honors chemistry: chapter 12 solutions

of 22 /22
SOLUTIONS (Made of solute and solvent) Honors Chemistry: Chapter 12

Author: others

Post on 16-Oct-2021




0 download

Embed Size (px)


SOLUTIONS & SOLUBILITIESHonors Chemistry: Chapter 12
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
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.
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.
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.
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!
2. How do ionic cpds “dissociate” in solvents?
3. How do molecular (covalent) cpds dissolve in
substance dissolves in a solvent to
make a solution (usually expressed
as grams of solute per liter of
Answer these questions from
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…
the solubility of GASES
4. CuSO4
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.
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.
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)
100 g salt solution which contains 20 g salt.
20 g NaCl / 100 g solution x 100 = 20% NaCl solution
X 100
Volume solute
in the store that is prepared by taking 700 mL
of isopropyl alcohol and adding sufficient
water to obtain 1000 mL of solution.
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
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
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)
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
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
1. 11 g CaCl2 x (1 mol CaCl2/110 g CaCl2) = 0.10 mol CaCl2 2. 100 mL x (1 L / 1000 mL) = 0.10 L
3. molarity = 0.10 mol / 0.10 L = 1.0 M
moles of solute per kilogram of solvent.
m = moles solute
Example:What is the molality of a solution of 10 g
NaOH in 500 g water? (Molecular weight of NaOH is 40)
1. 10 g NaOH / (40 g NaOH / 1 mol NaOH) = 0.25 mol
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
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
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
1 L solution
(54 g NaOH) (1mole NaOH) = 1.35 moles NaOH 1 40 g NaOH
M = 1.35 moles NaOH = 1.35 M 1.0 L soln.
X 100
X 100
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