# Solubility by 6en78I

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```									Solubility
Friday, October 5

DMA: Why do we call water a “universal”
solvent?

Today: Solubility notes
C1-C2 vocabulary
well plate testing
Water-a Universal Solvent?

Everything is “soluble,” or will dissolve in
water to some extent given enough time.
As a class…

What happens when oil is added to water?

What happens when food coloring is

What happens when salt or sugar is added
to water?
Solubility
Soluble: one substance dissolves in
another

Insoluble: one substance does NOT
dissolve in another

“Like dissolves like”
Solubility Data

Copy data table into notebook.
Write Soluble or Insoluble in each box.
Dye#2 is a teacher demonstration.
Solubility Vocabulary
Solubility: how much of something will
dissolve in a substance
Solute: (see p.24)

Solvent: (see p.24)

Solution: _________ mixture of two or more
substances
Monday, October 8

DMA:
What does “soluble” mean?
Make sure you have the definitions for
Solubility (p. 39)
Solute (p. 24)
Solvent (p. 24)
 Solubility: how much of something will
dissolve in a substance
 Solute: the dissolved substance in a solution; the
component present in the smaller amount

 Solvent: component of a solution present in the
largest amount

 Solution: homogeneous mixture of two or more
substances
Solution Concentration

Concentration is crucial to all life:

--red/white blood cell counts in blood
--oxygen gas dissolved in air and water

Common way of expressing this:
g solute / g solution usually 100 or 1,000 g
solution
Ex. Problem: solution concentration

One teaspoon of sucrose (table sugar) ,
which has a mass of 10g, is dissolved in
240g of water.
What is the concentration of the solution,
expressed as grams sucrose per 100 g
solution? (% sucrose by mass, or pph)

What is the concentration in ppt?
pph= parts per hundred, ppt = per thousand
Example solution concentration

10 g sugar +240 g water = 250 g solution

10 g sugar/250g solution = 0.04 or 4%

To get ppt, multiply the % by 10
and divide by 1000

40 /1000 or 40 ppt
this is the same ratio as the original solution

Read C1 and C2 individually (p.39-43)
(skip “Your Turn” sections for now)

We will go over the bold vocabulary words
together and fill in the definitions in 8 min.
Solubility Vocabulary #2
Saturated: the solvent has dissolved as
much solute as possible at a certain
temperature

Unsaturated: (book)
Supersaturated: (book)
Solubility Curve: plotted line of the amount
of solute that will dissolve vs. temperature

Solution Concentration: (book)
 Saturated: the solvent has dissolved as much
solute as possible at a certain temperature
 Unsaturated: contains a lower concentration of
solute than is possible at a certain temperature
 Supersaturated: contains a higher concentration
of solute than a saturated solution at a temp.

 Solubility Curve: compares the mass of solute
that will dissolve in solute at many temperatures
 Solution Concentration: quantity of solute
dissolved in a specific quantity of solvent
C1 and C2 Your Turn problems

Your Turn p. 42: Group works on #1-4.

Your Turn p.44: Group works on # 2, 4.

Do #1=6 on worksheet.
HOMEWORK if not finished in class.
Tuesday, October 9

DMA:
Fill in the blanks:
5 = ______ = _______
100       1,000    1,000,000

Same numbers go in these blanks:
5% or pph = _____ ppt = ________ ppm
Fill in the blanks:

5 = 50 =          50,000
100   1,000       1,000,000
Check for Understanding

You need to make a solution of fertilizer and
water that has a 5 % concentration.

1. What is the solvent?
2. What is the solute?
3. To make 1,000 mL (1 Liter) of solution,
How many grams of fertilizer should you use?
How much water?

 1a. 80 grams
 1b. 42 grams
 2a. 42 g- 25 g = 17 g
 2b. 42/100 = 25/x so x = 58.1 g
 3a. 7 mg
 3b. 9 mg
 4. 0.9 mg because it is only100 g water vs. 1000

1a. solute = sugar (the smaller amount)
1b. solvent = water (the larger amount)
2a. 17 g sugar/200 g solution
= 0.085 or 8.5%
2b. 34 g sugar/400 g solution =0.085 or
8.5%
4a. the dissolved oxygen (DO) concentration
will be less in the warmer water b. because
less oxygen dissolves in warmer water
Solution concentration problems

#7-12 on the worksheet can be solved with
algebra using cross-multiplication.
1. Find the saturated line on the graph at
that temperature.
2. Write this number divided by 100
3. Set this equal to x and the given amount.
For example, #7:
46g KCl/100 g water = 40 g KCl/ x g water
x = 86.95 g water
Wednesday, October 10

DMA:
Looking at the 2 graphs on p. 40 and 41,
explain how to change the temperature if:
a. You want more solid to dissolve in water.
b. You want more gas to dissolve in water.
DMA

To get more solid to dissolve in water, I
would raise the temperature.

To get more gas to dissolve in water, I
would lower the temperature.

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