Additional Aqueous Equilibria CHAPTER 16

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Additional Aqueous Equilibria CHAPTER 16
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I. Buffers

• A. Definitions
• Buffer- solutions that resist changes in pH
  when acid or base are added to it.
• - they are composed of a weak acid and
  its conjugate base or a weak base and its
  conjugate acid.
• acetic acid / sodium
  acetate
• carbonic acid / carbonate

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• B. Explanation of Buffers
• 1. Shifting of Equilibrium
• For some weak acid and its conjugate
  base
• HA ? H A-
• a. What happens if you add excess acid (H)?
• b. What happens if you add excess base
  (OH-)?
• Best buffering capacity when approximately
  equal amounts of both acid and conjugate
  base.
• Why strong acids and bases not good
  buffers?

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• 2. Titration Curve For Buffer
• Acetic acid / acetate (CH3COOH /
  CH3COO-)

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• 3. Mathematical Relationship
• a. Henderson-Hasselbalch Equation
• For HA ?H A-
• When HA A-, it is true that pH
  pKa
• Best buffering region 1 unit of pKa
  

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• C. Buffer Problem Solving
• Similar to other equilibrium problems,
  except we will use the Henderson Hasselbalch
  Equation
• Always utilizes assumption that we can
  neglect small amount of dissociation of weak acid
  or base!!!!!!!!
• Different from Tro text, all our buffer problem
  solving will be done using the Henderson-Hasselbal
  ch!!!!

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1. Determining the pH of a Buffer Solution

• What is the pH of a buffered solution that is
• 0.015 M acetic acid and 0.045 M sodium
  acetate?
• For acetic acid, Ka 1.8 x 10-5 ?

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• 2. Preparation of a Buffer Solution
• a. Which acid-base pair should be used to
  prepare a buffer pH 5.0?
• Acetic acid / acetate Ka 1.8 x 10-5
• Ammonium / ammonia Ka 5.6 x 10-10
• b. A buffer solution pH 5.0 is to be prepared
  using acetic acid (Ka 1.8 x 10-5)
  and sodium acetate . How many grams of
  sodium acetate (NaC2H3O2)
• should you add to 250.0 mL of a 0.0500 M
  acetic acid (HC2H3O2) solution?

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• 3. Addition of Acid or Base To A Buffer
• A buffer is prepared by adding 0.30 mol of
  lactic acid and 0.40 mol of sodium lactate to
  sufficient water to make 2.00 L of buffer
  solution. The Ka of lactic acid is 1.4 x 10-4.
  (Assume Ka small enough to neglect
  dissociation of weak acid).
• a) Calculate the pH of the buffer.
• b) Calculate the pH of the buffer after the
  addition
• of 0.050 moles of NaOH.

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When Adding Strong Acid or Base To Buffer

• 1. Addition of strong acid to buffer ?
  converts a stoichiometric amount of the base to
  the conjugate acid.
• 2. Addition of strong base to buffer ? converts
  a stoichiometric amount of the acid to the
  conjugate base.
• (Adding acid creates more acid, adding base
  creates more base)

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CONCEPT QUESTION

• A 1.00 liter buffer solution is 0.10 M in HF
  and 0.050 M in NaF. Which action will destroy
  the buffer? Why?
• a. addition of 0.050 mol HCl
• b. addition of 0.050 mol NaOH
• c. addition of 0.050 mol NaF
• d. addition of 0.05 mol HF

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• D. Buffer Capacity
• 1. Definition- amount (moles) of acid
  or base required to change the pH of a given
  volume of buffer by 1 pH unit.
• A measure of the ability of a buffer to resist
  changes in pH.
• Addition of more acid or base will cause the
  buffer to fail, meaning inability to resist
  changes in pH.

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II. Acid-Base Titrations

• A. Titration Process
• 1. Discuss Process (acid base ? salt
  water)
• 2. Equivalence Point point in titration
  when stoichiometric amounts of titrant have been
  added to neutralize acid or base titrated.
• 3. Endpoint point where titration is
  actually ended because of indicator color
  change.
• (Endpoint is used to visually determine
  equivalence pt.)

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• B. Explanation of Acid-Base Indicators
• 1. Weak organic acids which undergo color
  change when there is a shift from acid ? base at
  complete neutralization.
• HIn (aq) H2O (l) ? H3O (aq) In-(aq)
• acid base
• one color another color
• 2. Different indicators change colors at
  different pHs
• Bromocresol green pH 3.8 to 5.4
• Phenolphthalein pH 8.2 to 10

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• C. Titration of Strong Acid with Strong Base
• Example HCl NaOH ? H2O NaCl

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Strong Acid-Strong Base Titration Problem

• A 50.00 mL sample of 0.100 M HCl is titrated
  with a 0.100 M NaOH solution.
• a. Determine the pH of the HCl solution before
  the addition of any NaOH solution.
• a. Determine the pH after the addition of 25.00
  mL
• of the NaOH solution.
• b. Determine the pH after the addition of 50.00
  mL of the NaOH solution.

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• D. Titration of Weak Acid with Strong Base

Example HC2H3O2 NaOH ? C2H3O2- Na H2O
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Weak Acid-Strong Base Titration Problem

• A 50.00 mL sample of 0.100 M acetic acid
  (HC2H3O2, Ka 1.8 x 10-5) is titrated with a
  0.100 M NaOH solution.
• a. Determine the pH of the solution before the
  addition of the NaOH solution.
• b. Determine the pH after the addition of 50.00
  mL of the NaOH solution.

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• E. Titration of Weak Base with Strong Acid
• Example NH3 HCl ? NH4 Cl-
• 1. Is the equivalence point less than or
  greater than pH 7? Explain you answer.
• 2. Draw a titration curve for the titration.
• (titrate NH3 using HCl as the titrant)

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• F. Titration of Polyprotic Acid with Strong Base
• Example H3PO4

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III. Solubility Equilibria

• A. Equilibrium Expressions for Sparingly
  
• Soluble Salts
• 1. Ksp Solubility Product Constant
• (Write like a normal Kc and recall it does
  not include solids in the equilibrium
  expression).
• Meaning idea of extent of solubility of salt
• Larger Ksp ? Greater solubility of salt

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• Write the chemical equations and Ksp
  expressions for the slightly soluble salts
• AgCl
• CaCO3
• Ag3PO4
• Mg(OH)2

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• 2. Solubility amount of substance
  dissolved based on grams / volume.
• 3. Molar Solubility maximum amount of
  substance dissolved based on moles / liter.
• Molar solubility is not the same as Ksp !!!!!
• The molar solubility of BaF2 is 6.27 x 10-3 M.
  The Ksp of BaF2 is 9.8 x 10-7. What is the
  concentration of F- ions in a BaF2 saturated
  solution?

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• B. Problem Solving
• 1. Calculating Ksp from Solubility Data
• One liter of water is able to dissolve 2.15 x
  10-3 mol of PbF2 at 250C.
• What is the molar solubility of PbF2?
• What is the Ksp for PbF2?

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• 2. Calculating Molar Solubility From Ksp
• Determine the molar solubility of barium
  phosphate, Ba3(PO4)2, given a Ksp value of
• 3.4 x 10-23.
• What are the equilibrium concentrations of each
  ion in solution?

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• 3. Common Ion Effect (Factor Affecting
  Solubility)
• Ag2CO3 has a Ksp of 8.1 x 10-12 at 250C. The
  molar solubility of Ag2CO3 in water is 1.3 x 10-4
  M. What is the molar solubility of Ag2CO3 in a
  0.10 M NaCO3 solution. (NaCO3 is very soluble).
• a. Qualitative Explanation Le Chateliers
  Principle
• Ag2CO3 (s) ? 2 Ag (aq) CO32- (aq)
• What is the effect of adding excess CO32- ?
• What is the effect on the molar solubility of
  Ag2CO3 ?

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• b. Terminology Used
• Common Ion the ion which is common to
  both salts being considered.
• Common Ion Effect the solubility of a
  compound is always lowered due to the
  addition of the common ion.
• Now solving problem

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• 4. Predicting If Precipitation Occurs
• Comparison of known Ksp to the calculated
  ion
• product (Q) for a given mixture can serve to
  indicate whether or not a precipitate will be
  formed.
• Q
• Ion Product gt Ksp ppt. forms
• Ion Product Ksp ppt. forms
• Ion Product lt Ksp no ppt. forms

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• Will a precipitate form at equilibrium when
  50.00 mL of 0.00100 M BaCl2 is added to
• 50.00 mL of 0.000100 M Na2SO4 ? The solubility
  product constant for barium sulfate is 1.1 x
  10-10.