Transcript Hein and Arena - Solano Community College

Solutions
Chapter 13
Tro, 2nd ed.
1
SOLUTIONS & PHYSICAL PROPERTIES
A solution is a system in which one or more substances
are homogeneously mixed or dissolved in another
substance.
The solute is the component that is dissolved or is the
least abundant component of the solution.
The solvent is the dissolving agent or the most abundant
component in the solution.
If the solution has only one phase, it is homogeneous.
(A mixture which has two phases is heterogeneous.)
2
Similar to table 13.1
Metals in
metals =
alloys
3
Properties of True Solutions
The solute remains uniformly distributed throughout
the solution and will not settle out with time.
The solute can generally be separated from the
solvent by purely physical means such as
evaporation, usually of the solvent.
The solute particles of a true solution are molecular
or ionic in size.
The solution can be either colored or colorless but is
always clear or transparent.
4
Formation of a Solution
solute
solvent
solution
5
Solvated Ions
When materials dissolve, the solvent molecules surround the solvent
particles due to the solvent’s attractions for the solute. The process
is called solvation. Solvated ions are effectively isolated from each
other. (Also called hydrated ions if the solvent is water.)
6
TERMS THAT DESCRIBE THE
SOLUBILITY OF LIQUIDS:
Solubility (limit) is the maximum amount of solute that
can be dissolved in a given amount of solvent at a given
fixed T, at equilibrium.
Saturated solutions contain the maximum amount of
solute, so any additional solute appears as a precipitate
or a gas, or a separate liquid phase.
Unsaturated means that more solute can be added to the
solution.
Supersaturated is a temporary condition where more
solute has dissolved, but add just 1 crystal to this, many
crystals will precipitate from solution.
7
Supersaturated Solution
A supersaturated solution has more dissolved solute than
the solvent can hold. When disturbed, all the solute above
the saturation level comes out of solution.
8
TERMS THAT DESCRIBE
THE SOLUBILITY OF LIQUIDS
Miscible: liquids that are capable of mixing and
forming solutions.
methanol and water
Immiscible: liquids that are insoluble in each
other.
oil and water
9
Solubilities of substances vary widely.
10
Solubility of Various Common Ions in Cold Water
See back of blue
Periodic Table for
rules and table.
14.2
Similar to Table 7.2
11
Electrolytes
Electrolytes: substances whose
aqueous solution is a conductor of
electricity
Strong electrolytes: all the electrolyte
units are dissociated into ions (salts,
strong bases, strong acids)
Nonelectrolytes: none of the units are
dissociated into ions (molecular
solutes)
Weak electrolytes: a small percentage
of the units are dissociated into ions
(weak bases, weak acids)
12
Factors affecting solubility:
1. Natural inclination of the universe towards
disorder so substances do mix
2. Strength of Force of Attraction between
solute particles, between solvent particles and
between solute & solvent (nature of solute
and solvent)
3. Temperature
4. Pressure
NOTE: Gases are always completely miscible!
13
Factors affecting solubility:
Nature/Intermolecular Forces
Liquids:
Similar liquid molecules will dissolve in each other heptane in octane - both have only London forces
involved at about the same strength, both nonpolar.
Different IP forces - octane and water. H-bonding very
strong in water, very different from London forces in
octane; water would have to give up H-bonding for
weaker force. THIS WON'T HAPPEN. The less
dense liquid will rise and stay in separate phase on top
of water.
14
Factors affecting solubility:
Nature/Intermolecular Forces
Polar compounds tend to be more soluble in polar
solvents than nonpolar solvents. NaCl (sodium
chloride) is an ionic compound which is:
Solvent
Polarity
• soluble in water
• slightly soluble in ethanol
• insoluble in ether and benzene
General rule is “like dissolves like”
15
Dissolution of sodium chloride in water.
The hydrated ions slowly
diffuse away from the crystal
to
dissolved
in
Asbecome
the attraction
between
Polar
water molecules are
solution.
the ions weakens,
the ions
+
attracted to Na and Clmove apart and become
ions in the salt or crystal,
surrounded by water
weakening the attraction
dipoles.
between the ions.
14.3
16
Factors affecting solubility:
Nature/Intermolecular Forces
Nonpolar compounds tend to be more soluble
in nonpolar solvents than in polar solvents.
Benzene is a nonpolar organic compound
which is:
• insoluble in water
Solvent
• soluble in ether
Polarity
17
Classifying Solvents
Solvent
Water, H2O
Class
polar
Structural
Feature
O-H
Ethyl Alcohol, C2H5OH
polar
O-H
Acetone, C3H6O
polar
C=O
Benzene, C6H6
nonpolar
C-C & C-H
Hexane, C6H14
nonpolar
C-C & C-H
Diethyl Ether, C4H10O
nonpolar
C-C, C-H &
C-O
Similar to Table 13.2
18
Practice with solutes and solvents
1.
Which solvent will ethanol dissolve in more readily,
water or octane? Why?
2.
What about liquid C6H13OH? Water or octane?
3.
What about solid glucose? (Draw it.) Water or
octane?
4.
What about solid I2? Will it dissolve more readily in
water or octane?
19
Practice with solutes and solvents
1.
2.
3.
4.
Water, because of the polarity of both with their –
OH groups.
Octane, because most of the molecule is nonpolar
like octane is.
Solids must also be "like" the solvent.
Glucose dissolves in water because of extensive Hbonding w/ its many -OH groups. (See Lewis
structure)
Solid I2 - held by nonpolar London forces; octane
also London forces. I2 dissolves in octane better than
water.
20
Effect of Temperature on Solubility
For most solids dissolved in a liquid, an increase
in temperature results in increased solubility.
The solubility of a gas in water usually
decreases with temperature.
21
large increase in solubility
with temperature
decrease in solubility with
increasing temperature
FIGURE 13.4
slight increase in
solubility with
temperature
22
Effect of Pressure on Solubility
Small pressure changes have little effect on the
solubility of solids in liquids or liquids in liquids
Small pressure changes have a great effect on the
solubility of gases in liquids
- The solubility of a gas in a liquid is directly
proportional to the pressure of that gas above
the liquid
23
Rate of Dissolving Solids: 4 factors
1. Particle Size
A solid can dissolve only at the surface that is
in contact with the solvent.
Smaller crystals have a larger surface to volume
ratio than large crystals.
Smaller crystals dissolve faster than larger
crystals.
24
Rate of Dissolving Solids: 4 factors
2. Temperature
In most cases, the rate of dissolving of a solid
increases with temperature.
This occurs because solvent molecules strike the
surface of the solid more frequently and harder,
causing the solid to dissolve more rapidly.
25
Rate of Dissolving Solids: 4 factors
3. Concentration of the Solution
Δc
Δt
Δc
Δt
As solution
concentration
increases, the rate of
dissolving decreases.
The rate of dissolving
is at a maximum when
solute and solvent are
first mixed.
26
Rate of Dissolving Solids: 4 factors
4. Agitation or stirring.
When a solid is first put into water, it comes in
contact only with water. The rate of dissolving is a
maximum.
As the solid dissolves, the amount of dissolved solute
around the solid increases and the rate of dissolving
decreases.
Stirring distributes the dissolved solute throughout
the water; more water is in contact with the solid
causing it to dissolve more rapidly.
27
Concentration of Solutions
The concentration of a solution expresses the
amount of solute dissolved in a given quantity
of solvent or solution.
The terms dilute and concentrated are qualitative
expressions of the amount of solute present in a
solute.
You will need to learn the mathematical
representations of concentration: mass-percent,
mass/volume percent, Molarity, etc.
28
Mass Percent Solution
Mass percent expresses the concentration of
solution as the percent of solute in a given mass
of solution.
%-mass = (g solute/(g solute + g solvent)) * 100
= (g solute/g solution) * 100
29
Example of Mass Percent
How do you prepare 425 g of a 2.40%-wt solution of
sodium acetate?
Solve for grams of solute, then grams of water:
(X g/425 g) x 100 = 2.40%
x = 10.2 g NaC2H3O2
425 – 10.2 = 414.8 g H2O
You try: What is the mass-percent of a solution that has
13.6 g of NaCl in 250.0 g of solution?
What is the mass-percent of a solution that has 15.0 g of
ethanol in 35.0 g of water?
30
Mass/Volume Percent (m/v)
Mass /volume percent expresses the concentration
as grams of solute of solute per 100 ml solution.
%-mass/vol = g solute/mL solution * 100
31
A 3.00 %-mass/vol H2O2 solution is commonly used as
a topical antiseptic to prevent infection. What volume
of this solution will contain 10.0 g of H2O2?
Solve the mass/volume equation for grams of
solute.
%-mass/vol = (g solute/mL solution) * 100
mL solution = (g solute/%-mass/vol) * 100
= (10.0 g/3.00%) * 100
= 333 mL
32
Volume Percent
Solutions that are formulated from liquids are
often expressed as volume percent with respect
to the solute.
The volume percent is the volume of a liquid in
100 ml of solution.
%-vol = (solute vol/solution vol) * 100
33
Volumes are not necessarily additive.
A bottle of rubbing alcohol reads 70% by volume.
The alcohol solution could be prepared by mixing
70 mL of alcohol with water to make a total
volume of 100 mL of solution.
30 m L of water could not be added to 70 mL of
alcohol because the volumes are not necessarily
additive.
34
Practice
Find both the mass-percent and volume percent
of a solution that has 10.0 g of ethanol (D =
0.7893 g/mL) and 90.0 g of water (D = 0.9987
g/mL). Assume volumes are additive.
35
Molarity
Molarity of a solution is the number of moles of
solute per liter of solution.
Molarity = M = moles solute/liter of solution
= mol/L
Sometimes n represents moles, then M = n/V.
36
Molarity Calculations
How many moles of NaOH are present in 25.0mL
of 0.555 M NaOH?
M = mol/vol, therefore
mol = M * V (must be in Liters)
0.555 mol/L * (25.0mL * 1L/103mL)
= 0.0138 mol NaOH
37
Molarity Calculations
What is the Molarity of a solution that has 13.6 g
of NaCl in 250.0 mL of solution?
What is the volume of 1.25 M HCl solution that
will provide 0.4414 moles of HCl?
38
Solubility
solvent
Normality
g/100g
grams of solute/100 grams of
N
molarity * (# of H+ or OH-)
Yes, you have to know all of these!
39
NORMALITY: EXAMPLES
1.0 M HCl * 1 H+ = 1.0 N
1.0 M H2SO4 * 2 H+ = 2.0 N
1.0 M NaOH * 1 OH- = 1.0 N
1.0 M Ba(OH)2 * 2 OH- = 2.0 N
40
Dilution Problems
If a solution is diluted by adding pure solvent:
- the volume of the solution increases.
- the number of moles of solute remain the same.
41
Dilution Problems
When the moles of a solute in a solution before
and after dilution are the same, the moles before
and after dilution may be set equal to each other.
Remember M = n/V, or n = M * V. If V changes,
M will change to keep moles, n, constant.
Dilution Equation: M1V1 = M2V2
MEMORIZE THIS!!!
42
Practice
How would you make 500.0 mL of a 0.500 M
solution of NaOH from 6.00 M NaOH?
0.500 M * 500.0 mL = 6.00 M * V2
V2 = 41.7 mL
43
Practice Making Solutions
How would you make 250.0ml of a 0.555 M standard
NaOH solution?
Use two methods: (1) Weigh out dry chemical NaOH
solid and dissolve it.
(2) Dilute a more concentrated solution of 6.00 M
NaOH.
A. Weighing a solid and using M = n/V. Find n first,
then grams of solid.
250.0mL(1L/103mL)(0.555 mol/L) = 0.13875 mol NaOH
0.13875 mol * 39.997 g/mol = 5.55 g NaOH
Get a vol. flask and add water about halfway up. Add
solid. Mix til dissolved, then dilute to mark on flask.
44
Practice Making Solutions
B. Dilution: Given 6.00 M NaOH. Use the
Dilution Equation, M1V1=M2V2
6.00 mol/L * V1 = 0.555 mol/L * 0.2500 L
V1 = O.O231 L or 23.1 mL
Obtain 23.1 mL in a graduated cylinder. Get a vol.
flask and add water about halfway up. Add the
correct amount of 6.00 M NaOH. Mix until
dissolved, then dilute to mark on flask.
45
Preparation of a 1 molar solution
Start with
some water
in the flask!
14.7
46
Molarity, Dissociation and
Solution Inventories
When strong electrolytes dissolve, all the solute
particles dissociate into ions.
By knowing the formula of the compound and the
molarity of the solution, it is easy to determine
the molarity of the dissociated ions simply
multiply the salt concentration by the number of
ions.
It’s like doing stoichiometry!
47
Find the solution inventory for each of
these strong electrolytes
1.
2.
3.
0.25 M MgBr2
0.33 M Na2CO3
0.0750 M Fe2(SO4)3
1. MgBr2(aq) → Mg2+(aq) + 2 Br-(aq)
I: 0.25 M
0
0
R: -0.25 M
+0.25 +0.50
F: 0
0.25 M 0.50 M So we say: [Mg2+] = 0.25 M, [Br-] = 0.50 M
2. Na2CO3(aq) → 2 Na+(aq) + CO32-(aq)
I: 0.33 M
0
0
R: -0.33
+0.66 +0.33
F: 0
0.66 M 0.33 M [Na+] = 0.66 M, [CO32-] = 0.33 M
3. [Fe3+] = 0.150 M, [SO42-] = 0.225 M
48
STOICHIOMETRY WITH AQUEOUS
SOLUTIONS:
A. GAS-FORMING: use regular stoichiometry and
concentration information
B. PRECIPITATION: use solubility rules (see
examples on following pages)
C. NEUTRALIZATION: use concentration
information and stoichiometry steps. (Do NOT use
dilution equation! Use stoichiometry.)
acid + base  salt + water
TAKE GOOD NOTES – THE TEXT ASSUMES YOU
KNOW HOW TO DO THIS!
49
STOICHIOMETRY OF SOLUTION
PRECIPITATION REACTIONS:
How many grams of silver will precipitate if excess copper is added
to 500.0 ml OF 0.100 M AgNO3?
1: Balanced chemical equation
Cu(s) + 2 AgNO3(aq)  2 Ag(s) + Cu(NO3)2(aq)
What you know & what you want to know
500.0 ml OF 0.100 M AgNO3. Want grams Ag.
2: 500.0 ml * 1 L * 0.100 moles = 0.0500 mol AgNO3
103 ml
1.00 L
3: 0.0500 mol AgNO3 * 2 mol Ag/2 mol AgNO3 = 0.0500 mol Ag
4: 0.0500 mol Ag * 107.9 g/mol = 5.40 g Ag
50
Group Practice:
Stoichiometry/Precipitation
What was the concentration in Molarity of a silver
nitrate solution, if 25.00 mL of it required 23.31
mL of 0.3161 M NaCl solution to completely
precipitate all the silver as silver chloride?
NaCl(aq) + AgNO3(aq)  AgCl(s) + NaNO3(aq)
23.31mL 25.00mL
.3161 M ??M
51
C. NEUTRALIZATION
Acid-base reactions are analyzed by titration. This
is volumetric analysis using a standardized
solution to find mass or concentration of
another substance.
Burettes - contain the titrant, usually standardized
Flask - contains analyte, the substance we want to
analyze
Endpoint - when the indicator changes color to
say reaction is done
52
C. NEUTRALIZATION
Standardized solution: exact concentration of a reagent known.
Two methods to standardize – (1) use a pure dry solid or (2) use a
purchased aqueous standard.
Method 1: Use 0.250 grams of sodium carbonate to determine
exact concentration of hydrochloric acid. (NOTE: this is also a
gas-producing reaction.) It took 25.76 mL of HCl to reach
"endpoint" - where the chemical indicator changed color
indicating a pH change.
Na2CO3(s) + 2 HCl(aq)  2 NaCl(aq) + H2O(l) + CO3(g)
Moles solid base = 0.250 g/(106.0 g/mol) = 0.0023585 mol
Moles acid = mol base (2 HCl/1 Na2CO3) = 0.004717 mol
Molarity of acid = mol/vol = 0.004717 mol/0.02576 L = 0.183 M
53
C. NEUTRALIZATION
Method 2: Use a purchased standardized solution of
0.100 M HCl to standardize a NaOH solution. Exactly
25.00 mL of base took 32.56 mL of acid.
HCl(aq) + NaOH(aq)  NaCl(aq) + H2O(l)
Moles acid = 0.100mol/L x 0.03256 L = 0.003256 mol
Moles base = mol acid (1/1) = 0.003256 mol base
Molarity of base = mol/vol = 0.003256mol/0.02500L
= 0.0130 M
54
STEPS TO PERFORMING CALCS FOR
ACID-BASE REACTIONS:
1. List species as reactants, decide what products
will form, and balance the equation.
2. Calculate moles of reactants. (Use molarity and
volume.)
3. Calculate moles of desired product or other
reactant using mole/mole ratio.
4. Convert to volume or molarity or grams, as
required.
55
Example of steps
A 1.034 g sample of clover has oxalic acid extracted from
it into a small volume of water. The endpoint of
titration is 34.47 mL of 0.100 M NaOH. What is the
mass-percent of oxalic acid in clover?
1.)
H2C2O4(aq) + 2 NaOH(aq)  2 H20(l) + Na2C2O4(aq)
2 & 3.)
0.03447 L *(0.100mol NaOH/L)(1 mol acid/2 mol base)
= 0.0017235 mol acid
4.)
0.0017235 mol (90.04g/mol) = 0.15518 g
(0.15518 g/1.034g)*100 = 15.01 %-mass oxalic acid
56
Colligative Properties of Solutions
When a nonvolatile solute is added to a solvent,
four physical properties of the solvent, called
Colligative Properties, will change:
Freezing Point Depression
Boiling Point Elevation
Vapor Pressure Lowering
Osmotic Pressure
57
Colligative Properties of Solutions
Colligative properties of a solution depend only
on the number of solute particles in a solution
and not on the nature of those particles.
You already know something about these!
Consider?
Why does salt water boil at a higher temp?
Why do we throw salt on snow and ice on
roads?
58
Each solvent shows a characteristic:
freezing point depression constant
boiling point elevation constant
59
A liquid
solution
boils
willwhen
haveitsa vapor
lower pressure
vapor pressure.
equals
atmospheric
and
consequently
pressure.
a higher boiling point.
Vapor Pressure Curve of
Pure Water and Water
Solution: Boiling Point
Elevation
60
A
solution
will have
pressure.
A liquid
freezes
whenaitslower
vaporvapor
pressure
and
consequently
a lower freezing
point.
equals
the vapor pressure
of its solid.
Vapor Pressure Curve of
Pure Water and Water
Solution: Freezing Point
Depression
water vapor
pressure curve
solution vapor
pressure curve
ice vapor
pressure curve
61
Molality
The freezing point depression and the boiling
point elevation are directly proportional to the
number of moles of solute per kilogram of
solvent, in a concentration unit called molality.
molality = moles solute/kg of solvent = molal
62
What is the molality (molal) of a solution prepared
by dissolving 2.70 g CH3OH in 25.0 g H2O?
The concentration unit of molality is moles
solute/kg solvent, so convert mass of methanol
to moles and mass of water to kg:
2.70 g/32.042g/mol = 0.084264 mol CH3OH
25.0 g/1000g/kg = 0.0250 kg of water
molality = 0.084264 mol/0.0250 kg
= 3.37 molal
63
Symbols and equations used in the calculation
of colligative properties for FP and BP
m = molality (molal)
ΔTf = freezing point depression in oC
ΔTb = boiling point elevation in oC
Kf = freezing point depression constant
Kb = boiling point elevation constant
ΔTf = Kf * molal
new Tf = solvent Tf
– ΔTf
ΔTb = Kb * molal
new Tb = solvent Tb + ΔTb
64
A solution is made by dissolving 100. 0 g of
ethylene glycol (CH2OHCH2OH) in 200. g of
water. What is the freezing point of the solution?
The calculation takes two steps:
1. Determine the freezing point depression.
2. Subtract the depression value from the freezing point
of the pure solvent.
Find the molality and the ΔTf:
100.0 g/62.068 g/mol = 1.611 mol EG
molality = 1.611 mol/0.2000 kg = 8.056 molal
ΔTf = Kf * molal
new Tf = solvent Tf – ΔTf
ΔTf = 1.86oC/molal * 8.056 molal = 15.0oC
new Tf = solvent Tf – ΔTf = 0.0oC – 15.0oC = -15.0oC
65
Practice
Calculate both the new freezing point and the new
boiling point for both 1.0 molal and 2.0 molal
glucose solutions.
66
Osmosis and Osmotic Pressure
Osmosis is the diffusion of water, either from a
dilute solution or from pure water, through a
semipermeable membrane into a solution of
higher concentration.
Semipermeable membrane: A semipermeable
membrane allows the passage of water (solvent)
molecules through it in either direction, but it
prevents the passage of larger solute molecules.
67
Some Definitions
Osmotic pressure is the pressure that must be exerted to
prevent osmosis of solvent particles through a
semipermeable membrane that is separating two
solutions of different solute concentrations
Isotonic: Having the same concentration of solute
particles as the blood
Hypotonic: Having a lower concentration of solute
particles than blood plasma
Hypertonic: Having a higher concentration of solute
particles than blood plasma
68
In osmosis,the net transfer of
solvent is always from the more
concentrated to the less
concentrated solution.
water passes
through the
cellophane
14.9
69
Hemolysis & Crenation
red blood cell in
red blood cell in
normal red blood
cell in an isotonic hypotonic solution hypertonic solution
– water flows into – water flows out
solution
the cell –
of the cell –
eventually causing eventually causing
the cell to burst
the cell to distort
and shrink
70
COLLIGATIVE PROPERTIES
Know the definitions, try the homework. Be able
to do the freezing point depression and boiling
point elevation.
There is an equation for vapor pressure lowering,
but you don’t want to know it!
The osmotic pressure will not be a major deal on
any quiz or test, but you need to understand it
for biology.
71