Chapter 15 Review
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Transcript Chapter 15 Review
Chapter 15 Review
Quiz
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
400 g NaOH/(400g + 600g) x 100 = 40%
Pour the NaOH slowly into the water while stirring (2
pts). It gets very HOT!
Saponification
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?
Ammonia contains an “Ammonium” ion
What is an ammonium ion like?
Why do we need positive ions after the mixing?
To balance any remaining OH- ions
Things to Remember
What is “Solvation”?
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
Text uses “solvation”. Test uses “dissolution”.
Vocabulary Terms
Solute, solvent
Saturated, unsaturated, supersaturated
Solubility, soluble, insoluble, miscible, immiscible
Things to Remember
Like Dissolves Like
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:
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
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
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
Examples
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
Examples
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