Post on 28-Dec-2015
Buffers and
Acid/Base Titration
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Common Ion
• Suppose we have a solution containing hydrofluoric acid (HF) and its salt sodium fluoride (NaF).
• Major species (HF, Na+, F-, and H2O)• HF slightly dissociated, therefore common ion
is F-
• How does this effect equilibrium of HF dissociation?
Common Ion Effect
HF(aq) H+(aq) + F-
(aq)
Added fluoride ions from NaFShifts reaction left.Less H+ ions present.
• This effect makes a solution of NaF and HF less acidic than a solution of HF alone.
Buffered Solutions
• A solution that resists a change in pH when either hydroxide ions or protons are added.
• Buffered solutions contain either:• A weak acid and its salt• A weak base and its salt
Acid/Salt Buffering Pairs
Weak AcidFormula
of the acidExample of a salt of the
weak acid Hydrofluoric HF KF – Potassium fluoride
Formic HCOOH KHCOO – Potassium formate
Benzoic C6H5COOH NaC6H5COO – Sodium benzoate
Acetic CH3COOH NaH3COO – Sodium acetate
Carbonic H2CO3 NaHCO3 - Sodium bicarbonate
Propanoic HC3H5O2 NaC3H5O2 - Sodium propanoate
Hydrocyanic HCN KCN - potassium cyanide
The salt will contain the anion of the acid, and the cation of a strong base (NaOH, KOH)
Base/Salt Buffering Pairs
The salt will contain the cation of the base, and the anion of a strong acid (HCl, HNO3)
BaseFormula of the base
Example of a salt of the weak acid
Ammonia NH3 NH4Cl - ammonium chloride
Methylamine
CH3NH2 CH3NH3Cl – methylammonium
chloride
Ethylamine C2H5NH2 C2H5NH3NO3 - ethylammonium
nitrate
Aniline C6H5NH2 C6H5NH3Cl – aniline hydrochloride
Pyridine C5H5N C5H5NHCl – pyridine hydrochloride
Henderson-Hasselbalch Equation
][
][log
][
][log
acid
basepK
HA
ApKpH aa
][
][log
][
][log
base
acidpK
B
BHpKpOH bb
This is an exceptionally powerful tool, and it’s use will be emphasized in our problem solving.
Calculate the pH of the following mixtures:• 0.75 M lactic acid (HC3H5O3) and 0.25 M sodium
lactate (Ka = 1.4 x 10-4)
• 0.25 M NH3 and 0.40 M NH4Cl (Kb = 1.8 x 10-5)
Buffer capacity
• The pH of a buffered solution is determined by the ratio [A-]/[HA].
• As long as this doesn’t change much the pH won’t change much.
• The more concentrated these two are the more H+ and OH- the solution will be able to absorb.
• Larger concentrations bigger buffer capacity.
Buffer Capacity
• Calculate the change in pH that occurs when 0.010 mol of HCl(g) is added to 1.0L of each of the following:
• 5.00 M HC2H3O2 and 5.00 M NaC2H3O2
• 0.050 M HC2H3O2 and 0.050 M NaC2H3O2
• Ka= 1.8x10-5
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0.00 5.00 10.00 15.00 20.00 25.00 30.00 35.00 40.00 45.00m illiliters NaOH (0.10 M)
pH
Weak Acid/Strong Base Titration
A solution that is 0.10 M CH3COOH is titrated with 0.10 M NaOH
Equivalence is above pH 7
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0.00 5.00 10.00 15.00 20.00 25.00 30.00 35.00 40.00 45.00
m illiliters NaOH (0.10 M)
pH
Strong Acid/Strong Base Titration
A solution that is 0.10 M HCl is titrated with 0.10 M NaOH
Equivalance is at pH 7
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m illiliters HCl (0.10 M)
pH
Strong Acid/Strong Base Titration
A solution that is 0.10 M NaOH is titrated with 0.10 M HCl
Equivalence is at pH 7 It is important to
recognize that titration curves are not always increasing from left to right.
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m illiliters NH3 (0.10 M)
pH
Strong Acid/Weak Base Titration
A solution that is 0.10 M HCl is titrated with 0.10 M NH3
Equivalence is below pH 7
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0.00 5.00 10.00 15.00 20.00 25.00 30.00 35.00 40.00 45.00m illiliters NaOH (0.10 M)
pH
Titration of an Unbuffered Solution
A solution that is 0.10 M CH3COOH is titrated with 0.10 M NaOH
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0.00 5.00 10.00 15.00 20.00 25.00 30.00 35.00 40.00 45.00
m illiliters NaOH (0.10 M)
pH
Titration of a Buffered Solution
A solution that is 0.10 M CH3COOH and 0.10 M NaCH3COO is titrated with 0.10 M NaOH
Comparing ResultsGraph
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0 5 10 15 20 25 30 35 40 45
mL 0.10 M NaOH
pH
Buffered
Unbuffered
Selection of Indicators
The equivalence point of a titration, defined by the stoichiometry, is not necessarily the same as the end point, careful selection of the indicator will ensure that the error is negligible.
Indicator TransitionsIndicator Low pH color Transition pH range High pH colorGentian violet (Methyl violet 10B) yellow 0.0–2.0 blue-violet
Leucomalachite green (first transition) yellow 0.0–2.0 greenLeucomalachite green (second transition) green 11.6–14 colorless
Thymol blue (first transition) red 1.2–2.8 yellow
Thymol blue (second transition) yellow 8.0–9.6 blue
Methyl yellow red 2.9–4.0 yellowBromophenol blue yellow 3.0–4.6 purpleCongo red blue-violet 3.0–5.0 redMethyl orange red 3.1–4.4 orangeBromocresol green yellow 3.8–5.4 blueMethyl red red 4.4–6.2 yellowMethyl red red 4.5–5.2 greenAzolitmin red 4.5–8.3 blueBromocresol purple yellow 5.2–6.8 purpleBromothymol blue yellow 6.0–7.6 bluePhenol red yellow 6.8–8.4 redNeutral red red 6.8–8.0 yellowNaphtholphthalein colorless to reddish 7.3–8.7 greenish to blueCresol Red yellow 7.2–8.8 reddish-purplePhenolphthalein colorless 8.3–10.0 fuchsiaThymolphthalein colorless 9.3–10.5 blueAlizarine Yellow R yellow 10.2–12.0 redLitmus red 4.5-8.3 blue
Source: Wikipedia
Indicators• Weak acids that change color when they
become bases.• weak acid written HIn• Weak base• HIn H+ + In- clear
red• Equilibrium is controlled by pH• End point - when the indicator changes color.
Solubility
Equilibria
Lead (II) iodide precipitates when potassium iodide is mixed with lead (II) nitrate.
Graphic: Wikimedia Commons user PRHaney
Ksp Values for Some Salts at 25C
Name Formula
Ksp
Barium carbonate BaCO3 2.6 x 10-9
Barium chromate BaCrO4 1.2 x 10-10
Barium sulfate BaSO4 1.1 x 10-10
Calcium carbonate CaCO3 5.0 x 10-9
Calcium oxalate CaC2O4 2.3 x 10-9
Calcium sulfate CaSO4 7.1 x 10-5
Copper(I) iodide CuI 1.3 x 10-12
Copper(II) iodate Cu(IO3)2 6.9 x 10-8
Copper(II) sulfide CuS 6.0 x 10-37
Iron(II) hydroxide Fe(OH)2 4.9 x 10-17
Iron(II) sulfide FeS 6.0 x 10-19
Iron(III) hydroxide Fe(OH)3 2.6 x 10-39
Lead(II) bromide PbBr2 6.6 x 10-6
Lead(II) chloride PbCl2 1.2 x 10-5
Lead(II) iodate Pb(IO3)2 3.7 x 10-13
Lead(II) iodide PbI2 8.5 x 10-9
Lead(II) sulfate PbSO4 1.8 x 10-8
Name Formula
Ksp
Lead(II) bromide PbBr2 6.6 x 10-6
Lead(II) chloride PbCl2 1.2 x 10-5
Lead(II) iodate Pb(IO3)2 3.7 x 10-13
Lead(II) iodide PbI2 8.5 x 10-9
Lead(II) sulfate PbSO4 1.8 x 10-8
Magnesium carbonate MgCO3 6.8 x 10-6
Magnesium hydroxide Mg(OH)2 5.6 x 10-12
Silver bromate AgBrO3 5.3 x 10-5
Silver bromide AgBr 5.4 x 10-13
Silver carbonate Ag2CO3 8.5 x 10-12
Silver chloride AgCl 1.8 x 10-10
Silver chromate Ag2CrO4 1.1 x 10-12
Silver iodate AgIO3 3.2 x 10-8
Silver iodide AgI 8.5 x 10-17
Strontium carbonate SrCO3 5.6 x 10-10
Strontium fluoride SrF2 4.3 x 10-9
Strontium sulfate SrSO4 3.4 x 10-7
Zinc sulfide ZnS 2.0 x 10-25
Solving Solubility ProblemsFor the salt AgI at 25C, Ksp = 1.5 x 10-16
AgI(s) Ag+(aq) + I-(aq)
I
C
E
OO
+x +x
x x
1.5 x 10-16 = x2
x = solubility of AgI in mol/L = 1.2 x 10-8 M
Solving Solubility ProblemsFor the salt PbCl2 at 25C, Ksp = 1.6 x 10-5
PbCl2(s) Pb2+(aq) + 2Cl-(aq)
I
C
E
OO
+x +2x
x 2x
1.6 x 10-5 = (x)(2x)2 = 4x3
x = solubility of PbCl2 in mol/L = 1.6 x 10-2 M
Solving Solubility with a Common IonFor the salt AgI at 25C, Ksp = 1.5 x 10-16
What is its solubility in 0.05 M NaI?
AgI(s) Ag+(aq) + I-(aq)
I
C
E
0.05O
+x 0.05+x
x 0.05+x
1.5 x 10-16 = (x)(0.05+x) (x)(0.05)
x = solubility of AgI in mol/L = 3.0 x 10-15 M
Precipitation
• Ion Product, Q =[M+]a[Nm-]b • If Q>Ksp a precipitate forms.• If Q<Ksp No precipitate.• If Q = Ksp equilibrium.
Precipitation
A solution of 750.0 mL of 4.00 x 10-3M Ce(NO3)3
is added to 300.0 mL of 2.00 x 10-2M KIO3. Will
Ce(IO3)3 (Ksp= 1.9 x 10-10) precipitate and if so, what is the concentration of the ions?
Precipitation and Qualitative Analysis
Complex IonsA Complex ion is a charged species composed of:
1. A metallic cation
2. Ligands – Lewis bases that have a lone electron pair that can form a covalent bond with an empty orbital belonging to the metallic cation
NH3, CN-, and H2O are Common Ligands
O
H
HNH
HH
C N-
Coordination Number Coordination number refers to the number of ligands attached to the cation 2, 4, and 6 are the most common coordination numbers
Coordination
number
Example(s)
2 Ag(NH3)2+
4 CoCl42- Cu(NH3)4
2+
6 Co(H2O)62+ Ni(NH3)6
2+
Complex Ions and Solubility
AgCl(s) Ag+ + Cl- Ksp = 1.6 x 10-10
Ag+ + NH3 Ag(NH3)+ K1 = 2.1 x 103
Ag(NH3)+ NH3 Ag(NH3)2+ K2 = 8.2 x 103
AgCl + 2NH3 Ag(NH3)2+ + Cl- K = KspK1K2
23
233
][
]][)([108.2
NH
ClNHAgxK
• Calculate the concentrations of Ag+, Ag(S2O3)-,
and Ag(S2O3)2-3 in a solution made by mixing
150.0 mL of 1.00x10-3M AgNO3 with 200.0 mL of 5.00 M Na2S2O3.
Ag+ + S2O3-2 Ag(S2O3)-
K1=7.4 x 108
Ag(S2O3)- + S2O3-2 Ag(S2O3)2
-3
K2=3.9 x 104