Transcript Chapter 15 Review

Chapter 15 Review
Quiz
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To make the soap, I prepared a 40% by mass
solution of NaOH. How did I make 1000g of
solution? Describe measurements and procedures.
NaOH is a solid.
Wear Lab Coat and Goggles (2 pts)
 Measure out 600 ml of distilled water (600g) into a large
flask (2 pts)
 Weigh 400 g of NaOH (2 Pts)
 This gives me
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400 g NaOH/(400g + 600g) x 100 = 40%
Pour the NaOH slowly into the water while stirring (2
pts). It gets very HOT!
Saponification
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Definition: Usually, a process by which
triglycerides are reacted with sodium or
potassium hydroxide to produce glycerol and a
fatty acid salt, called 'soap'.
Saponification
Why Do We Need the Ammonia?
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Ammonia contains an “Ammonium” ion
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What is an ammonium ion like?
Why do we need positive ions after the mixing?
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To balance any remaining OH- ions
Things to Remember
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What is “Solvation”?
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How does it occur?
What has to happen for it to occur?
How do you change the rate of solvation?
“Solvation” and “Dissolution” are the same
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Text uses “solvation”. Test uses “dissolution”.
Vocabulary Terms
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Solute, solvent
Saturated, unsaturated, supersaturated
Solubility, soluble, insoluble, miscible, immiscible
Things to Remember
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Like Dissolves Like
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Polar solvents dissolve polar or ionic solutes
Non-polar solvents dissolve non-polar solutes
Alcohols and soaps dissolve both because they have both
parts
Abbreviations:
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M means Molar = moles/liter of solution
m means Molal = moles/Kg solvent
% by mass = mass solute/mass of solution
% by volume = volume of solute/volume of solution
Examples
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What is the molality of a solution that contains
31.0 g HCl in 5.00 kg of water?
Molality = moles/kg solvent
31.0g HCl x 1 mol/35.5g = 0.873 mol HCl
m = 0.873 mol HCl/5.00 kg = 0.175 m
Examples
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How many moles of HCl are present in 0.70 L
of a 0.33 M HCl solution
M means Molar = Moles/Liter solution
 0.33M = ? moles/0.70L
 ?moles = 0.33mol/liter x 0.70L
 = 0.23 moles HCl
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Examples
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The concentration of a water solution of NaCl
is 2.48 m and it contains 806 g of water. How
many grams of NaCl is in the solution?
m = molality = moles/kg solvent
 806 g x 1 kg/1000 g = 0.806kg water
 2.48 m = ?moles/0.806kg
 ?moles = 2.48 m x 0.806 kg = 2.00 moles
 2.00 moles NaCl x 58.5g NaCl/mole = 117g NaCl
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Examples
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A solution contains 85.0 g of NaNO3 and has a
volume of 750 ml. What is the molarity of the
solution?
Molarity = moles/L
 750 ml x 1 L/1000 ml = 0.750 L solution
 85.0 g NaNO3 x 1 mole/85 g = 1 mole NaNO3
 Molarity = 1 mole/0.750 L = 1.33 M NaNO3
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