ch. 16: aqueous ionic equilibrium dr. namphol sinkaset chem 201: general chemistry ii
TRANSCRIPT
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Ch. 16: Aqueous Ionic Ch. 16: Aqueous Ionic EquilibriumEquilibrium
Dr. Namphol Sinkaset
Chem 201: General Chemistry II
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I. Chapter OutlineI. Chapter Outline
I. Introduction
II. Buffers
III. Titrations and pH Curves
IV. Solubility Equilibria and Ksp
V. Complex Ion Equilibria
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I. Last Aspects of EquilibriaI. Last Aspects of Equilibria
• In this chapter, we cover some final topics concerning equilibria.
• Buffers are designed to take advantage of Le Châtelier’s Principle.
• Solubility can be reexamined from an equilibrium point of view.
• Complex ions are introduced and their formation explained using equilibrium ideas.
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II. pH Resistive SolutionsII. pH Resistive Solutions
• A solution that resists changes in pH by neutralizing added acid or base is called a buffer.
• Many biological processes can only occur within a narrow pH range.
• In humans, the pH of blood is tightly regulated between 7.36 and 7.42.
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II. Creating a BufferII. Creating a Buffer
• To resist changes in pH, any added acid or base needs to be neutralized.
• This can be achieved by using a conjugate acid/base pair.
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II. How the Buffer WorksII. How the Buffer Works
• For a buffer comprised of the conjugate acid/base pair of acetic acid/acetate: OH-
(aq) + CH3COOH(aq) H2O(l) + CH3COO-(aq)
H+(aq) + CH3COO-
(aq) CH3COOH(aq)
• As long as we don’t add too much OH- or H+, the buffer solution will not change pH drastically.
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II. Calculating the pH of BuffersII. Calculating the pH of Buffers
• The pH of a buffer can be calculated by approaching the problem as an equilibrium in which there are two initial concentrations.
• Since an acid and its conjugate base are both in solution, problem can be solved from a Ka or Kb point of view.
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II. Sample ProblemII. Sample Problem
• A solution was prepared in which [CH3COONa] = 0.11 M and [CH3COOH] = 0.090 M. What is the pH of the solution? Note that the Ka for acetic acid is 1.8 x 10-5.
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II. Sample ProblemII. Sample Problem
• A student dissolves 0.12 mole NH3 and 0.095 mole NH4Cl in 250 mL of water. What’s the pH of this buffer solution? Note that the Kb for ammonia is 1.8 x 10-5.
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II. A Special Buffer EquationII. A Special Buffer Equation
• Buffers are used so widely that an equation has been developed for it.
• This equation can be used to perform pH calculations of buffers.
• More importantly, it can be used to calculate how to make solutions buffered around a specific pH.
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II. Henderson-Hasselbalch Eqn.II. Henderson-Hasselbalch Eqn.
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II. The H-H EquationII. The H-H Equation
• The Henderson-Hasselbalch equation works for a buffer comprised of conjugate acid/base pairs.
• It works provided the “x is small” approximation is valid.
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II. Sample ProblemII. Sample Problem
• A student wants to make a solution buffered at a pH of 3.90 using formic acid and sodium formate. If the Ka for formic acid is 1.8 x 10-4, what ratio of HCOOH to HCOONa is needed for the buffer?
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II. Sample ProblemII. Sample Problem
• A researcher is preparing an acetate buffer. She begins by making 100.0 mL of a 0.010 M CH3COOH solution. How many grams of CH3COONa does she need to add to make the pH of the buffer 5.10 if the Ka for acetic acid is 1.8 x 10-5?
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II. Upsetting the BufferII. Upsetting the Buffer
• A buffer resists changes to pH, but it is not immune to change.
• Adding strong acid or strong base will result in small changes of pH.
• We calculate changes in pH of a buffer by first finding the stoichiometric change and then performing an equilibrium calculation.
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II. Illustrative ProblemII. Illustrative Problem
• A 1.00 L buffer containing 1.00 M CH3COOH and 1.00 M CH3COONa has a pH of 4.74. It is known that a reaction carried out in this buffer will generate 0.15 mole H+. If the pH must not change by more than 0.2 pH units, will this buffer be adequate?
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II. Stoichiometric CalculationII. Stoichiometric Calculation
• When the H+ is formed, it will react stoichiometrically with the base.
• We set up a different type of table to find new equilibrium concentrations.
• We use ≈0.00 mol because [H+] is negligible.
H+ + CH3COO- CH3COOH
Initial 0.15 mol 1.00 mol 1.00 mol
Change -0.15 mol -0.15 mol +0.15 mol
Final ≈0.00 mol 0.85 mol 1.15 mol
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II. Equilibrium CalculationII. Equilibrium Calculation
• We use the new buffer concentrations in an equilibrium calculation to find the new pH.
• Again, use ≈0.00 M because [H3O+] is negligible.
CH3COOH + H2O H3O+ + CH3COO-
Initial 1.15 M ---- ≈0.00 M 0.85 M
Change -x ---- +x +x
Equil. 1.15 – x ---- x 0.85 + x
Solve to get [H3O+] = 2.44 x 10-5 M, and pH = 4.613.
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II. How the Buffer WorksII. How the Buffer Works
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II. A Simplification for BuffersII. A Simplification for Buffers
• In buffer problems, # of moles can be used in place of concentration.
• This can be done because all components are in the same solution, and hence have the same volume.
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II. Sample ProblemII. Sample Problem
• Calculate the pH when 10.0 mL of 1.00 M NaOH is added to a 1.0-L buffer containing 0.100 mole CH3COOH and 0.100 mole CH3COONa. Note that the Ka for acetic acid is 1.8 x 10-5.
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II. Buffers Using a Weak BaseII. Buffers Using a Weak Base
• Up until now, we’ve been creating buffers with a weak acid and it’s conjugate base.
• Can also make a buffer from a weak base and its conjugate acid.
• Henderson-Hasselbalch still applies, but we need to get Ka of the conjugate acid.
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II. pKII. pKaa/pK/pKbb Relationship Relationship
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II. Sample ProblemII. Sample Problem
• Calculate the pH of a 1.0-L buffer that is 0.50 M in NH3 and 0.20 M in NH4Cl after 30.0 mL of 1.0 M HCl is added. Note that the Kb for NH3 is 1.8 x 10-5.
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II. Making Effective BuffersII. Making Effective Buffers
• There are a few parameters to keep in mind when making a buffer: The relative [ ]’s of acid and conjugate base
should not differ by more than factor of 10. The higher the actual [ ]’s of acid and
conjugate base, the more effective the buffer.
The effective range for a buffer system it +/- 1 pH unit on either side of pKa.
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II. Buffer CapacityII. Buffer Capacity
• Buffer capacity is defined as the amount of acid or base that can be added to a buffer without destroying its effectiveness.
• A buffer is destroyed when either the acid or conjugate base is used up.
• Buffer capacity increases w/ higher concentrations of buffer components.