honors chemistry: chapter 12 solutions
TRANSCRIPT
SOLUTIONS(Made of solute and solvent)
Honors Chemistry: Chapter 12
SOLUTION AND SOLUBILITIES 2
Learning Targets - Unit 12: Solutions (PLEASE READ)
1. Identify solute and solvent in a given solution.
2. Use solubility curves and laboratory methods to determine whether
solutions are unsaturated, saturated, or supersaturated.
3. Differentiate between various ways of expressing concentration,
including appropriate units. (i.e. molarity, molality, percent, mole
fraction).
4. Solve concentration problems.
5. Determine how to prepare molarity solutions.
6. Perform dilution problems.
7. Define colligative properties.
8. Explain how concentration and colligative properties are related.
9. List and describe two types of colligative properties: boiling point,
freezing point.
10. Calculate changes expected (in freezing point, boiling point) at
various concentrations.
11. Perform solution stoichiometry problems.
TERMS YOU NEED TO KNOW!
Solution: a homogeneous mixture. Examples are air, brass, saline, bronze.
Colloid: a homogeneous mixture containing slightly larger particles. Examples include fog, smoke, dust.
Suspensions: are mixtures with even larger particles, but they are not considered true solutions because they separate upon standing. An example is Italian Dressing that eventually settles out after sitting.
Solute: the dissolved substance in a solution
Solvent: the major component in a solution
A solution is saturated when no additional solute can be dissolved at a particular temperature
A supersaturated solution can form when more than the equilibrium amount of solute is dissolved at an elevated temperature, and then the supersaturated solution is slowly cooled.
An unsaturated solution is formed when more of the solute can dissolve in it at a particular temperature.
4
TERMS (cont.)
How to Make a
Supersaturated Solution(this is on your chapter 12 test!)
5
https://youtu.be/FcxZ9DyOaUk
See how unstable a supersaturated solution is:
1. Place a beaker of measured water onto a hotplate and add MORE than the
amount of solute than would normally dissolve into the solvent. Though
you stir and stir you will not get it all to go into solution and will see it
swirling around.
2. Heat the solution until all of the solute dissolves. By heating it you FORCE
the solute to dissolve into the solvent.
3. This step is important as it is not yet a supersaturated solution. You must
first cool it back down to room temperature before we consider it
supersaturated. The solution is very unstable at this point and agitating it
just a bit will cause all of the solute to come back out.
Be sure you understand these steps as how to make this is on your test. See
the link below for what happens when you agitate a supersaturated solution.
Now watch the video describing different
characteristics of solutions, but before you do,
these are the things you will want to look for
on the video because every one of these
questions is also on your chapter 12 test!
6
1. What are solutes and solvents?
2. How do ionic cpds “dissociate” in solvents?
3. How do molecular (covalent) cpds dissolve in
solvents?
4. What do the terms solubility and concentration
mean?
https://youtu.be/i6EDUQSWWgo
WHAT IS SOLUBILITY?
Solubility is the degree to which a
substance dissolves in a solvent to
make a solution (usually expressed
as grams of solute per liter of
solvent).
SOLUBILITY GRAPH OF SALTS IN WATER
SOLUTION AND SOLUBILITIES 8
Answer these questions from
the previous slide…
1. What is the solubility of NH4Cl at 20 C? _________
2. Ce2(SO4)3 is the only gas on the chart. What can you
gather from the curve of this substance as T increases. This
is actually true for all gases. __________________________
3. Which substance has the least change in solubility as the
temperature increases? __________________________
4. At 60 C, which substance has a solubility of
approximately 40 g/100 mL of water? ________________
Answers on next slide…
Solutions
1. 20 g/ 100 mL
2. As the temperature increases,
the solubility of GASES
decreases.
3. NaCl has the smallest slope.
4. CuSO4
How did you do????
Things we Need to Know About
SOLUBILITY The amount of solute per unit solvent required to
form a saturated solution is called the solute's Solubility.
When two liquids are completely soluble in each other they are said to be Miscible.
Solubility is effected by Temperature. With increase in temperature solubility of most of the substances increases.
Most gases become less soluble in water as the temperature increases. See the graph on the next slide.
SOLUBILITY GRAPH OF GASES IN WATER
Pressure has little effect on the solubility of liquids and
solids. The solubility of gases is strongly influenced by
pressure. Gases dissolve more at high pressure.
Concentration
In chemistry, a solution’s concentration is how much
of a dissolvable substance, known as a solute, is
mixed with another substance, called the solvent.
There are many ways to do this and the following
slides will go over each method and include
examples which will help you with your homework
problems. We will learn 6 of the ways to do this!
1. Percent Composition by Mass
Mass solute (g) x 100
Mass Solution (g)
Example:
Determine the percent composition by mass of a
100 g salt solution which contains 20 g salt.
Solution:
20 g NaCl / 100 g solution x 100 = 20% NaCl solution
X 100
2. Volume Percent (% v/v) Volume
Volume solute
Volume solutionX 100
Example:
What is the percent of rubbing alcohol bought
in the store that is prepared by taking 700 mL
of isopropyl alcohol and adding sufficient
water to obtain 1000 mL of solution.
Solution:
700 mL
1000 mLX 100 = 70%
3. Parts per Hundred, per Thousand,
per Million (can be in any unit V/V, m/m, m/V)All of these are simply exactly what you do to calculate your grade. They are:
little number (solute)
big number (solution)
Part per Hundred is really just “percent”! The others are done the same way but
parts per thousand (ppt) and parts per million (ppm) are just multiplied by those
numbers!
Example: If 5.4 g of solute is dissolved in 1300 grams of solvent,
calculate the concentration in ppm.
Solution: 5.4 g solute
(1300 g solvent + 5.4 g solute)
***remember, the solution is made of the solute and the solvent!
= 4137 ppm = 4100 ppm (with sig figs)
X 100 or 1,000 or 1,000,000
X 1,000,000
4. Mole Fraction (X)Mol solute 1
Total moles of solutes and solvents
***Keep in mind, the sum of all mole fractions in a
solution always equals 1.
Example: What are the mole fractions of the components of the solution
formed when 92 g glycerol is mixed with 90 g water? (molecular weight
water = 18; molecular weight of glycerol = 92)
Solution:
90 g water = 90 g x (1 mol / 18 g) = 5 mol water
92 g glycerol = 92 g x (1 mol / 92 g) = 1 mol glycerol
total mol = 5 + 1 = 6 mol
xwater = 5 mol / 6 mol = 0.833
x glycerol = 1 mol / 6 mol = 0.167
X 100
5. Molarity (M) Molarity is probably the
most commonly used unit of concentration.
Example: What is the molarity of a solution made
when water is added to 11 g CaCl2 to make 100 mL
of solution? (The molecular weight of CaCl2 = 110)
Moles solute
L solution
Solution:
1. 11 g CaCl2 x (1 mol CaCl2/110 g CaCl2) = 0.10 mol CaCl22. 100 mL x (1 L / 1000 mL) = 0.10 L
3. molarity = 0.10 mol / 0.10 L = 1.0 M
6. Molality (m) Molality is the number of
moles of solute per kilogram of solvent.
m = moles solute
Kg solvent
Example:What is the molality of a solution of 10 g
NaOH in 500 g water? (Molecular weight of NaOH is 40)
Solution:
1. 10 g NaOH / (40 g NaOH / 1 mol NaOH) = 0.25 mol
NaOH
2. 500 g water x (1 kg/1000 g) = 0.50 kg water
3. molality = 0.25 mol / 0.50 kg = 0.05 mol/kg = 0.50 m
Summin’ It Up…a
You must make notecards of all of these
formulas. The bulk of this chapter deals with
these units of concentration and many of your
homework problems have you calculate them.
So be sure to write them all down and learn
them. They are simply plug-n-chug so if you
know the formulas you will have no issues!!!a
HINT: KNOW THE DIFFERENCE
BETWEEN SOLUTE, SOLVENT AND
SOLUTION!!!Now let’s practice a few of them from your HW packet
HW 12.2 (Note: some of the answers for 12.2 in the packet are incorrect
so ignore them!)
1. 5.0 g = 4.8 pph or %(100 g + 5.0 g)
4. (done on slide #16)
5. 2.8 % = X g Solve for x = 1.4 g
50 mL
9a. M = moles solute …first convert g to moles
1 L solution
(54 g NaOH) (1mole NaOH) = 1.35 moles NaOH1 40 g NaOH
M = 1.35 moles NaOH = 1.35 M1.0 L soln.
X 100
X 100
Continue to work through these problems using
the formulas given from the slides. We will go
over them in our Monday Zoom session or you
can join tomorrow’s office hours if you need
help with a question.
The rest of this powerpoint will be uploaded
as soon as I finish it. I wanted to get you
what I had completed. Sorry it is taking
so long. It takes me FOREVER to
write using superscripts and
subscripts!