1 solutions properties of water solutions 2 predict the % water in the following foods
TRANSCRIPT
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Solutions
Properties of Water
Solutions
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Predict the % water in the following foods
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Predict the % water in the following foods
88%
water
94% water
85%
water
86% water
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Water in the Body
water gain water loss
liquids1000 mL urine 1500 mL
food 1200 mL perspiring 300 mL
cells 300 mL exhaling 600 mL
feces 100 mL
Calculate the total water gain and water loss
Total ______ mL _____ mL
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Water
Most common solventA polar molecule
O -
a hydrogen bond
H +
H +
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Hydrogen Bonds Attract Polar Water Molecules
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Explore: Surface Tension HW
Fill a glass to the brim with water
How many pennies can you add to the glass without causing any water to run over?
Predict _________________
Actual _________________Explain your results
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Explore
1. Place some water on a waxy surface. Why do drops form?
2. Carefully place a needle on the surface of water. Why does it float? What happens if you push it through the water surface?
3. Sprinkle pepper on water. What does it do? Add a drop of soap. What happens?
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Surface Tension
Water molecules within water hydrogen bond in all directions
Water molecules at surface cannot hydrogen bond above the surface, pulled inward
Water surface behaves like a thin, elastic membrane or “skin”
Surfactants (detergents) undo hydrogen bonding
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Solute and Solvent
Solutions are homogeneous mixtures of two or more substances
Solute
The substance in the lesser amountSolvent
The substance in the greater amount
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Nature of Solutes in Solutions
Spread evenly throughout the
solution
Cannot be separated by filtration
Can be separated by evaporation
Not visible, solution appears
transparent
May give a color to the solution
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Types of Solutions
air O2 gas and N2 gas gas/gas
soda CO2 gas in water gas/liquid
seawater NaCl in water solid/liquid
brass copper and zinc solid/solid
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Discussion
Give examples of some solutions and explain why they are solutions.
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Learning Check SF1
(1) element (2) compound (3) solution
A. water 1 2 3
B. sugar 1 2 3
C. salt water 1 2 3
D. air 1 2 3
E. tea 1 2 3
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Solution SF1
(1) element (2) compound (3) solution
A. water 2
B. sugar 2
C. salt water 3
D. air 3
E. tea 3
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Learning Check SF2
Identify the solute and the solvent.
A. brass: 20 g zinc + 50 g copper
solute = 1) zinc 2) copper
solvent = 1) zinc 2) copper
B. 100 g H2O + 5 g KCl
solute = 1) KCl 2) H2O
solvent = 1) KCl 2) H2O
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Solution SF2
A. brass: 20 g zinc + 50 g copper
solute = 1) zinc solvent = 2) copper
B. 100 g H2O + 5 g KCl
solute = 1) KCl
solvent = 2) H2O
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Learning Check SF3
Identify the solute in each of the following solutions:
A. 2 g sugar (1) + 100 mL water (2)
B. 60.0 mL ethyl alcohol(1) and 30.0 mL
of methyl alcohol (2)
C. 55.0 mL water (1) and 1.50 g NaCl (2)
D. Air: 200 mL O2 (1) + 800 mL N2 (2)
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Solution SF3
Identify the solute in each of the following solutions:
A. 2 g sugar (1)
B. 30.0 mL of methyl alcohol (2)
C. 1.5 g NaCl (2)
D. 200 mL O2 (1)
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Like dissolves like
A ____________ solvent such as water is
needed to dissolve polar solutes such as
sugar and ionic solutes such as NaCl.
A ___________solvent such as hexane
(C6H14) is needed to dissolve nonpolar
solutes such as oil or grease.
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Learning Check SF4
Which of the following solutes will dissolve in water? Why?
1) Na2SO4
2) gasoline
3) I2
4) HCl
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Solution SF4
Which of the following solutes will dissolve in water? Why?
1) Na2SO4 Yes, polar (ionic)
2) gasoline No, nonnpolar
3) I2 No, nonpolar
4) HCl Yes, Polar
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Formation of a Solution
Cl-
Na+ Cl-Na+
H2O
H2O
Na+
Cl-
solute
Dissolvedsolute
Hydration
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Electrolyte and Non-electrolyte
• Electrolyte: a substance that conducts electricity when dissolved in water. – Acids, bases and soluble ionic
solutions are electrolytes.
• Non-electrolyte: a substance that does not conduct electricity when dissolved in water. – Molecular compounds and insoluble
ionic compounds are non-electrolytes.
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Electrolytes
• Some solutes can dissociate into ions.
• Electric charge can be carried.
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Types of solutes
Na+
Cl-
Strong Electrolyte -100% dissociation,all ions in solution
high conductivity
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Types of solutes
CH3COOH
CH3COO-
H+
Weak Electrolyte -partial dissociation,molecules and ions in solution
slight conductivity
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Types of solutes
SugarC6H12O6
Non-electrolyte -No dissociation,all molecules in solution
no conductivity
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Types of Electrolytes
• Weak electrolyte partially dissociates.– Fair conductor of electricity.
• Non-electrolyte does not dissociate. – Poor conductor of electricity.
• Strong electrolyte dissociates completely.– Good electrical conduction.
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Representation of Electrolytes using Chemical Equations
MgCl2(s) → Mg2+(aq) + 2 Cl- (aq)
A strong electrolyte:
A weak electrolyte:
CH3COOH(aq) ← CH3COO -(aq) +H+(aq)→
CH3OH(aq)
A non-electrolyte:
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Electrolytes in Action
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Strong Electrolytes
Strong acids: HNO3, H2SO4, HCl, HClO4
Strong bases: MOH (M = Na, K, Cs, Rb etc)
Salts: All salts dissolving in water are completely ionized.
Stoichiometry & concentration relationship
NaCl (s) Na+ (aq) + Cl– (aq)
Ca(OH)2 (s) Ca2+(aq) + 2 OH– (aq)
AlCl3 (s) Al3+ (aq) + 3 Cl– (aq)
(NH4)2SO4 (s) 2 NH4 + (aq) + SO42– (aq)
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Writing An Equation for a Solution
When NaCl(s) dissolves in water, the reaction can be written as
H2O
NaCl(s) Na+ (aq) + Cl- (aq)
solid separation of ions in water
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Learning Check SF5
Solid LiCl is added to some 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
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Solution SF5
Solid LiCl is added to some water. It dissolves because
A. The Li+ ions are attracted to the
1) oxygen atom(-) of water
B. The Cl- ions are attracted to the
2) hydrogen atom(+) of water
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Rate of Solution
You are making a chicken broth using a bouillon cube. What are some things you can do to make it dissolve faster?
Crush it
Use hot water (increase temperature)
Stir it
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How do I get sugar to dissolve faster in my iced tea?
Stir, and stir, and stir
Add sugar to warm tea then add ice
Grind the sugar to a powder
Fresh solvent contact and interaction with solute
Greater surface area, more solute-solvent interaction
Faster rate of dissolution at higher temperature
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Learning Check SF6
You need to dissolve some gelatin in water. Indicate the effect of each of the following on the rate at which the gelatin dissolves as (1) increase, (2) decrease,
(3) no change
A. ___Heating the water
B. ___Using large pieces of gelatin
C. ___Stirring the solution
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Learning Check SF6
You need to dissolve some gelatin in water. Indicate the effect of each of the following on the rate at which the gelatin dissolves as (1) increase, (2) decrease,
(3) no change
A. 1 Heating the water
B. 2 Using large pieces of gelatin
C. 2 Stirring the solution
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SolubilityPercent Concentration
Colloids and Suspensions
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Solubility
The maximum amount of solute that can dissolve in a specific amount of solvent usually 100 g.
g of solute
100 g water
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Saturated and Unsaturated
A saturated solution contains the maximum
amount of solute that can dissolve.
Undissolved solute remains.
An unsaturated solution does not contain all
the solute that could dissolve
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SolubilitySATURATED SOLUTION
no more solute dissolves
UNSATURATED SOLUTIONmore solute
dissolves
SUPERSATURATED SOLUTION
becomes unstable, crystals form
increasing concentration increasing concentration
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Polarity
Factors Affecting Solid Solubility
Temperature
Surface Area
Stirring
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Polarity
Factors Affecting Solubility
Temperature
Pressure
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Intramolecular Bonding• Intramolecular bonding refers to the chemical
bonding that holds atoms together within a molecule of a compound
Covalent bonding and ionic bonding are the two main types of intramolecular bonding
Covalent bonding involves the sharing of valence electrons involves the sharing of valence electrons between two atoms. POLAR- unequal sharing of electrons
NON POLAR – equal sharing of electrons
Ionic bonding involves the transference of valence electrons
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SOLUTE POLAR SOLVENT
NONPOLAR SOLVENT
Ionic Soluble Insoluble
Polar Soluble Insoluble
Nonpolar Insoluble soluble
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Learning Check S1
At 40C, the solubility of KBr is 80 g/100 g
H2O. Indicate if the following solutions are
(1) saturated or (2) unsaturated
A. ___60 g KBr in 100 g of water at 40C
B. ___200 g KBr in 200 g of water at 40C
C. ___25 KBr in 50 g of water at 40C
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Solution S1
At 40C, the solubility of KBr is 80 g/100 g H2O.
Indicate if the following solutions are
(1) saturated or (2) unsaturated
A. 2 Less than 80 g/100 g H2O
B. 1 Same as 100 g KBr in 100 g of water
at 40C, which is greater than its solubility
C. 2 Same as 60 g KBr in 100 g of water,
which is less than its solubility
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Temperature and Solubility of Solids
Temperature Solubility (g/100 g H2O)
KCl(s) NaNO3(s)
0° 27.6 74
20°C 34.0 88
50°C 42.6 114
100°C 57.6 182
The solubility of most solids (decreases or increases ) with an increase in the temperature.
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Temperature and Solubility of Solids
Temperature Solubility (g/100 g H2O)
KCl(s) NaNO3(s)
0° 27.6 74
20°C 34.0 88
50°C 42.6 114
100°C 57.6 182
The solubility of most solids increases with an increase in the temperature.
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Temperature and Solubility of Gases
Temperature Solubility (g/100 g H2O)
CO2(g) O2(g)
0°C 0.34 0.0070
20°C 0.17 0.0043
50°C 0.076 0.0026
The solubility of gases (decreases or increases) with an increase in temperature.
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Temperature and Solubility of Gases
Temperature Solubility (g/100 g H2O)
CO2(g) O2(g)
0°C 0.34 0.0070
20°C 0.17 0.0043
50°C 0.076 0.0026
The solubility of gases decreases with an increase in temperature.
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Learning Check S2
A. Why would a bottle of carbonated drink
possibly burst (explode) when it is left out
in the hot sun ?
B. Why would fish die in water that gets too warm?
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Solution S2
A. Gas in the bottle builds up as the gas becomes less soluble in water at high temperatures, which may cause the bottle to explode.
B. Because O2 gas is less soluble in warm
water, the fish may not obtain the needed
amount of O2 for their survival.
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Gas Solubility
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CH4
O2
CO
He
Sol
ubili
ty (
mM
)
2.0
1.0
0 10 20 30 40 50
Higher Temperature
…Gas is LESS Soluble
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Solubility Curves
Show the conditions that affect states of the solution: unsaturated, saturated, supersaturated.
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Solubility Table
LeMay Jr, Beall, Robblee, Brower, Chemistry Connections to Our Changing World , 1996, page 517
0 10 20 30 40 50 60 70 80 90 100
Solubility vs. Temperature for Solids
Sol
ubili
ty (
gram
s of
sol
ute/
100
g H
2O)
KI
KCl
20
10
30
40
50
60
70
80
90
110
120
130
140
100
NaNO3
KNO3
HCl NH4Cl
NH3
NaCl KClO3
SO2
shows the dependence
of solubility on temperature
gases
solids
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How to determine the solubility of a given substance?
• Find out the mass of solute needed to make a saturated solution in 100 cm3 of water for a specific temperature(referred to as the solubility).
• This is repeated for each of the temperatures from 0ºC to 100ºC. The data is then plotted on a temperature/solubility graph, and the points are connected. These connected points are called a solubility curve.
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How to use a solubility graph?A. IDENTIFYING A SUBSTANCE
( given the solubility in g/100 cm3 of water and the temperature)
• Look for the intersection of the solubility and temperature.
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Learning Check SG1:What substance has a solubility of 90 g/100 cm3 in water at a temperature of 25ºC ?
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Learning Check SG2:What substance has a solubility of 200 g/100 cm3 of water at a temperature of 90ºC ?
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Look for the temperature or solubility
•Locate the solubility curve needed and see for a given temperature, which solubility it lines up with and visa versa.
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Learning Check SG3:
What is the solubility of potassium nitrate at 80ºC ?
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• What is the solubility of potassium nitrate at 80ºC ?
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Learning Check SG4:
At what temperature will sodium nitrate have a solubility of 95 g/100 cm3 ?
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Learning Check SG4:
At what temperature will sodium nitrate have a solubility of 95 g/100 cm3 ?
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Learning Check SG5:
At what temperature will potassium iodide have a solubility of 230 g/100 cm3 ?
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Learning Check SG5:
At what temperature will potassium iodide have a solubility of 130 g/100 cm3 ?
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Using Solubility Curves:
What is the solubility of sodium chloride at 25ºC in 150 cm3 of water ?
From the solubility graph we see that sodium chlorides solubility is 36 g.
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Solubility in grams = unknown solubility in grams
100 cm3 of water other volume of water
___36 grams____ = unknown solubility in grams
100 cm3 of water 150 cm3 water
Place this in the proportion below and solve for the unknown solubility. Solve for the unknown quantity by cross multiplying.
The unknown solubility is 54 grams. You can use this proportion to solve for the other volume of water
if you're given the other solubility.
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C. Determine if a solution is saturated,unsaturated,or supersaturated.
• If the solubility for a given substance places it anywhere on it's solubility curve line it is saturated.• If it lies above the solubility curve line, then it's supersaturated,• If it lies below the solubility curve line it's an unsaturated solution. Remember though, if the volume of water isn't 100 cm3 to use a proportion first as shown above.
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Temp. (oC)
Solubility(g/100 g
H2O)
KNO3 (s)
KCl (s)
HCl (g)
SOLUBILITYCURVE
Solubility how much solute dissolves in a given amt.of solvent at a given temp.
unsaturated: solution could hold more solute; belowbelow linesaturated: solution has “just right” amt. of solute; onon linesupersaturated:solution has “too much” solute dissolved in
it;aboveabove the line
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To
Sol.
To
Sol.
Solids dissolved in liquids Gases dissolved in liquids
As To , solubility As To , solubility
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Sometimes you'll need to determine how much additional solute needs to be added to a unsaturated solution in order to make it saturated.
For example,30 grams of potassium nitrate has been added to 100 cm3 of water at a temperature of 50ºC. How many additional grams of solute must be added in order to make it saturated?
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How many additional grams of solute must be added in order to make it saturated?
From the graph you can see that the solubility for potassium nitrate at 50ºC is 84 grams
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If there are already 30 grams of solute in the solution, all you need to get to 84 grams is 54 more grams ( 84g-30g )
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Solubility Table
LeMay Jr, Beall, Robblee, Brower, Chemistry Connections to Our Changing World , 1996, page 517
shows the dependence
of solubility on temperature
0 10 20 30 40 50 60 70 80 90 100
Solubility vs. Temperature for Solids
Sol
ubili
ty (
gram
s of
sol
ute/
100
g H
2O)
KI
KCl
20
10
30
40
50
60
70
80
90
110
120
130
140
100
NaNO3
KNO3
HCl NH4Cl
NH3
NaCl KClO3
SO2
gases
solids
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Classify as unsaturated, saturated, or supersaturated.Classify as unsaturated, saturated, or supersaturated.
per100
gH2O
80 g NaNO3 @ 30oC
45 g KCl @ 60oC
50 g NH3 @ 10oC
70 g NH4Cl @ 70oC
=unsaturated
=saturated
=unsaturated
=supersaturated
0 10 20 30 40 50 60 70 80 90 100
Solubility vs. Temperature for Solids
Sol
ubili
ty (
gram
s of
sol
ute/
100
g H
2O)
KI
KCl
20
10
30
40
50
60
70
80
90
110
120
130
140
100
NaNO3
KNO3
HCl NH4Cl
NH3
NaCl KClO3
SO2
gases
solids
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(A) Per 100 g H2O, 100 g Unsaturated; all
soluteNaNO3 @ 50oC. dissolves; clear
solution.
(B) Cool solution (A) very Supersaturated; extraslowly to 10oC. solute remains in solution;
still clear.
Describe each situation below.
(C) Quench solution (A) in Saturated; extra solute an ice bath to 10oC. (20 g) can’t remain in
solution, becomes visible.
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Soluble and Insoluble Salts
A soluble salt is an ionic compound that dissolves in water.
An insoluble salt is an ionic compound that does not dissolve in water
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How do we know ions are How do we know ions are present in aqueous present in aqueous solutions?solutions?
The solutions:_________The solutions:_________They are called They are called
ELECTROLYTESELECTROLYTESHCl, MgClHCl, MgCl22, and NaCl are , and NaCl are
strong electrolytesstrong electrolytes. . They dissociate They dissociate completely (or nearly so) completely (or nearly so) into ions.into ions.
Aqueous SolutionsAqueous Solutions
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Solubility Rules
1. A salt is soluble in water if it contains any one of the following ions:
NH4+ Li+ Na+ K+ or NO3
-
Examples:
soluble salts
LiCl Na2SO4 KBr Ca(NO3)2
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Cl- Salts
2. Salts with Cl- are soluble, but not if the
positive ion is Ag+, Pb2+, or Hg22+.
Examples:
soluble not soluble(will not dissolve)
MgCl2 AgCl
PbCl2
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SO42- Salts
3. Salts with SO42- are soluble, but not if
the positive ion is Ba2+, Pb2+, Hg2+ or Ca2+.
Examples:
soluble not soluble
MgSO4 BaSO4
PbSO4
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Other Salts
4. Most salts containing CO32-, PO4
3-, S2-
and OH- are not soluble.
Examples:
soluble not soluble
Na2CO3 CaCO3
K2S CuS
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90
Learning Check S3
Indicate if each salt is (1)soluble or (2)not soluble:
A. ______ Na2SO4
B. ______ MgCO3
C. ______ PbCl2
D. ______ MgCl2
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Solution S3
Indicate if each salt is (1) soluble or
(2) not soluble:
A. _1_ Na2SO4
B. _2_ MgCO3
C. _2_ PbCl2
D. _1_ MgCl2
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Solutions
Molarity
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Molarity (M)
A concentration that expresses the
moles of solute in 1 L of solution
Molarity (M) = moles of solute
1 liter solution
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Units of Molarity
2.0 M HCl = 2.0 moles HCl
1 L HCl solution
6.0 M HCl = 6.0 moles HCl
1 L HCl solution
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Molarity Calculation
NaOH is used to open stopped sinks, to treat cellulose in the making of nylon, and to
remove potato peels commercially.
If 4.0 g NaOH are used to make 500. mL of NaOH solution, what is the molarity (M) of the solution?
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Calculating Molarity
1) 4.0 g NaOH x 1 mole NaOH = 0.10 mole NaOH
40.0 g NaOH
2) 500. mL x 1 L _ = 0.500 L
1000 mL
3. 0.10 mole NaOH = 0.20 mole NaOH
0.500 L 1 L
= 0.20 M NaOH
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Learning Check M1
A KOH solution with a volume of 400 mL contains 2 mole KOH. What is the molarity of the solution?
1) 8 M
2) 5 M
3) 2 M Dra
no
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Solution M1
A KOH solution with a volume of 400 mL contains 2 moles of KOH. What is the molarity of the solution?
2) 5 M
M = 2 mole KOH = 5 M
0.4 L Dra
no
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Learning Check M2
A glucose solution with a volume of 2.0 L contains 72 g glucose (C6H12O6). If glucose
has a molar mass of 180. g/mole, what is the molarity of the glucose solution?
1) 0.20 M
2) 5.0 M
3) 36 M
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Solution M2
A glucose solution with a volume of 2.0 L contains 72 g glucose (C6H12O6). If glucose
has a molar mass of 180. g/mole, what is the molarity of the glucose solution?
1) 72 g x 1 mole x 1 = 0.20 M
180. g 2.0 L
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Molarity Conversion Factors
A solution is a 3.0 M NaOH.. Write the molarity in the form of conversion factors.
3.0 moles NaOH and 1 L NaOH soln
1 L NaOH soln 3.0 moles NaOH
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Learning Check M3
Stomach acid is a 0.10 M HCl solution. How many moles of HCl are in 1500 mL of stomach acid solution?
1) 15 moles HCl
2) 1.5 moles HCl
3) 0.15 moles HCl
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Solution M3
3) 1500 mL x 1 L = 1.5 L
1000 mL
1.5 L x 0.10 mole HCl = 0.15 mole HCl
1 L
(Molarity factor)
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Learning Check M4
How many grams of KCl are present in 2.5 L
of 0.50 M KCl?
1) 1.3 g
2) 5.0 g
3) 93 g
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Solution M4
3)
2.5 L x 0.50 mole x 74.6 g KCl = 93 g KCl
1 L 1 mole KCl
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Learning Check M5
How many milliliters of stomach acid, which is 0.10 M HCl, contain 0.15 mole HCl?
1) 150 mL
2) 1500 mL
3) 5000 mL
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Solution M5
2) 0.15 mole HCl x 1 L soln x 1000 mL
0.10 mole HCl 1 L
(Molarity inverted)
= 1500 mL HCl
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Learning Check M6
How many grams of NaOH are required to prepare 400. mL of 3.0 M NaOH solution?
1) 12 g
2) 48 g
3) 300 g
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Solution M6
2) 400. mL x 1 L = 0.400 L 1000 mL
0.400 L x 3.0 mole NaOH x 40.0 g NaOH 1 L 1 mole NaOH
(molar mass)
= 48 g NaOH
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Solution
Percent Concentration
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Percent Concentration
Describes the amount of solute dissolved in 100 parts of solution
amount of solute
100 parts solution
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Mass-Mass % Concentration
mass/mass % = g solute x 100% 100 g solution
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Mixing Solute and Solvent
Solute + Solvent
4.0 g KCl 46.0 g H2O
50.0 g KCl solution
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Calculating Mass-Mass %
g of KCl = 4.0 g
g of solvent = 46.0 g
g of solution = 50.0 g
%(m/m) = 4.0 g KCl (solute) x 100 = 8.0% KCl
50.0 g KCl solution
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Learning Check PC1
A solution contains 15 g Na2CO3 and 235 g of
H2O? What is the mass % of the solution?
1) 15% (m/m) Na2CO3
2) 6.4% (m/m) Na2CO3
3) 6.0% (m/m) Na2CO3
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Solution PC1
mass solute = 15 g Na2CO3
mass solution = 15 g + 235 g = 250 g
%(m/m) = 15 g Na2CO3 x 100
250 g solution
= 6.0% Na2CO3 solution
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Mass-Volume %
mass/volume % = g solute x 100% 100 mL solution
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Learning Check PC2
An IV solution is prepared by
dissolving 25 g glucose
(C6H12O6) in water to make 500.
mL solution. What is the
percent (m/v) of the glucose in
the IV solution?
1) 5.0% 2) 20.% 3) 50.%
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Solution PC2
1) 5.0%
%(m/v) = 25 g glucose x 100
500. mL solution
= 5.0 %(m/v) glucose solution
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Writing Factors from %
A physiological saline solution is a 0.85% (m/v) NaCl solution.
Two conversion factors can be written for the % value.
0.85 g NaCl and 100 mL NaCl soln
100 mL NaCl soln 0.85 g NaCl
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% (m/m) Factors
Write the conversion factors for a 10 %(m/m) NaOH solution
NaOH and NaOH soln
NaOH soln NaOH
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% (m/m) Factors
Write the conversion factors for a 10 %(m/m) NaOH solution
NaOH and NaOH soln
NaOH soln NaOH
10 g
10 g100 g
100 g
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Learning Check PC 3
Write two conversion factors for each of the following solutions:
A. 8 %(m/v) NaOH
B. 12 %(v/v) ethyl alcohol
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Solution PC 3
Write conversion factors for the following:
A. 8 %(m/v) NaOH
8 g NaOH and 100 mL 100 mL 8 g NaOH
B. 12 %(v/v) ethyl alcohol
12 mL alcohol and 100 mL 100 mL 12 mL alcohol
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Using % Factors
How many grams of NaCl are needed to prepare 250 g of a 10.0% (m/m) NaCl solution?
Complete data:____________ g solution
____________% or (______/_100 g_) solution
____________ g solute
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Clculation Using % Factors
250 g solution
10.0% or (10.0 g/100 g) solution
? g solute
250 g NaCl soln x 10.0 g NaCl = 25 g NaCl100 g NaCl soln
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Learning Check PC4
How many grams of NaOH do you need to measure out to prepare 2.0 L of a 12%(m/v) NaOH solution?
1) 24 g NaOH
2) 240 g NaOH
3) 2400 g NaOH
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Solution PC4
2.0 L soln x 1000 mL = 2000 mL soln
1 L
12 % (m/v) NaOH = 12 g NaOH
100 mL NaOH soln
2000 mL x 12 g NaOH = 240 g NaOH
100 mL NaOH soln
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Learning Check PC5
How many milliliters of 5 % (m/v) glucose solution are given if a patient receives 150 g of glucose?
1) 30 mL
2) 3000 mL
3) 7500 mL
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Solution PC5
5% m/v factor
150 g glucose x 100 mL = 3000 mL
5 g glucose
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Preparing a Solution by Dilution
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Units of Concentrations
amount of solute per amount of solvent or solution
Percent (by mass) =g solute
g solutionx 100
g solute
g solute + g solvent
x 100=
Molarity (M) =
moles of solute
volume in liters of solution
moles = M x VL
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Solutions
Colloids and Suspensions
Osmosis and Dialysis
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Solutions
Have small particles (ions or molecules)
Are transparent
Do not separate
Cannot be filtered
Do not scatter light.
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Colloids
Have medium size particles
Cannot be filtered
Separated with semipermeable membranes
Scatter light (Tyndall effect)
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Examples of Colloids
Fog
Whipped cream
Milk
Cheese
Blood plasma
Pearls
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Suspensions
Have very large particles
Settle out
Can be filtered
Must stir to stay suspended
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Examples of Suspensions
Blood platelets
Muddy water
Calamine lotion
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Osmosis
In osmosis, the solvent water moves
through a semipermeable membrane
Water flows from the side with the lower solute concentration into the side with the higher solute concentration
Eventually, the concentrations of the two
solutions become equal.
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Osmosis
semipermeable membrane
4% starch 10% starch
H2O
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Equilibrium is reached.
water flow becomes equal
7% starch
7% starch
H2OO
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Osmotic Pressure
Produced by the number of solute particles
dissolved in a solution
Equal to the pressure that would prevent
the flow of additional water into the more
concentrated solution
Increases as the number of dissolved
particles increase
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Osmotic Pressure of the Blood
Cell walls are semipermeable membranes
The osmotic pressure of blood cells
cannot change or damage occurs.
The flow of water between a red blood
cell and its surrounding environment
must be equal
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Isotonic solutions
• Exert the same osmotic pressure as red blood cells.
• Medically 5% glucose and 0.9% NaCl are used their solute concentrations provide an osmotic pressure equal to that of red blood cells
H2O
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Hypotonic Solutions
Lower osmotic pressure than red blood cells
Lower concentration of particles than RBCs
In a hypotonic solution, water flows into the
RBC
The RBC undergoes hemolysis; it swells and
may burst.
H2O
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Hypertonic Solutions
Has higher osmotic pressure than RBCHas a higher particle concentration In hypertonic solutions, water flows out
of the RBCThe RBC shrinks in size (crenation)
H2O
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Dialysis
Occurs when solvent and small solute
particles pass through a semipermeable
membrane
Large particles retained inside
Hemodialysis is used medically (artificial
kidney) to remove waste particles such as
urea from blood
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Colligative Colligative PropertiesPropertiesOn adding a solute to a solvent, the On adding a solute to a solvent, the
properties of the solvent are modified.properties of the solvent are modified.• Vapor pressure Vapor pressure decreasesdecreases• Melting point Melting point decreasesdecreases• Boiling point Boiling point increasesincreases• Osmosis is possible (osmotic pressure)Osmosis is possible (osmotic pressure)
These changes are called These changes are called COLLIGATIVE COLLIGATIVE PROPERTIESPROPERTIES. .
They depend only on the They depend only on the NUMBERNUMBER of solute of solute particles relative to solvent particles, not on particles relative to solvent particles, not on the the KINDKIND of solute particles. of solute particles.
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Change in Freezing Change in Freezing Point Point
The freezing point of a solution is The freezing point of a solution is
LOWERLOWER than that of the pure than that of the pure solventsolvent
Pure waterPure waterEthylene glycol/water Ethylene glycol/water
solutionsolution
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Change in Freezing Point Change in Freezing Point Common Applications of Common Applications of
Freezing Point Freezing Point DepressionDepression
Propylene glycol
Ethylene glycol – deadly to small animals
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Common Applications of Common Applications of Freezing Point DepressionFreezing Point DepressionWhich would you use for the streets of
Bloomington to lower the freezing point of ice and why? Would the temperature make any difference in your decision?
a) sand, SiO2
b) Rock salt, NaCl
c) Ice Melt, CaCl2
Change in Freezing Change in Freezing Point Point
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Change in Boiling Change in Boiling Point Point
Common Applications of Common Applications of Boiling Point ElevationBoiling Point Elevation