: pH change inthe equivalence-point region becomes smaller asthe acid become weaker (the reaction between theacid and the base becomes less complete)

Choosing an Indicator: the color change must occur in the equivalence-point region


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How do Buffer Solutions

Change as a Function of pH?

Slides 25-27


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Alpha ValuesDef.: the relative equilibrium concentration of theweak acid/base and its conjugate base/acid (titratingwith HOAc with NaOH):

*at any point in a titration, c T = c HOAc + c NaOAc *0 = [HOAc]

c T1 = [OAc -]

c T*alpha values are unitless and are equal to one*

0 + 1 = 1


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