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Take out your purple sheet, 17.2 worksheet, and
problem set 1
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Questions on 17.2 worksheet?
Questions on problem set 1?
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The Henderson–Hasselbalch equation is generally good
enough when the “x is small” approximation is applicable.
Generally, the “x is small” approximation will work when
both of the following are true:
a) The initial concentrations of acid and salt are not
very dilute
b) The Ka is fairly small.
For most problems, this means that the initial acid and salt
concentrations should be over 100 to 1000 times larger
than the value of Ka.
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Calculating the new pH after adding acid or base requires
breaking the problem into two parts:
1. A stoichiometry calculation for the reaction of the added
chemical with one of the ingredients of the buffer to
reduce its initial concentration and increase the
concentration of the other.
 Added acid reacts with the A− to make more HA
 Added base reacts with the HA to make more A−
2. An equilibrium calculation of [H3O+] using the new
initial values of [HA] and [A−]
Figure 16.3 pg 762
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Buffers can also be made by mixing a weak base,
(B:), with a soluble salt of its conjugate acid, H:B+Cl−
H2O(l) + NH3 (aq)  NH4+(aq) + OH−(aq)
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The Henderson–Hasselbalch equation is written for a
chemical reaction with a weak acid reactant and its
conjugate base as a product.
The chemical equation of a basic buffer is written with a
weak base as a reactant and its conjugate acid as a
product.
B: + H2O  H:B+ + OH−
To apply the Henderson–Hasselbalch equation, the
chemical equation of the basic buffer must be looked at like
an acid reaction.
H:B+ + H2O  B: + H3O+
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This does not affect the concentrations, just the way we are looking
at the reaction.
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Problem Set 2
Review Questions 1-9
Quiz tomorrow on 17.1-17.2
Thanksgiving Homework:
◦ Bare minimum
 Notes on 17.3-17.4
 Problem set 3
 5.1-5.7 notes
◦ Keep in mind
 Review questions 10-18 due 12/1
 Ch 5 packet due 12/7 (14 pages long )