chapter 15: applications of aqueous equilibria
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Chapter 15: Applications of Aqueous Equilibria. Dr. Aimée Tomlinson. Chem 1212. Section 15.1. Neutralization Reactions. Four Types of Neutralization. Strong Acid + Strong Base. Strong Acid + Weak Base. Always lead to acidic solution Only cation of acid is a spectator ion. - PowerPoint PPT PresentationTRANSCRIPT
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CHAPTER 15: APPLICATIONS OF AQUEOUS EQUILIBRIA
Dr. Aimée TomlinsonChem1212
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Neutralization Reactions
Section 15.1
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Four Types of Neutralization
Strong Acid + Strong BaseStrong Acid + Weak Base
Weak Acid + Strong Base Weak Acid + Weak Base
3 2 3HNO KOH H O K NO
2net ionic equation: H OH H O Always lead to neutral solution Salt is present but a spectator ion
3 4NH HCl NH Cl
3 4net ionic equation: NH H NH Always lead to acidic solution Only cation of acid is a spectator ion
2HCN KOH H O K CN
2net ionic equation: HCN OH H O CN
Always lead to basic solution Only anion of base is a spectator ion
3 4NH HCN NH CN net ionic equation is same as above
Must compare Ka & Kb to determine pH There are no spectator ions
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Common-Ion Effect
Section 15.2
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Common Ions
When two different species give rise to the same ion
I.E. both NaCl & HCl give rise to Cl-
Placing NaCl in HCl will repress the dissociation of the acid
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Common-Ion Effect• Defn: a shift based on equilibrium due to the addition of
a common ion
• It is based on Le Châtelier’s Principle
• Example:
Adding HCO3- will shift the eq to the left
As a consequence less H3O+ is produced The pH is higher than it would have been
2 3( ) 2 ( ) 3( ) 3 ( )aq l aq aqH CO H O HCO H O
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Common-Ion Effect ExampleCalculate the pH of a solution prepared by mixing equal volumes of 0.20 M CH3NH2 and 0.60 M CH3NH3Cl (Kb = 3.7 x 10-4). What is the pH of 0.20 M CH3NH2 without addition of CH3NH3Cl?
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Buffer Solutions
Section 15.3
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pH Buffer
A solution that resists changes in pH
Generated from an acid with its conjugate base or a base with its conjugate acid pair
Overall it is the result of the common-ion effect
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Forming Good Buffers
Need a large buffer capacity: the quantity of acid/base needed to significantly change the pH of a buffer
Two ways to make a good buffer: Equal volumes – where more concentrated solutions
will give rise to a larger capacity Equal moles – where larger volumes will give rise to
larger capacities
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Henderson-Hasselbalch
Section 15.4
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Henderson-Hasselbalch EquationThis is how we figure out how to make our buffers
𝐻𝐴+𝐻2𝑂⇌ 𝐴−+𝐻3𝑂+¿ ¿
𝐾 𝑎=[ 𝐴− ]¿¿
− log 𝐾𝑎=− log [ 𝐴− ]¿¿
− log 𝐾𝑎=− log[ 𝐴− ]
[𝐻𝐴 ]− log ¿
𝑝𝐻=𝑝 𝐾𝑎+ log ( [ 𝐴− ][𝐻𝐴 ] )
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Buffer Example ICalculate the pH and pOH of a 500.0 mL solution containing 0.225 M HPO4
-2 and 0.225M PO43- at
25⁰C where Ka(HPO4-)= 4.2 x 10-13.
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Buffer Example IIHow would we prepare a pH = 4.44 buffer using CH3CO2H and CH3CO2Na (Ka = 1.8 x 10-5)?
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Comprehensive ExampleThis set of questions will include questions from both chapter 14 & chapter 15
Determine the pH for each of the following:a.) 0.100 M HC3H5O2 with Ka = 1.3 x 10-5
b.) 0.100 M NaC3H5O2c.) Mixture of a.) and b.)d.) Mixture of c.) with 0.020 mol NaOHe.) Mixture of c.) with 0.020 mol HCl
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Part a.)0.100 M HC3H5O2 with Ka = 1.3 x 10-5
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Part a.)0.100 M HC3H5O2 with Ka = 1.3 x 10-5
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Part b.)0.100 M NaC3H5O2 with Ka = 1.3 x 10-5
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Part b.)0.100 M NaC3H5O2 with Ka = 1.3 x 10-5
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Part c.)0.100 M HC3H5O2 & 0.100 M NaC3H5O2 with Ka = 1.3 x 10-5
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Part d.)0.100 M HC3H5O2 & 0.100 M NaC3H5O2 & 0.020 mol NaOH with Ka = 1.3 x 10-5
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Part e.)0.100 M HC3H5O2 & 0.100 M NaC3H5O2 & 0.020 mol HCl with Ka = 1.3 x 10-5
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Comprehensive ExamplePutting it all together:
Determine the pH for each of the following:a.) 0.100 M HC3H5O2 with Ka = 1.3 x 10-5 pH = 2.96b.) 0.100 M NaC3H5O2
pH = 8.94c.) Mixture of a.) and b.)
pH = 4.89d.) Mixture of c.) with 0.020 mol NaOH pH = 5.06e.) Mixture of c.) with 0.020 mol HCl pH = 4.71
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Titrations
Sections 15.5-15.8
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TitrationUsed to determine the concentration of an unknown
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Titration CurveA plot of pH versus volume of titrant
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Equivalence PointPoint at which moles of acid = moles of base
Strong acid with weak base: pH < 7.0 acidic
Strong acid with strong base: pH = 7.0 neutral
Weak acid with strong base: pH > 7.0 basic
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pH IndicatorsChange color at the equivalence point
Strong acid with weak base
thymol blue
Strong acid with strong base
phenolphthalein
Weak acid with strong base
alizarin yellow
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Titration Example IWhat is [NH3] if 22.35mL of 0.1145 M HCl were needed to titrate a 100.0mL sample?
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Titration Example IIStrong with Strong: A 15.0 mL sample of 0.200 M NaOH is titrated with 0. 250 M of HCl. Calculate the pH of the mixture after 10.0, and 20.0 mL of acid have been added.
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Titration Example IIStrong with Strong: A 15.0 mL sample of 0.200 M NaOH is titrated with 0. 250 M of HCl. Calculate the pH of the mixture after 10.0, and 20.0 mL of acid have been added.
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Titration Example IIIStrong with Weak: A 25.0 mL sample of 0.100 M acetic acid (HC2H3O2) is titrated with 0.125 M of NaOH. Calculate the pH of the mixture after 0.0,10.0, 20.0, and 30.0 mL of base have been added. (Ka = 1.8 x 10-5)
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Titration Example IIIStrong with Weak: A 25.0 mL sample of 0.100 M acetic acid (HC2H3O2) is titrated with 0.125 M of NaOH. Calculate the pH of the mixture after 0.0,10.0, 20.0, and 30.0 mL of base have been added. (Ka = 1.8 x 10-5)
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Titration Example IIIStrong with Weak: A 25.0 mL sample of 0.100 M acetic acid (HC2H3O2) is titrated with 0.125 M of NaOH. Calculate the pH of the mixture after 0.0,10.0, 20.0, and 30.0 mL of base have been added. (Ka = 1.8 x 10-5)
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Titration Example IIIStrong with Weak: A 25.0 mL sample of 0.100 M acetic acid (HC2H3O2) is titrated with 0.125 M of NaOH. Calculate the pH of the mixture after 0.0,10.0, 20.0, and 30.0 mL of base have been added. (Ka = 1.8 x 10-5)
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Titration Example IIIStrong with Weak: A 25.0 mL sample of 0.100 M acetic acid (HC2H3O2) is titrated with 0.125 M of NaOH. Calculate the pH of the mixture after 0.0,10.0, 20.0, and 30.0 mL of base have been added. (Ka = 1.8 x 10-5)
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Titration Example IIIStrong with Weak: A 25.0 mL sample of 0.100 M acetic acid (HC2H3O2) is titrated with 0.125 M of NaOH. Calculate the pH of the mixture after 0.0,10.0, 20.0, and 30.0 mL of base have been added. (Ka = 1.8 x 10-5)
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Titration Example IIIStrong with Weak: A 25.0 mL sample of 0.100 M acetic acid (HC2H3O2) is titrated with 0.125 M of NaOH. Calculate the pH of the mixture after 0.0,10.0, 20.0, and 30.0 mL of base have been added. (Ka = 1.8 x 10-5)
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Titration Example IIIStrong with Weak: A 25.0 mL sample of 0.100 M acetic acid (HC2H3O2) is titrated with 0.125 M of NaOH. Calculate the pH of the mixture after 0.0,10.0, 20.0, and 30.0 mL of base have been added. (Ka = 1.8 x 10-5)
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Polyprotic Acid Titrations - SKIP
Section 15.9
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Solubility Equilibria
Section 15.10
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Solubility ProductIt is the equilibrium constant for solids in solution, Ksp
( ) ( ) [ ] [ ]m xm x aq aq spM X mM xX K M X
2 112( ) ( ) ( )
2 2( ) ( )
2 3.9 10
[ ][ ]s aq aq sp
sp aq aq
CaF Ca F K
K Ca F
An actual example looks like:
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Measuring Ksp & Calculating Solubility
Section 15.11
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Solubility ExampleDetermine the equilibrium concentrations (and solubilities) of BaF2(s), Ksp = 1.7x10-6.
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Section 15.12
Factors that Affect Solubility
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Three Solubility Factors
Common-Ion Effect
pH
Complex ion formation – we are skipping this one
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Common-Ion Solubility ExampleCalculate the solubility of calcite (CaCO3) in 0.00100 M of Na2CO3 and in just plain water (Ksp = 4.5x10-9 at 25C).
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pH Solubility ExampleIf the salt added possesses a conjugate acid or conjugate base then pH will impact the solubility of the salt
Addition of acid will pull carbonate out of the solution Recall LCP when we remove product the eq will push forward This leads to more CaCO3 being dissolved Overall this trend demonstrates why so many compounds are more
soluble in acidic solutions
2 23( ) ( ) 3( )
2( ) 3( ) 3( )
23( ) ( ) ( ) 3( )
s aq aq
aq aq aq
s aq aq aq
CaCO Ca CO
H CO HCO
CaCO H Ca HCO
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Sections 15.13-15.15
Skip these sections