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No Fear Translations of Shakespeare’s plays (along with audio!) and other classic works
Flashcards
Mastery Quizzes
Infographics
Graphic Novels
AP® Test Prep PLUS
AP® Practice & Lessons
My PLUS Activity
Note-taking
Bookmarking
Dashboard
Testimonials from SparkNotes
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No Fear
provides access to Shakespeare for students who normally couldn’t (or wouldn’t) read his plays.
It’s also a very useful tool when trying to explain Shakespeare’s wordplay!
Erika M.
I
tutor high school students in a variety of subjects. Having access to the literature
translations helps me to stay informed about the various assignments. Your summaries and
translations are invaluable.
Kathy B.
Teaching Shakespeare to today's generation can be challenging. No Fear helps a ton with
understanding the crux of the text.
Kay
H.
Testimonials from SparkNotes Customers
No Fear provides access to Shakespeare for students who normally couldn’t (or wouldn’t) read his plays. It’s also a very useful tool when trying to explain Shakespeare’s wordplay!
Erika M.
I tutor high school students in a variety of subjects. Having access to the literature translations helps me to stay informed about the various assignments. Your summaries and translations are invaluable.
Kathy B.
Teaching Shakespeare to today's generation can be challenging. No Fear helps a ton with understanding the crux of the text.
Kay H.
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Note that the sample species HA and A- in the above
Expression are generalized to the terms
acid and base, respectively. To use
the equation, place the concentration of the acidic buffer species where
the equation says "acid" and
place the concentration of the basic buffer species where the equation
calls for "base". It is essential
that you use the pKa of the acidic species and not the
pKb of the basic
species when working with basic buffers--many students forget this point
when doing buffer
problems.
A buffer problem can be fairly simple to solve, provided you don't get
confused by all the other
chemistry you know. For example, let's calculate the pH of a
solution that is 0.5 M acetic acid and 0.5 sodium acetate both before and after
enough SO3
gas is dissolved to
make the solution 0.1 M in sulfuric acid. Before the acid is added, we can
use the
Henderson-Hasselbalch equation to calculate the pH.
Figure %: Note that the pKa of acetic acid is 4.75.
This part of the problem does not require us to do the sort of equilibrium
calculations that we must use for Non-Buffered
Solutions, but many students still
try to do it the hard way. The hard way is a correct way of doing the
problem, but it may cost you valuable time on a test.
To calculate the pH after the acid is added, we assume that the
acid reacts with the base in
solution and that the reaction has a 100% yield. Therefore, we say that
0.1 moles per liter of acetate
ion reacts with 0.1 moles per liter of sulfuric acid to give 0.1 moles per
liter of acetic acid and
hydrogen sulfate. Here, we ignore the second dissociation of sulfuric acid
because it is minor in
comparison to the first. So the final concentration of acetic acid
is 0.6 M and acetate is 0.4M.
Plugging those values into the Henderson-Hasselbalch equation gives a pH of
4.57. Note that a
0.1 M solution of strong acid would give to a pH of 1 but the buffer gives a
pH of 4.57 instead.
To probe the useful range of the buffer, let's calculate the pH of the
solution resulting from the same
situation above but with different concentrations of the buffer. If the
buffer is 1.0 M in both acetate
and acetic acid, then the pH of the resulting solution after the
introduction of acid is 4.66. However,
if we make the solution only 0.11 M in acetic acid and acetate, then we
calculate a pH of 3.45!
Therefore, if you want a more effective buffer, make sure that the
concentration of the buffering
agents is large in comparison to the added acid or base.