Transcript Chapter 7 Solutions

CHAPTER 7
SOLUTIONS
7.5
Molarity and Dilution
1
MOLARITY (M)
Molarity (M)
• is a concentration term for solutions.
• gives the moles of solute in 1 L of solution.
• moles of solute
liter of solution
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PREPARING A 1.0 MOLAR SOLUTION
A 1.00 M NaCl solution is
prepared
• by weighing out 58.5 g of NaCl
(1.00 mole) and
• adding water to make 1.00 liter
of solution.
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CALCULATION OF MOLARITY
What is the molarity of 0.500 L of NaOH solution if it
contains 6.00 g of NaOH?
STEP 1: Given 6.00 g of NaOH in 0.500 L of solution
Need molarity (mole/L)
STEP 2: Plan g NaOH
mole NaOH
molarity
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EXAMPLE
What is the molarity of 325 mL of a solution
containing 46.8 g of NaHCO3?
1)
0.557 M
2)
1.44 M
3)
1.71 M
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EXAMPLE
What is the molarity of 225 mL of a KNO3
solution containing 34.8 g of KNO3?
1)
2)
3)
0.344 M
1.53 M
15.5 M
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MOLARITY CONVERSION FACTORS
The units of molarity are used as conversion factors
in calculations with solutions.
Molarity
3.5 M HCl
Equality
1 L = 3.5 moles of HCl
Written as Conversion Factors
3.5 moles HCl
and
1L
1L
3.5 moles HCl
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CALCULATIONS USING MOLARITY
How many grams of KCl are needed to prepare 125
mL
of a 0.720 M KCl solution?
STEP 1: Given 125 mL (0.125 L) of 0.720 M KCl
Need g of KCl
STEP 2: Plan L KCl
moles KCl
g KCl
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EXAMPLE
How many grams of AlCl3 are needed to prepare
125 mL of a 0.150 M solution?
1)
20.0 g of AlCl3
2)
16.7 g of AlCl3
3)
2.50 g of AlCl3
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EXAMPLE
How many milliliters of 2.00 M HNO3 contain 24.0 g of
HNO3?
1) 12.0 mL
2) 83.3 mL
3) 190. mL
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DILUTION
In a dilution
• water is added.
• volume increases.
• concentration decreases.
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COMPARING INITIAL AND DILUTED
SOLUTIONS
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
initial
diluted
For molarity:
M1V1 = M2V2
initial
diluted
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DILUTION CALCULATIONS WITH
PERCENT
What volume of a 2.00% (m/v) HCl solution can be
prepared by diluting 25.0 mL of 14.0% (m/v) HCl solution?
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
=
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EXAMPLE
What is the percent (% m/v) of a solution prepared
by diluting 10.0 mL of 9.00% NaOH to 60.0 mL?
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DILUTION CALCULATIONS WITH
MOLARITY
What is the molarity (M) of a solution prepared
by diluting 0.180 L of 0.600 M HNO3 to 0.540 L?
Prepare a table:
M1= 0.600 M
V1 = 0.180 L
M 2= ?
V2 = 0.540 L
Solve dilution equation for unknown and enter values:
M 1V 1 = M 2V 2
M2
=
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EXAMPLE
What is the final volume (mL) of 15.0 mL of a 1.80 M
KOH diluted to give a 0.300 M solution?
1) 27.0 mL
2) 60.0 mL
3) 90.0 mL
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7.6 MOLARITY IN CHEMICAL
REACTIONS
In a chemical reaction,
• the volume and molarity of a solution are used to
determine the moles of a reactant or product.
molarity ( mole ) x volume (L) = moles
1L
• if molarity (mole/L) and moles are given, the
volume (L) can be determined.
moles x
1L
moles
=
volume (L)
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USING MOLARITY OF REACTANTS
How many mL of 3.00 M HCl are needed to completely
react with 4.85 g of CaCO3?
2 HCl(aq) + CaCO3(s)
CaCl2(aq) + CO2(g) + H2O(l)
STEP 1: Given 3.00 M HCl; 4.85 g of CaCO3
Need
volume in mL
STEP 2: Plan
g CaCO3
mole CaCO3
mole HCl
mL HCl
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EXAMPLE
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?
MgCl2(aq) + 2AgNO3(aq)
2AgCl(s) + Mg(NO3)2(aq)
1) 0.0760 M
2) 0.152 M
3) 0.304 M
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EXAMPLE
How many liters of H2 gas at STP are produced
when Zn reacts with 125 mL of 6.00 M HCl?
Zn(s) + 2HCl(aq)
ZnCl2 (aq) + H2(g)
1) 4.20 L of H2
2) 8.40 L of H2
3) 16.8 L of H2
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SOL7.7 PROPERTIES OF SOLUTIONS
UTIONS
Solutions
• contain small particles (ions or molecules).
• are transparent.
• do not separate.
• cannot be filtered.
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C7.7 PROPERTIES OF SOLUTIONS
OLLOIDS
Colloids
• have medium-size particles.
• cannot be filtered.
• can be separated by semipermeable membranes.
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EXAMPLES OF COLLOIDS
Examples of colloids include
• fog
• whipped cream
• milk
• cheese
• blood plasma
• pearls
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7.7 PROPERTIES OF SOLUTIONS
SUSPENSIONS
Suspensions
• 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
Copyright © 2009 by Pearson Education, Inc.
25
EXAMPLENING CHECK
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.
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OSMOSIS
In osmosis,
• water (solvent) flows from
the lower solute
concentration into the
higher solute
concentration.
• the level of the solution
with the higher solute
concentration rises.
• the concentrations of the
two solutions become
equal with time.
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OSMOSIS
Suppose 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?
semipermeable
membrane
4% starch
10% starch
H2 O
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WATER FLOW EQUALIZES
• The 10% starch solution is diluted by the flow of water
out of the 4% and its volume increases.
• The 4% solution loses water and its volume decreases.
• Eventually, the water flow between the two becomes
equal.
7% starch
7% starch H O
2
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OSMOTIC PRESSURE
Osmotic pressure is
• produced by the solute particles dissolved in a
solution.
• equal to the pressure that would prevent the flow of
additional water into the more concentrated solution.
• greater as the number of dissolved particles in the
solution increases.
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EXAMPLECHECK
A semipermeable membrane separates a 10% sucrose
solution A from a 5% sucrose solution B. If sucrose is a
colloid, fill in the blanks in the statements below.
1. Solution ____ has the greater osmotic pressure.
2. Water initially flows from ___ into ___.
3. The level of solution ____will be lower.
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OSMOTIC PRESSURE OF THE
BLOOD
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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ISOTONIC SOLUTIONS
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.
H2O
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HYPOTONIC 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.
H2O
• causes hemolysis: RBCs
swell and may burst.
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HYPERTONIC SOLUTIONS
A hypertonic solution
• has a higher osmotic
pressure than RBCs.
• has a higher
concentration than
physiological solutions.
• causes water to flow out
of RBCs.
• cause crenation: RBCs
shrink in size.
H2O
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DIALYSIS
In dialysis,
• solvent and small solute particles pass through an
artificial membrane.
• large particles are retained inside.
• waste particles such as urea from blood are removed
using hemodialysis (artificial kidney).
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EXAMPLES
Indicate if each of the following solutions is
1) isotonic, 2) hypotonic, or 3) hypertonic.
A.____ 2% NaCl solution
B.____ 1% glucose solution
C.____ 0.5% NaCl solution
D.____ 5% glucose solution
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EXAMPLES
When placed in each of the following, indicate if a red
blood cell will
1) not change, 2) hemolyze, or 3) crenate.
A.____ 5% glucose solution
B.____ 1% glucose solution
C.____ 0.5% NaCl solution
D.____ 2% NaCl solution
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EXAMPLES
Each of the following mixtures is placed in a
dialyzing bag and immersed in pure water. Which
substance, if any, will be found in the water outside
the bag?
A. 10% KCl solution
B. 5% starch solution
C. 5% NaCl and 5% starch solutions
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