Transcript Chapter 7 Solutions

Chapter 7
Solutions
7.1
Solutions
Copyright © 2009 by Pearson Education, Inc.
1
Solute and Solvent
Solutions
• are homogeneous
mixtures of two or
more substances.
• consist of a solvent
and one or more
solutes.
2
Nature of Solutes in Solutions
Solutes
• spread evenly
throughout the
solution.
• cannot be separated
by filtration.
• can be separated by
evaporation.
• are not visible, but
can give a color to
the solution.
Copyright © 2009 by Pearson Education, Inc.
3
Examples of Solutions
The solute and solvent in a solution can be a
solid, liquid, and/or a gas.
Copyright © 2009 by Pearson Education, Inc.
4
Learning Check
Identify the solute in each of the following
solutions.
A. 2 g of sugar and 100 mL of water
B. 60.0 mL of ethyl alcohol and 30.0 mL of
methyl alcohol
C. 55.0 mL of water and 1.50 g of NaCl
D. Air: 200 mL of O2 and 800 mL of N2
5
Water
Water
• is the most
common solvent.
• is a polar molecule.
• forms hydrogen
bonds between the
hydrogen atom in
one molecule and
the oxygen atom in
a different water
molecule.
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6
Formation of a Solution
Na+ and Cl- ions
• on the surface of a
NaCl crystal are
attracted to polar
water molecules.
• are hydrated in
solution with many
H2O molecules
surrounding each ion.
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7
Equations for Solution Formation
When NaCl(s) dissolves in water, the
reaction can be written as
NaCl(s)
(aq)
solid
H2O
Na+(aq) + Clseparation of ions
8
Learning Check
Solid LiCl is added to water. It dissolves
because
A. the Li+ ions are attracted to the
1) oxygen atom ( -) of water.
2) hydrogen atom (+) of water.
B.
the Cl- ions are attracted to the
1) oxygen atom ( -) of water.
2) hydrogen atom (+) of water.
9
Like Dissolves Like
Two substances form a solution
• when there is an attraction between the
particles of the solute and solvent.
• when a polar solvent such as water dissolves
polar solutes such as sugar, and ionic solutes
such as NaCl.
• when a nonpolar solvent such as hexane
(C6H14) dissolves nonpolar solutes such as oil
or grease.
10
Water and a Polar Solute
11
Like Dissolves Like
Solvents
Solutes
Water (polar)
Ni(NO3)2
(polar)
CH2Cl2(nonpolar)
I2 (nonpolar)
12
Learning Check
Which of the following solutes will
dissolve in water? Why?
1) Na2SO4
2) gasoline (nonpolar)
3) I2
4) HCl
13
7.2
Electrolytes and Nonelectrolytes
14
Solutes and Ionic Charge
In water,
• strong electrolytes produce ions and
conduct an electric current.
• weak electrolytes produce a few ions.
• nonelectrolytes do not produce ions.
15
Strong Electrolytes
Strong electrolytes
• dissociate in water, producing positive and negative ions.
• conduct an electric current in water.
• in equations show the formation of ions in aqueous(aq) solutions.
H2 O
100% ions
NaCl(s)
Na+(aq) + Cl−(aq)
H2O
CaBr2(s)
Ca2+(aq) + 2Br−(aq)
16
Learning Check
Complete each for strong electrolytes in water.
H2O
A. CaCl2(s)
1) CaCl2(s)
2) Ca2+(aq) + Cl2−(aq)
3) Ca2+(aq) + 2Cl−(aq)
H2O
B. K3PO4(s)
1) 3K+(aq) + PO43−(aq)
2) K3PO4(s)
3) K3+(aq) + P3−(aq) + O4−(aq)
17
Weak Electrolytes
A weak electrolyte
• dissociates only slightly in water.
• in water forms a solution of a few ions and
mostly undissociated molecules.
HF(g) +
(aq)
H2O(l)
NH3(g) + H2O(l)
OH-(aq)
H3O+(aq) + FNH4+(aq) +
18
Nonelectrolytes
Nonelectrolytes
• dissolve as
molecules in water.
• do not produce
ions in water.
• do not conduct an
electric current.
19
Equivalents
An equivalent (Eq) is the amount of an electrolyte
or an ion that provides 1 mole of electrical charge
(+ or -).
1 mole of Na+
=
1 mole of Cl− =
1 equivalent
1 mole of Ca2+
=
1 mole of Fe3+ =
1 equivalent
2 equivalents
3 equivalents
20
Electrolytes in Body Fluids
In replacement solutions for body fluids, the
electrolytes
are given in milliequivalents per liter (mEq/L).
Ringer’s Solution
Na+
Cl−
K+
Ca2+
147 mEq/L
155 mEq/L
4 mEq/L
4 mEq/L
The milliequivalents per liter of cations must
equal the milliequivalents per liter of anions.
21
Electrolytes in Body Fluids
22
Learning Check
A.
B.
C.
In 1 mole of Fe3+, there are
1) 1 Eq.
2) 2 Eq.
3) 3 Eq.
In 2.5 moles of SO42−, there are
1) 2.5 Eq.
2) 5.0 Eq.
3) 1.0 Eq.
An IV bottle contains NaCl. If the Na+ is
34 mEq/L, the Cl− is
1) 34 mEq/L.
2) 0 mEq/L.
3) 68 mEq/L.
23
Chapter 7
Solutions
7.3
Solubility
24
Solubility
Solubility is
• the maximum amount of solute that
dissolves in a specific amount of
solvent.
• expressed as grams of solute in 100
grams of solvent, usually water.
g of solute
100 g water
25
Unsaturated Solutions
Unsaturated
solutions
• contain less than the
maximum amount of
solute.
• can dissolve more
solute.
26
Saturated Solutions
Saturated solutions
• contain the
maximum amount of
solute that can
dissolve.
• have undissolved
solute at the bottom
of the container.
27
Learning Check
At 40 C, the solubility of KBr is 80 g/100 g of H2O.
Identify the following solutions as either
1) saturated or 2) unsaturated. Explain.
A. 60 g KBr added to 100 g of water at 40 C.
B. 200 g KBr added to 200 g of water at 40 C.
C. 25 g KBr added to 50 g of water at 40 C.
28
Effect of Temperature on Solubility
Solubility
• depends on
temperature.
• of most solids
increases as
temperature
increases.
• of gases decreases
as temperature
increases.
29
Solubility and Pressure
Henry’s law states
• the solubility of a
gas in a liquid is
directly related to
the pressure of that
gas above the liquid.
• at higher pressures,
more gas molecules
dissolve in the
liquid.
30
Chapter 7 Solutions
7.4
Percent Concentration
31
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
32
Mass Percent
Mass percent (% m/m) is the
• concentration by mass of solute in a
solution.
mass percent =
100
g of solute
x
g of solute + g of solvent
• amount in g of solute in 100 g of solution.
mass percent =
g of solute
a
100 g of solution
33
Mass of Solution
Add water to
give 50.00 g of
solution
8.00 g KCl
50.00 g of KCl
solution
34
Calculating Mass Percent
The calculation of mass percent (% m/m)
requires the
• grams of solute (g KCl) and
• grams of solution (g KCl solution).
g of KCl
g of solvent (water)
g of KCl solution
=
=
=
8.00 g
42.00 g
50.00 g
8.00 g KCl (solute)
x 100 = 16.0% (m/m)
50.00 g KCl solution
35
Learning Check
A solution is prepared by mixing 15.0 g of
Na2CO3 and 235 g of H2O. Calculate the
mass percent
(% m/m) of the solution.
1) 15.0% (m/m) Na2CO3
2) 6.38% (m/m) Na2CO3
3) 6.00% (m/m) Na2CO3
36
Volume Percent
The volume percent (% v/v) is
• percent volume (mL) of solute (liquid) to
volume (mL) of solution.
volume % (v/v) = 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
37
Mass/Volume Percent
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
• solute (g) in 100 mL of solution.
mass/volume % (m/v) =
g of solute
100 mL of solution
38
Percent Conversion Factors
Two conversion factors can be written for
each type of % value.
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39
Using Percent Concentration (m/m)
as Conversion Factors
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 using the factor that cancels g solution.
40
Learning Check
How many milliliters of a 5.75% (v/v)
ethanol solution can be prepared from
2.25 mL of ethanol?
1) 2.56 mL
2) 12.9 mL
3) 39.1 mL
41
Using Percent Concentration (m/v)
as Conversion Factors
How many mL of a 4.20% (m/v) will contain 3.15 g of KCl?
STEP 1: Given: 3.15 g of KCl(solute); 4.20% (m/v) KCl
Need: mL of KCl solution
STEP 2: Plan: g of KCl
mL of KCl solution
STEP 3: Write conversion factors.
4.20 g KCl
and
100 mL solution
STEP 4: Set up the problem
100 mL solution
4.20 g KCl
42
Learning Check
How many grams of NaOH are needed to
prepare
125 mL of a 8.80% (m/v) NaOH solution?
43