chapter 12: principles of neutralization titrations by: andie aquilato
Post on 19-Dec-2015
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TRANSCRIPT
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 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) 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)
Indicator pH 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
Commonly Used Indicators
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 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) = 3.88pH = 14 – pOH = 10.12
Calculating pH (cont’d) – Ex.
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
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
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- + HAKb = [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)
[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
What is the pH of a solution that is 0.400 M in formic acid and 1.00M 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
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
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 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 “BuffersFormed 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 volumecNH3 = (400 x 0.200) + (100 x 0.300) = 0.170 M
500 cNH4Cl = original mol acid – mol base added
total volumecNH3 = (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
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
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 CalculationDetermine 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 volumecHOAc = (50.00 x 0.100) – (5.00 x 0.100) = 0.075
60.0cNaOAc = mol base added
total volumecNaOAc = (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) 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 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
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*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