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CHEM162S08DC
Assignment 5
Due at 3:00am on Friday, March 28, 2008
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Introduction to Solubility and the Solubility Product Constant
Learning Goal:
To learn how to calculate the solubility from
and vice versa.
Consider the following equilibrium between a solid salt and its dissolved form (ions) in a saturated
solution:
At equilibrium, the ion concentrations remain constant because the rate of dissolution of solid
equals
the rate of the ion crystallization. The equilibrium constant for the dissolution reaction is
is called the
solubility product
and can be determined experimentally by measuring the
solubility
,
which is the amount of compound that dissolves per unit volume of saturated solution.
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View Full Document Part A
A saturated solution of lead(II) fluoride,
, was prepared by dissolving solid
in water. The
concentration of
ion in the solution was found to be
. Calculate
for
.
Part A.1
Find the concentration of F
–
using stoichiometry
If solid
dissolves to produce
, what concentration of
ion is produced?
Hint A.1.a
How to approach the problem
Hint not displayed
Express the molar concentration numerically.
ANSWER:
=
4.16×10
3
−
Now you can plug the concentrations into the
expression.
Part A.2
Identify the
K
sp
expression
Part not displayed
Express your answer numerically.
ANSWER:
=
3.60×10
8
−
Part B
The value of
for silver sulfate,
, is
. Calculate the solubility of
in
grams per liter.
Part B.1
Identify the
K
sp
expression
The balanced equation for the complete dissociation of
is
What is the expression for the solubility product?
ANSWER:
Part B.2
Express
K
sp
in terms of the molar solubility of Ag
2
SO
4
If
represents the number of moles per liter of
that dissolves to form a saturated solution,
then we can construct the following table:
excess
0
0
What is
in terms of
?
Express
in terms of
.
ANSWER:
=
Knowing that
, you can now solve this expression for
and find the molar
solubility of
.
Part B.3
Calculate the molar solubility of Ag
2
SO
4
The solubility product is
. Calculate
, the molar solubility of
.
Express the molar solubility numerically.
ANSWER:
=
1.44×10
2
−
Now you can convert the solubility from moles per liter to grams per liter using the molar mass of
.
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View Full Document Precipitation
Learning Goal:
To understand the relationship between precipiatation and the solubility product. To be
able to predict whether a substance will precipitate or not.
Precipitation is the formation of an insoluble substance. For the equation
,
precipitation represents a shift to the left, and the production of a solid. From Le Chatelier's principle, we
know that when the product of the concentrations of
and
get above a certain level, the reaction will
respond by shifting left to decrease the concentrations of
and
. This critical level,
, is a contstant
at a certain temperature. In this case,
.
When
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This note was uploaded on 04/03/2008 for the course CHEM 162 taught by Professor Siegal during the Spring '08 term at Rutgers.
 Spring '08
 siegal

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