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01:160:161 Chapter Notes - Chapter 18: Solubility Equilibrium, Inflection Point, Bayerischer RundfunkExam


Department
Chemistry
Course Code
01:160:161
Professor
Asbed Vassilia
Study Guide
Final

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1
CHAPTER 18
ACID & BASE EQUILIBRIA
(BUFFERS & TITRATIONS)
OTHER EQUILIBRIA
Problems to prepare students for hourly exam II.
Buffer concepts
Buffer calculations
Titrations and Indicators
Solubility Product
Complex Ion Equilibria
E. Tavss, PhD
2/15/12

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2
BUFFER CONCEPTS
44
Chem 162-2011 Final exam
Chapter 15b – Applic. of Acid & Base Equilibria (Buffers & Titrations)
Buffer concepts
Adding the appropriate amount of which of the following to a 0.1 M solution of CH3COONa
could produce a buffer?
CH3COOH HCl CH3COOK NaCl NH4Cl
A. CH3COOH only
B. CH3COOH or HCl
C. NaCl or CH3COOK
D. HCl only
E. NH4Cl only
A buffer, in this case, would be the mixture of the conjugate base (CH3COONa) and its weak
acid (CH3COOH).
A. False. If CH3COOH is added to the conjugate base, it would form a buffer. However, since
HCl would do the same thing, the word “only” makes this option false.
B. True. If CH3COOH is added to CH3COONa it would form a buffer of CH3COOH and
CH3COONa. Also if HCl is added to CH3COONa it would form CH3COOH; therefore, a buffer
would be formed of CH3COOH and CH3COONa.
C. False. If CH3COOK were added to CH3COONa, the mixture would only contain a mixture
of conjugate bases, but no significant amount of weak acid. Addition of NaCl would not
produce a buffer.
D. False. If HCl is added to the conjugate base, it would form a buffer. However, since
CH3COOH would do the same thing, the word “only” makes this option false.
E. False. A buffer is defined as a weak acid and its conjugate base. NH4Cl is not the weak acid
necessary to combine with this conjugate base.

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3
Chem 162-2011 Hourly Exam II + Answers
Chapter 15B - Applic. Of Acid & Base Equilibria (Buffers & Titrations)
Buffer concepts
7. Acetic acid, CH3COOH is a weak acid with Ka = 1.8 x 10-5
Ammonia, NH3, is a weak base with Kb = 1.8 x 10-5
Which one of the following would produce a buffer with pH close to 5.0?
A. Add 1 mol CH3COOH and 1 mol NH3 to a liter of water.
B. Add 1 mol CH3COOH and 1 mol NaOH to a liter of water.
C. Add 1 mol CH3COOH and 0.5 mol NaOH to a liter of water.
D. Add 1 mol NH3 and 1 mol HCl to a liter of water.
E. Add 1 mol NH3 and 0.5 mol HCl to a liter of water.
CH3COOH:
The pKa of CH3COOH = -log(1.8x10-5) = 4.74.
NH3:
Kw = Kca x Kb
Kca = (1 x 10-14)/(1.8x10-5) = 5.56 x 10-10
pKca = -log(5.56 x 10-10) = 9.25
A.
HA + NH3 ⇄ NH4+ + A-
False. For the buffer to have a pH of close to 5.0 means that the pH of the buffer must be close to 5.0.
However, since the Ka of acetic acid is 1.8x10-5, and the Kb of NH3 is 1.8x10-5, then they are equal in
terms of acid-base strength. This means that a mixture of the two would exactly neutralize each other,
providing a pH of 7.
B. False
HA + OH- H
2O + A-
Initial 1M 1M 0 0
Change
Equilibrium
Large K rule
HA + OH- H
2O + A-
Initial 1M 1M 0 0
Change -1 -1 +1
Equilibrium 0 0 +1
This contains a conjugate base, but no weak acid. Hence, it isn’t a buffer.
C. True
HA + OH- H
2O + A-
Initial 1M 0.5M 0 0
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