thinking like a chemist about solubility equilibrium...
TRANSCRIPT
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Thinking Like a Chemist About
Solubility Equilibrium
UNIT 5 DAY 6
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What are we going to learn today?
Thinking Like a Chemist in the
Context of the Solution Equilbria Concept of Solubility Modeling Ionic Reactions Solubility Product Constant
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Independent Quiz: Clicker Question 13, NO TALKING
When comparing the free energy of the pure solvent to the free energy of a solution formed by dissolving a solid solute in the solvent, the free energy of the solution is:
A. Higher
B. No difference
C. Lower
D. Follows no trend, you need to calculate
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A. 4 M sugar aqueous solution B. 0.75 M KCl aqueous solution C. 0.25 M CaCl2 aqueous solution D. 1.5 M MgCl2 aqueous solution
Which would you expect to have the lowest vapor pressure?
Independent Quiz: Clicker Question 14, NO TALKING
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Imagine you have a beaker that contains water and lots of NaCl. You stir for hours to get the salt to dissolve. In the end, the solution still contains some amount of solid NaCl that won’t dissolve.
We call such a solution a saturated solution.
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Poll: Clicker Question 15
What will happen if I add even more solid salt to an already saturated solution?
a. A little more of the salt will dissolve. b. The solution will become less saturated c. The concentration of the salt will remain the same. d. Can’t answer without knowing the solubility of the salt.
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Solubility
The amount of solute that will dissolve in a given amount of solvent or solution
We will discuss almost exclusively
the solubility of compounds in water
Solubility: Often given in grams of solute per liter solution For example: the solubility of KBr in water is 678g/L!
Molar Solubility: is the number of moles of solute that will dissolve in 1 L solvent. For example: the molar solubility of KBR is 5.7 M.
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Solubility
Solubilities are given for saturated solutions They depend on temperature
These systems are also in dynamic equilibrium
Rate of solute dissolving = Rate of solute recrystallizing
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Rock (CaCO3) in water versus salt (CaCl2)
START WORKING ON ACTIVITY
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A. None of the rock dissolved B. All of the rock dissolved C. A tiny amount dissolved
How much of the rock dissolved?
Poll: Clicker Question 16
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Compare solubilities… model with reaction:
CaCl2(s) Ca2+ (aq) + 2Cl-(aq) CaCO3(s) Ca2+ (aq) + CO3
2- (aq)
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A. 0.40 M B. 0.65 M C. 3.6 x 10-3 M D. 1.5 M E. 5.1 M
The molar solubility of CaCl2 is:
Poll: Clicker Question 17
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CaCO3(s) Ca2+ (aq) + CO32- (aq)
Ksp = 8.7 x 10-9
Set up the Ksp expression. Solve for molar solubility of CaCO3.
a. 9.3 x 10-5
b. 9.3 x 105 c. 4.4 x 10-9 d. Not enough information e. Insoluble means not soluble, so it is 0
Poll: Clicker Question 18
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CH302 Vanden Bout/LaBrake Spring 2012
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Given a generic formula, AX2, where A is the cation and X is the anion, and the molar solubility has been
determined to be 1 X 10-4 M. Calculate the value of the Ksp.
A. Ksp = 1 x 10-4 B. Ksp = 1 x 10-8 C. Ksp = 2 x 10-4 D. Ksp = 4 x 10-8 E. Ksp = 4 x 10-12
Poll: Clicker Question 19
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Which of the following compounds has the lowest molar solubility?
A. AgCl Ksp = 1.8 x 10-10
B. FeS Ksp = 8.0 x 10-19
C. LiF Ksp = 1.8 x 10-3|
D. ZnSe Ksp = 2.0 x 10-25
Poll: Clicker Question 20
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Which of the following compounds has the lowest molar solubility?
A. AgCl Ksp = 1.8 x 10-10
B. Cd3(PO4)2 Ksp = 2.5 x 10-30
C. Zn(OH)2 Ksp = 3.0 x 10-17
D. ZnSe Ksp = 2.0 x 10-25
Poll: Clicker Question 21
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Quick way to estimate using your exponent math skills.
How did you know that so fast?
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What did we learn today?
Solubility is an equilibrium condition. Quantify the solubility using equilibrium constant, K. K is “Ion Product” = product of the ions in solution
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Learning Outcomes
Understand the concept of the ion product. Write formula unit, total ionic and net ionic reactions, and identify spectator ions.
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IMPORTANT INFORMATION
LM09 and HW2 due this morning
LM11 and LM12 posted. HW 3 posted.
Looking ahead:
EXAM 1, TUESDAY Feb 4th 7 – 9 PM Details of room assignments will be posted on
website next week