Download Lesson 17 - Chapter 8_2

Chapter 8
Solutions
Concentration
8.4
Percent Concentration
The concentration of a solution
§  Is the amount of solute dissolved in a specific amount
of solution.
amount of solute
amount of solution
1
Mass Percent Concentration
2
Mass of Solution
Mass percent (%m/m) concentration is the
•  Percent by mass of solute in a solution.
mass percent (%m/m)
=
g of solute
x 100
8.00 g KCl
Add water to
give 50.00 g
solution
g of solute + g of solvent
•  Amount in g of solute in 100 g of solution.
mass percent =
g of solute
50.00 g KCl
solution
x 100
100 g of solution
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Calculating Mass Percent
Learning Check
The calculation of mass percent (%m/m) requires the
§  Grams of solute (g KCl) and
§  Grams of solution (g KCl solution).
g of KCl
=
8.00 g
g of solvent (water)
=
42.00 g
g of KCl solution
=
50.00 g
A solution is prepared by mixing 15.0 g Na2CO3 and
235 g of H2O. Calculate the mass percent (%m/m) of
the solution.
8.00 g KCl (solute)
50.00 g KCl solution
1) 15.0% (m/m) Na2CO3
2) 6.38% (m/m) Na2CO3
3) 6.00% (m/m) Na2CO3
x 100 = 16.0% (m/m)
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Solution
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Volume Percent
3) 6.00% (m/m) Na2CO3
The volume percent (%v/v) is
STEP 1 mass solute
§  Percent volume (mL) of solute (liquid) to volume (mL)
of solution.
= 15.0 g Na2CO3
mass solution = 15.0 g + 235 g = 250. g
§  Volume % (v/v) =
STEP 2 Use g solute/ g solution ratio
STEP 3 mass %(m/m) = g solute x 100
g solution
STEP 4 Set up problem
mass %(m/m) = 15.0 g Na2CO3 x 100 = 6.00% Na2CO3
250. g solution
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mL of solute x 100
mL of solution
§  Solute (mL) in 100 mL of solution.
volume % (v/v) = mL of solute
100 mL of solution
8
2
Mass/Volume Percent
Percent Conversion Factors
§  Two conversion factors can be written for each
type of % value.
The mass/volume percent (%m/v) is
§  Percent mass (g) of solute to volume (mL) of solution.
§  mass/volume % (m/v) =
g of solute x 100
mL of solution
TABLE 8.9
§  Solute (g) in 100 mL of solution.
mass/volume % (m/v) =
g of solute
100 mL of solution
5% (m/v) glucose
There are 5 g of glucose
in 100 mL of solution.
5 g glucose
100 mL solution
and
100 mL solution
5 g glucose
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Learning Check
Solution
Write two conversion factors for each solution:
A. 8.50 g NaOH
A. 8.50%(m/m) NaOH
and
100 g solution
B. 5.75 mL alcohol and
B. 5.75%(v/v) ethanol
100 mL solution
C. 4.8 g HCl
C. 4.8 %(m/v) HCl
100 mL solution
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100 g solution
8.50 g NaOH
100 mL solution
5.75 mL alcohol
and
100 mL HCl
4.8 g HCl
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3
Using Percent Concentration (m/
m) Factors
Learning Check
How many grams of NaCl are needed to prepare
225 g of a 10.0% (m/m) NaCl solution?
STEP 1 Given: 225 g solution; 10.0% (m/m) NaCl
Need: g of NaCl
STEP 2 g solution
g NaCl
STEP 3 Write the 10.0 %(m/m) as conversion factors.
10.0 g NaCl
and
100 g solution
100 g solution
10.0 g NaCl
STEP 4 Set up to cancel g solution.
225 g solution x 10.0 g NaCl
= 22.5 g NaCl
100 g solution
How many grams of NaOH are needed to prepare
75.0 g of 14.0%(m/m) NaOH solution?
1) 10.5 g NaOH
2) 75.0 g NaOH
3) 536 g NaOH
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Solution
Learning Check
1) 10.5 g NaOH
How many milliliters of a 5.75 % (v/v) ethanol
solution can be prepared from 2.25 mL
ethanol?
75.0 g solution x 14.0 g NaOH = 10.5 g NaOH
100 g solution
14.0 % (m/m) factor
1) 2.56 mL
2) 12.9 mL
3) 39.1 mL
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4
Using Percent Concentration(m/v)
Factors
Solution
How many mL of a 4.20%(m/v) will contain 3.15 g KCl?
3) 39.1 mL
STEP 1 Given: 3.15 g KCl(solute); 4.20% (m/v) KCl
Need: mL of KCl
solution
STEP 2 Plan: g KCl
mL KCl solution
STEP 3 Write conversion factors.
4.20 g KCl
and 100 mL solution
100 mL solution
4.20 g KCl
STEP 4 Set up the problem
3.15 g KCl x 100 mL KCl solution = 75.0 mL KCl
4.20 g KCl
2.25 mL ethanol x 100 mL solution
5.75 mL ethanol
5.75 %(v/v) inverted
= 39.1 mL solution
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Learning Check
Solution
How many grams of NaOH are needed to prepare 125
mL of a 8.80%(m/v) NaOH solution?
How many grams of NaOH are needed to prepare 125
mL of a 8.80%(m/v) NaOH solution?
125 mL solution x 8.80 g NaOH =
100 mL solution
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11.0 g NaOH
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Chapter 8
Solutions
Molarity (M)
8.5
Molarity and Dilution
Molarity (M)
§  Is a concentration term for solutions.
§  Gives the moles of solute in 1 L solution.
§  = moles of solute
liter of solution
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Preparing a 1.0 Molar Solution
Calculation of Molarity
A 1.00 M NaCl solution is prepared
§  By weighing out 58.5 g NaCl (1.00 mole) and
§  Adding water to make 1.00 liter of solution.
What is the molarity of 0.500 L NaOH solution if it
contains 6.00 g NaOH?
STEP 1 Given 6.00 g NaOH in 0.500 L solution
Need molarity (mole/L)
STEP 2 Plan g NaOH
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mole NaOH
molarity
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Calculation of Molarity (cont.)
Learning Check
What is the molarity of 325 mL of a solution containing
46.8 g of NaHCO3?
STEP 3 Conversion factors 1 mole NaOH = 40.0 g
1 mole NaOH
and 40.0 g NaOH
40.0 g NaOH
1 mole NaOH
STEP 4 Calculate molarity.
6.00 g NaOH x 1 mole NaOH = 0.150 mole
40.0 g NaOH
0.150 mole = 0.300 mole = 0.300 M NaOH
0.500 L
1L
1)
0.557 M
2)
1.44 M
3)
1.71 M
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Solution
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Learning Check
3) 1.71 M
46.8 g NaHCO3 x 1 mole NaHCO3 = 0.557 mole NaHCO3
84.0 g NaHCO3
What is the molarity of 225 mL of a KNO3 solution
containing 34.8 g KNO3?
1) 0.344 M
2) 1.53 M
3) 15.5 M
0.557 mole NaHCO3 = 1.71 M NaHCO3
0.325 L
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Solution
Molarity Conversion Factors
2) 1.53 M
34.8 g KNO3 x 1 mole KNO3
101.1 g KNO3
The units of molarity are used to write conversion
factors for calculations with solutions.
= 0.344 mole KNO3
TABLE 8.10
M = mole =
0.344 mole KNO3 = 1.53 M
L
0.225 L
In one setup
34.8 g KNO3 x 1 mole KNO3 x
1
= 1.53 M
101.1 g KNO3 0.225 L
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Calculations Using Molarity
Calculations Using Molarity
How many grams of KCl are needed to prepare 125 mL
of a 0.720 M KCl solution?
STEP 3 Conversion factors
1 mole KCl = 74.6 g
1 mole KCl
and 74.6 g KCl
74.6 g KCl
1 mole KCl
STEP 1 Given 125 mL (0.125 L) of 0.720 M KCl
Need Grams of KCl
STEP 2 Plan
L KCl
moles KCl
1 L KCl = 0.720 mole KCl
1L
and 0.720 mole KCl
0.720 mole KCl
1L
g KCl
STEP 4 Calculate g KCl
0.125 L x 0.720 mole KCl x 74.6 g KCl
1L
1 mole KCl
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= 6.71 g KCl
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Learning Check
Solution
3)
How many grams of AlCl3 are needed to prepare
125 mL of a 0.150 M solution?
1)
20.0 g AlCl3
2)
16.7g AlCl3
3)
2.50 g AlCl3
2.50 g AlCl3
0.125 L x 0.150 mole x 133.5 g = 2.50 g AlCl3
1L
1 mole
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Learning Check
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Solution
How many milliliters of 2.00 M HNO3 contain
24.0 g HNO3?
24.0 g HNO3 x 1 mole HNO3 x
63.0 g HNO3
1000 mL
=
2.00 moles HNO3
Molarity factor inverted
1) 12.0 mL
2) 83.3 mL
3) 190. mL
= 190. mL HNO3
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Dilution
Initial and Diluted Solutions
In a dilution,
In the initial and diluted solution,
§  The moles of solute are the same.
§  The concentrations and volumes are related
by the following equations:
For percent concentration
C1V1 = C2V2
§  Water is added.
§  Volume increases.
§  Concentration decreases.
initial
diluted
For molarity
M1V1 = M2V2
initial
diluted
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Dilution Calculations with Percent
Learning Check
What volume of a 2.00 %(m/v) HCl solution can be
What is the percent (%m/v) of a solution prepared
prepared by diluting 25.0 mL of 14.0%(m/v) HCl solution?
by diluting 10.0 mL of 9.00% NaOH to 60.0 mL?
Prepare a table:
C1= 14.0 %(m/v)
V1 = 25.0 mL
C2= 2.00%(m/v)
V2 = ?
Solve dilution equation for unknown and enter values:
C1V1 = C2V2
V2
= V1C1
C2
= (25.0 mL)(14.0%) = 175 mL
2.00%
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Solution
Dilution Calculations
What is the percent (%m/v) of a solution prepared
by diluting 10.0 mL of 9.00% NaOH to 60.0 mL?
Prepare a table:
C1= 9.00 %(m/v)
V1 = 10.0 mL
C 2= ?
V2 = 60.0 mL
Solve dilution equation for unknown and enter values:
C1V1 = C2V2
What is the molarity (M) of a solution prepared
by diluting 0.180L of 0.600 M HNO3 to 0.540 L?
Prepare a table:
M1= 0.600 M
V1 = 0.180 L
M2 = ?
V2 = 0.540 L
Solve dilution equation for unknown and enter values:
M1V1 = M2V2
C2
= C1 V1 = (10.0 mL)(9.00%) = 1.50 %(m/v)
V2
60.0 mL
M2
= M1V1
V2
= (0.600 M)(0.180 L) = 0.200 M
0.540 L
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Learning Check
Solution
What is the final volume (mL) of 15.0 mL of a 1.80 M
What is the final volume (mL) of 15.0 mL of a 1.80 M
KOH diluted to give a 0.300 M solution?
Prepare a table:
M1= 1.80 M
V1 = 15.0 mL
M2= 0.300M
V2 = ?
Solve dilution equation for V2 and enter values:
M1V1 = M2V2
KOH diluted to give a 0.300 M solution?
1) 27.0 mL
2) 60.0 mL
3) 90.0 mL
V2
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= M1V1
M2
= (1.80 M)(15.0 mL) = 90.0 mL
0.300 M
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Molarity in Chemical Reactions
Using Molarity of Reactants
In a chemical reaction,
§  The volume and molarity of a solution are used to
determine the moles of a reactant or product.
How many mL of 3.00 M HCl are needed to completely
react with 4.85 g CaCO3?
2HCl(aq) + CaCO3(s)
molarity ( mole ) x volume (L) = moles
1L
§  If molarity (mole/L) and moles are given, the volume
(L) can be determined
moles x
1L
=
volume (L)
moles
CaCl2(aq) + CO2(g) + H2O(l)
STEP 1 Given 3.00 M HCl; 4.85 g CaCO3
Need volume in mL
STEP 2 Plan
g CaCO3
mole CaCO3
mole HCl
mL HCl
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Using Molarity of Reactants
(cont.)
2HCl(aq) + CaCO3(s)
Learning Check
CaCl2(aq) + CO2(g) + H2O(l)
STEP 3 Equalitites
How many mL of a 0.150 M Na2S solution are needed to
completely react 18.5 mL of 0.225 M NiCl2 solution?
1 mole CaCO3 = 100.1 g; 1 mole CaCO3 = 2 mole HCl
1000 mL HCl = 3.00 mole HCl
STEP 4 Set Up
4.85 g CaCO3 x 1 mole CaCO3 x 2 mole HCl x 1000 mL HCl
100.1 g CaCO3
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1 mole CaCO3 3.00 mole HCl
NiCl2(aq) + Na2S(aq)
NiS(s) + 2NaCl(aq)
1) 4.16 mL
2) 6.24 mL
3) 27.8 mL
= 32.3 mL HCl required
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Solution
Learning Check
If 22.8 mL of 0.100 M MgCl2 is needed to completely
react 15.0 mL of AgNO3 solution, what is the molarity of
the AgNO3 solution?
3) 27.8 mL
0.0185 L x 0.225 mole NiCl2 x 1 mole Na2S x 1000 mL
1L
1 mole NiCl2 0.150 mole Na2S
MgCl2(aq) + 2AgNO3(aq)
= 27.8 mL Na2S solution
2AgCl(s) + Mg(NO3)2(aq)
1) 0.0760 M
2) 0.152 M
3) 0.304 M
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Solution
Learning Check
3) 0.304 M AgNO3
How many liters of H2 gas at STP are produced
when Zn react with 125 mL of 6.00 M HCl?
Zn(s) + 2HCl(aq)
ZnCl2 (aq) + H2(g)
0.0228 L x 0.100 mole MgCl2 x 2 moles AgNO3 x
1L
1 mole MgCl2
1
0.0150 L
1) 4.20 L H2
2) 8.40 L H2
3) 16.8 L H2
= 0.304 mole/L = 0.304 M AgNO3
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Solution
Chapter 8
2) 8.40 L H2 gas
Solutions
8.6
Properties of Solutions
0.125 L x 6.00 moles HCl x 1 mole H2 x 22.4 L
1L
2 moles HCl 1 mole H2
= 8.40 L H2 gas
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Solutions
Colloids
Solutions
Colloids
§  Contain small particles (ions or molecules).
§  Have medium size particles.
§  Are transparent.
§  Cannot be filtered.
§  Do not separate.
§  Can be separated by semipermeable membranes.
§  Cannot be filtered.
§  Scatter light (Tyndall effect).
§  Do not scatter light.
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Examples of Colloids
Suspensions
Suspensions
TABLE 8.11
§  Have very large particles.
§  Settle out.
§  Can be filtered.
§  Must be stirred to stay
suspended.
§  Examples include blood
platelets, muddy water,
and Calamine lotion.
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Solutions, Colloids, and
Suspensions
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Summary
TABLE 8.12
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Learning Check
Solution
A mixture that has solute particles that do not settle out,
but are too large to pass through a semipermeable
membrane is called a
1) solution
2) colloid
3) suspension
A mixture that has solute particles that do not settle out,
but are too large to pass through a semipermeable
membrane is called a
2) colloid
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Osmosis
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Osmosis
In osmosis, water (solvent)
flows from the lower solute
concentration into the
A semipermeable membrane separates a 4%
starch solution from a 10% starch solution. Starch is a
colloid and cannot pass through the membrane, but
water can. What happens?
higher solute concentration.
The level of the solution with
the higher concentration
rises.
The concentrations of the two
4% starch
10% starch
H 2O
solutions become equal with
time.
semipermeable membrane
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Water flow Equalizes
Osmotic Pressure
§  The 10% starch solution is diluted by the flow of
water out of the 4% and its volume increases.
Osmotic pressure is
§  The 4% solution loses water and its volume
decreases.
§  Produced by the solute particles dissolved in a
§  Eventually, the water flow between the two becomes
equal.
§  Equal to the pressure that would prevent the flow of
solution.
additional water into the more concentrated solution.
§  Greater as the number of dissolved particles in the
solution increases.
7% starch
7% starch
H 2O
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Learning Check
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Solution
A semipermeable membrane separates a 10% starch
solution (A) from a 5% starch solution (B). If starch is a
colloid, fill in the blanks in the statements below.
1. Solution ____ has the greater osmotic pressure.
A semipermeable membrane separates a 10% starch
solution (A) from a 5% starch solution (B). If starch is a
colloid, fill in the blanks in the statements below.
1. Solution A has the greater osmotic pressure.
2. Water initially flows from ___ into ___.
2. Water initially flows from B into A.
3. The level of solution ____will be lower.
3. The level of solution B will be lower.
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Osmotic Pressure of the Blood
Isotonic Solutions
Red blood cells
An isotonic solution
§  Exerts the same osmotic
pressure as red blood
cells.
§  Is known as a
physiological solution .
§  Of 5.0% glucose or 0.90%
NaCl is used medically
because each has a
solute concentration equal
to the osmotic pressure
equal to red blood cells.
§  Have cell walls that are semipermeable
membranes.
§  Maintain an osmotic pressure that cannot
change or damage occurs.
§  Must maintain an equal flow of water between
the red blood cell and its surrounding
environment.
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Hypotonic Solutions
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Hypertonic Solutions
A hypotonic solution
§  Has a lower osmotic
pressure than red blood
cells.
§  Has a lower concentration
than physiological
solutions.
§  Causes water to flow into
red blood cells.
§  Causes hemolysis: RBCs
swell and may burst.
A hypertonic solution
§  Has a higher osmotic
pressure than RBCs.
§  Has a higher
concentration than
physiological solutions.
§  Causes water to flow out
of RBCs.
§  Causes crenation: RBCs
shrinks in size.
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Dialysis
Learning Check
In dialysis,
Indicate if each of the following solutions is
1) isotonic 2) hypotonic 3) hypertonic
§  Solvent and small solute
particles pass through an
artificial membrane.
A.____ 2% NaCl solution
B.____ 1% glucose solution
C.____ 0.5% NaCl solution
D.____ 5% glucose solution
§  Large particles are
retained inside.
§  Waste particles such as
urea from blood are
removed using
hemodialysis (artificial
kidney).
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Solution
Learning Check
Indicate if each of the following solutions is
When placed in each of the following, indicate if a red
1) isotonic
2) hypotonic
blood cell will
3) hypertonic
1) not change 2) hemolyze
A._3_ 2% NaCl solution
A.____ 5% glucose solution
B._2_ 1% glucose solution
B.____ 1% glucose solution
C._2_ 0.5% NaCl solution
C.____ 0.5% NaCl solution
D._1_ 5% glucose solution
D.____ 2% NaCl solution
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3) crenate
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Solution
Hemodialysis
When placed in each of the following, indicate if a red
§  When kidneys fail, an artificial kidney uses
hemodialysis to remove waste particles such as
urea from blood.
blood cell will
1) not change 2) hemolyze
3) crenate
A._1_ 5% glucose solution
B._2_ 1% glucose solution
C._2_ 0.5% NaCl solution
D._3_ 2% NaCl solution
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Learning Check
Solution
Each of the following mixtures is placed in a dialyzing
Each of the following mixtures is placed in a dialyzing
bag and immersed in pure water. Which substance, if
bag and immersed in pure water. Which substance, if
any, will be found in the water outside the bag?
any, will be found in the water outside the bag?
A. 10% KCl solution
A. 10% KCl solution
B. 5% starch solution
B. 5% starch solution None; starch is retained.
KCl ( K+, Cl−)
C. 5% NaCl and 5% starch solutions
C. 5% NaCl and 5% starch solutions
NaCl (Na+, Cl−), but starch is retained.
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