Solutes

advertisement
Chapter# 4
Solution Chemistry
Solutions
• Solutions are homogeneous mixtures of
two or more substances.
• The solvent is the substance in greatest
quantity.
• Solutes are the other ingredients in the
mixture.
• Solutions can exist in all states of matter.
Solution Examples
•Margarine
•Tap Water
•Steel
•18 Carat Gold
•Air
•Sterling Silver
Solution Examples
Composition of 18 carat gold: 75% gold, 12.5% silver,
12.5% copper.
What is the solvent in 18 ct gold ?
Solution Examples
Composition of 18 carat gold: 75% gold, 12.5% silver,
12.5% copper.
What is the solvent in 18 ct gold ?
Gold
Solution Examples
Composition of 18 carat gold: 75% gold, 12.5% silver,
12.5% copper.
What is the solvent in 18 ct gold ?
Gold
What are the solutes?
Solution Examples
Composition of 18 carat gold: 75% gold, 12.5% silver,
12.5% copper.
What is the solvent in 18 ct gold ?
Gold
What are the solutes?
Silver and Copper
Solution Properties
Some general properties of solutions include:
 Solutions may be formed between solids, liquids or
gases.
 They are homogenous in composition
 They do not settle under gravity
 They do not scatter light (Called the Tyndall Effect)
Solute particles are too small to scatter light and therefore
light will go right through a solution like is shown on the next
slide.
Tyndall Effect
Laser light reflected by a colloid. In a solution you
would not see any red light.
Solutions
Aqueous Solutions
Water is the dissolving medium


104.5o

Some Properties of Water
•
Water is “bent” or V-shaped.
•
Water is a molecular compound.
•
Water is a polar molecule.
•
Hydration occurs when ionic compounds
dissolve in water.
SOLUTION CONCENTRATION
The ratio of the amount of solute to amount of solution, or
solvent is defined by the concentration.
Concentration = solute
solution
=
solute
solvent
There are various combinations of units that are used in these
rations.
Ratio
X 102 X 103 X 106 X 109
g solute
g solution
= % (w/w)
g solute
= % (w/v)
mL solution
mL solute
mL solution
=
% (v/v)
ppt (w/w) ppm (w/w) ppb (w/w)
ppt (w/v)
ppm (w/v) ppb (w/v)
ppt (v/v)
ppm (v/v)
ppb (v/v)
SAMPLE SOLUTION PROBLEMS
1. Find the % (w/w) when 25.2 g NaCl is combined with 33.6g
H2O.
25.2 g NaCl
1. Find the mass of water and salt required to make 333 g of a
44.6 % (w/w) solution.
SAMPLE SOLUTION PROBLEMS
1. Find the % (w/w) when 25.2 g NaCl is combined with 33.6g
H2O.
25.2 g NaCl
33.6g H2O + 25.2 g NaCl
2. Find the mass of water and salt required to make 333 g of a
44.6 % (w/w) solution.
SAMPLE SOLUTION PROBLEMS
1. Find the % (w/w) when 25.2 g NaCl is combined with 33.6g
H2O.
25.2 g NaCl
100
33.6g H2O + 25.2 g NaCl
2. Find the mass of water and salt required to make 333 g of a
44.6 % (w/w) solution.
SAMPLE SOLUTION PROBLEMS
1. Find the % (w/w) when 25.2 g NaCl is combined with 33.6g
H2O.
25.2 g NaCl
100
33.6g H2O + 25.2 g NaCl
= 43.0 % NaCl
2. Find the mass of water and salt required to make 333 g of a
44.6 % (w/w) solution.
SAMPLE SOLUTION PROBLEMS
1. Find the % (w/w) when 25.2 g NaCl is combined with 33.6g
H2O.
25.2 g NaCl
100
33.6g H2O + 25.2 g NaCl
= 43.0 % NaCl
2. Find the mass of water and salt required to make 333 g of a
44.6 % (w/w) solution.
44.6 g NaCl
100 g solution
SAMPLE SOLUTION PROBLEMS
1. Find the % (w/w) when 25.2 g NaCl is combined with 33.6g
H2O.
25.2 g NaCl
100
33.6g H2O + 25.2 g NaCl
= 43.0 % NaCl
2. Find the mass of water and salt required to make 333 g of a
44.6 % (w/w) solution.
44.6 g NaCl 333 g solution
100 g solution
SAMPLE SOLUTION PROBLEMS
1. Find the % (w/w) when 25.2 g NaCl is combined with 33.6g
H2O.
25.2 g NaCl
100
33.6g H2O + 25.2 g NaCl
= 43.0 % NaCl
2. Find the mass of water and salt required to make 333 g of a
44.6 % (w/w) solution.
44.6 g NaCl 333 g solution
100 g solution
SAMPLE SOLUTION PROBLEMS
1. Find the % (w/w) when 25.2 g NaCl is combined with 33.6g
H2O.
25.2 g NaCl
100
33.6g H2O + 25.2 g NaCl
= 43.0 % NaCl
2. Find the mass of water and salt required to make 333 g of a
44.6 % (w/w) solution.
44.6 g NaCl 333 g solution
= 149 g NaCl
100 g solution
SAMPLE SOLUTION PROBLEMS
1. Find the % (w/w) when 25.2 g NaCl is combined with 33.6g
H2O.
25.2 g NaCl
100
33.6g H2O + 25.2 g NaCl
= 43.0 % NaCl
2. Find the mass of water and salt required to make 333 g of a
44.6 % (w/w) solution.
44.6 g NaCl 333 g solution
= 149 g NaCl
100 g solution
Mass of water?
SAMPLE SOLUTION PROBLEMS
1. Find the % (w/w) when 25.2 g NaCl is combined with 33.6g
H2O.
25.2 g NaCl
100
33.6g H2O + 25.2 g NaCl
= 43.0 % NaCl
2. Find the mass of water and salt required to make 333 g of a
44.6 % (w/w) solution.
44.6 g NaCl 333 g solution
= 149 g NaCl
100 g solution
Mass of water? 333 g solution – 149 g NaCl = 184 g H2O
SAMPLE SOLUTION PROBLEMS
1. Find the % (w/w) when 25.2 g NaCl is combined with 33.6g
H2O.
25.2 g NaCl
100
33.6g H2O + 25.2 g NaCl
= 43.0 % NaCl
2. Find the mass of water and salt required to make 333 g of a
44.6 % (w/w) solution.
44.6 g NaCl 333 g solution
= 149 g NaCl
100 g solution
Mass of water? 333 g solution – 149 g NaCl = 184 g H2O
SAMPLE SOLUTION PROBLEMS
3. How many grams of NaCl are required to dissolve in 88.2 g
of water to make a 29.2% (w/w) solution.
4. A sugar solution is 35.2%(w/v) find the mass of sugar
contained in a 432 mL sample of this sugar solution.
SOLUTION CONCENTRATION
The solution concentration can also be defined using moles.
The most common example is molarity (M).
The molarity of a solution is defined as:
“The number of moles of solute in 1 L of solution”
and is given the formula:
Molarity (M) =
Moles solute
Liters solution
MOLARITY SAMPLE PROBLEMS
1. A student dissolves 25.8 g of NaCl in a 250 mL volumetric
flask. Calculate the molarity of this solution. (picture of
volumetric flask is on the next slide)
2. Find the mass of HCl required to form 2.00 L of a 0.500 M
solution of HCl.
3. A student evaporates the water form a 333 mL sample of a
0.136 M solution of NaCl. What mass of salt remains?
SOLUTION PREPARATION
In the lab we would use a piece of glassware called a
volumetric flask to prepare this solution.
SOLUTION PREPARATION
VOLUMETRIC FLASK
SOLUTION DILUTION
Often we will want to make a dilute solution from a more
concentrated one.
To determine how to do this we use the formula :
C1V1 = C2V2
Where:
C1 = concentration of more concentrated solution
V1 = volume required of more concentrated solution
C2 = concentration of more dilute solution
V2 = volume of more dilute solution
We can use any units in this equation but they must be the
same on both sides.
DILUTION PROBLEM
How would one prepare 50.0 mL of a 3.00 M solution of
NaOH using a 7.10 M stock solution?
C1V1 = C2V2
(7.10 M)V1 = (3.00 M) (50.0 mL)
This means that you add 21.1 mL of the concentrated stock solution to a
50.0 mL volumetric flask and add water until the bottom of the meniscus
touches the line on the volumetric flask.
DILUTION PROBLEM
How would one prepare 50.0 mL of a 3.00 M solution of
NaOH using a 7.10 M stock solution?
C1V1 = C2V2
(7.10 M)V1 = (3.00 M) (50.0 mL)
(7.10
M)V1
(7.10 M) =
(3.00 M) (50.0 mL)
(7.10 M)
V1 = 21.1 mL
This means that you add 21.1 mL of the concentrated stock solution to a
50.0 mL volumetric flask and add water until the bottom of the meniscus
touches the line on the volumetric flask.
Reaction Driving Forces
Five Driving Forces Favor Chemical
Change
1.
2.
3.
4.
5.
Formation of a solid
Formation of water
Transfer of electrons
Formation of a gas
Formation of a weak electrolyte
Types of Aqueous Solutions
Solutions are homogeneous mixtures of a
solute and a solvent.
• The solute is the solution component in the
smallest amount while the solvent is the larger
component of a solution.
• Solutes whose solutions conduct electricity are
called electrolytes
• Solutes whose solutions do not conduct electricity
are called nonelectrolytes
• Electrolytes are solutes that form ions when they
dissolve. Ionic solutes or acids usually form
solutions that conduct electricity.
Solution Conductivity
Strong electrolyte
Weak electrolyte
Nonelectrolyte
Solution Formation
Water is one of the best solvents known. It is
able to dissolve ionic solutes, such as
sodium chloride, to produce solutions that
conduct electricity. Molecules, containing a
positive and negative regions, are called
polar. Water is an example of a polar
molecule and can dissolve ionic solutes by the
positive region of water attracting to the
negative ion of an ionic solute thus separating
the crystal lattice in to a solution of solvated
ions.
Acid-Base Reactions
Acids undergo characteristic double replacement
reactions with oxides, hydroxides, carbonates and
bicarbonates.
2HCl (aq) + CuO (s)  CuCl2 (aq) + H2O (l)
2HCl (aq) + Ca(OH)2 (aq)  CaCl2 (aq) + 2H2O (l)
2HCl (aq) + CaCO3 (aq)  CaCl2 (aq) + H2O (l) + CO2 (g)
2HC l (aq) + Sr(HCO3)2 (aq)  SrCl2 (aq) + 2H2O (l) + 2CO2
(g)
Acid-Base Reactions
Bases undergo a double replacement reaction with
acids called neutralization:
NaOH (aq) + HCl (aq)  H2O (l) + NaC l (aq)
In words this well known reaction is often described as:
“acid plus base = salt plus water”
We previously discussed this reaction when describing
types of reactions.
Acid-Base Reactions
We have discussed the double replacement reactions
and ionic equations before. Since the acids and
bases undergo double replacement reactions called
neutralization reactions, then they can have ionic
equations too.
Molecular equation:
HCl (aq) + NaOH (aq)  NaCl (aq) + H2O (l)
Total ionic equation:
H+ (aq) + Cl- (aq) + Na+ (aq) + OH- (aq)  Na+ (aq) + Cl- (aq) + H2O (l)
Net ionic equation:
H+ (aq) + OH- (aq)  H2O (l)
Acid-Base Reactions
Another property of acids is their reaction with certain metals
to produce hydrogen gas, H2 (g).
Zn (s) + 2HC l (aq)  H2 (g) + ZnCl2 (aq)
This is an example of a single replacement reaction and is a
redox reaction.
Total ionic equation:
Zn (s) + 2H+ (aq) + 2Cl- (aq)  H2 (g) + Zn2+ (aq) + 2Cl(aq)
Net ionic equation:
Zn (s) + 2H+ (aq)  H2 (g) + Zn2+ (aq)
Solution Stoichiometry
Consider the following balanced equation:
CaCl2 (aq) + AgNO3 (aq) → AgCl (s) + Ca(NO3)2 (aq)
1. Find the mass of silver chloride formed from 33.2 mL of a
0.100 M solution of silver nitrate and an excess of calcium
chloride.
Solution Stoichiometry
Consider the following balanced equation:
CaCl2 (aq) + AgNO3 (aq) → AgCl (s) + Ca(NO3)2 (aq)
1. Find the mass of silver chloride formed from 33.2 mL of a
0.100 M solution of silver nitrate and an excess of calcium
chloride.
0.100 moles AgNO3
L solution
Solution Stoichiometry
Consider the following balanced equation:
CaCl2 (aq) + AgNO3 (aq) → AgCl (s) + Ca(NO3)2 (aq)
1. Find the mass of silver chloride formed from 33.2 mL of a
0.100 M solution of silver nitrate and an excess of calcium
chloride.
0.100 moles AgNO3 L solution
L solution
103 mL
Solution Stoichiometry
Consider the following balanced equation:
CaCl2 (aq) + AgNO3 (aq) → AgCl (s) + Ca(NO3)2 (aq)
1. Find the mass of silver chloride formed from 33.2 mL of a
0.100 M solution of silver nitrate and an excess of calcium
chloride.
0.100 moles AgNO3 L solution moles AgCl 143.45 g AgCl 33.2 mL
moles AgNO3 moles AgCl
L solution
103 mL
Solution Stoichiometry
Consider the following balanced equation:
CaCl2 (aq) + AgNO3 (aq) → AgCl (s) + Ca(NO3)2 (aq)
1. Find the mass of silver chloride formed from 33.2 mL of a
0.100 M solution of silver nitrate and an excess of calcium
chloride.
0.100 moles AgNO3 L solution moles AgCl 143.45 g AgCl 33.2 mL
moles AgNO3 moles AgCl
L solution
103 mL
Solution Stoichiometry
Consider the following balanced equation:
CaCl2 (aq) + AgNO3 (aq) → AgCl (s) + Ca(NO3)2 (aq)
1. Find the mass of silver chloride formed from 33.2 mL of a
0.100 M solution of silver nitrate and an excess of calcium
chloride.
0.100 moles AgNO3 L solution moles AgCl 143.45 g AgCl 33.2 mL
moles AgNO3 moles AgCl
L solution
103 mL
= 0.476 g AgCl
Solution Stoichiometry
Consider the following balanced equation:
CaCl2 (aq) + AgNO3 (aq) → AgCl (s) + Ca(NO3)2 (aq)
1. Find the mass of silver chloride formed from 33.2 mL of a
0.100 M solution of silver nitrate and an excess of calcium
chloride.
2. Find the mass of silver chloride formed from 33.2 mL of a
0.100 M solution of silver nitrate and 200.0 mL of a 0.200
M solution of calcium chloride solution.
3. Find the volume of the excess reactant.
Acid-Base Reactions
• Bronsted-Lowry acids are proton (H+)
donors.
• Bronsted-Lowry bases are proton
acceptors.
• Free hydrogen ions don’t exist in water
because they strongly associate with a
water molecule to create a hydronium
ion (H3O+).
Acid-Base Reactions
• A neutralization reaction takes place
when an acid reacts with a base and
produces a solution of a salt and water.
• A salt is made up of the cation
characteristic of the base and the anion
characteristic of the acid.
• Example: HCl + NaOH ---> NaCl + H2O
Strong Acids and Bases
• A strong acid or strong base is completely
ionized in aqueous solution.
• HCl, HBr, HI, HNO3, HClO4 and H2SO4 are all
strong acids. All other acids are assumed to
be weak acids.
• A weak acid or weak base only partially ionize
in aqueous solution.
• Amphiprotic substances can behave as
either a proton acceptor or a proton donor.
Types of Equations
• Molecular Equations have reactants and
products written as undissociated molecules.
HCl + NaOH ---> NaCl + H2O
• Overall Ionic Equations show all the species,
both ionic and molecular present in the
reaction.
H+ + Cl- + Na+ + OH-
Na+ + Cl- + H2O
Continued
• Strong acids and strong bases are written
as the corresponding ions in an overall
ionic equation.
• Net Ion Equations describe the actual
reaction taking place.
H+ + OH- ----> H2O
• The Na+ and Cl- ions are spectator ion is
this reaction, because they are unchanged
by reaction taking place.
Precipitation Reactions
• Reactions in which a solid product
forms from the reactants in solution.
• Solubility guidelines for ionic
compounds allows the prediction of the
formation of solid products.
Solubility Guidelines
All compounds containing the following ions are
soluble:
Cations:Group I ions (alkali metals) and NH4+
Anions:NO3- and CH3COO- (acetate)
Compounds containing the following anions are
soluble except as noted:
Group 17 ions (halides), except with Ag+, Cu+,
Hg22+, Pb2+, SO42All other compounds are insoluble except the
following group 2 (alkaline earth) hydroxides:
Ba(OH)2, Ca(OH)2, and Sr(OH)2
Precipitation
Precipitation is the formation of a solid
when two solutions are combined.
Precipitate Formation
Does a precipitate form when sodium chloride is
mixed with silver nitrate? If so what is the
precipitate?
NaCl
AgNO3
Na+ + Cl- (Salt water solution)
Ag+ + NO3- (Silver nitrate solution)
Net Ionic Equations
• Soluble ionic compounds are called strong
electrolytes and completely ionize in aqueous
solution.
• Write the balanced net ionic equation when sodium
sulfate reacts with barium acetate.
Types of Solutions
• Unsaturated solution does not contain the
maximum concentration
• A saturated solution contains the maximum
concentration of solute that can dissolve in it.
• A supersaturated solution contains more
than the quantity of a solute that is predicted
to be soluble in a given volume of solution at
a given temperature.
Solubility Curves of Various Solutes
Supersaturated Solution
Sodium acetate precipitates from a supersaturated solution.
Supersaturated Solutions
By forming a solution at a high temperature then slowly
cooling it we can form supersaturated solutions that contain
more solute than in a saturated solution.
These kinds of solutions are very unstable and tend to
separate out the excess solute with the slightest disturbance.
http://www.youtube.com/watch?v=uy6eKm8IRdI&NR=1
http://www.youtube.com/watch?v=aC-KOYQsIvU&feature=related
Titration
Titration is an experimental procedure to
determine the concentration of an
unknown acid or base.
The figure on the left shows the
glassware for a titration experiment. A
buret clamp holds the buret to a ring
stand and below the buret is a flask
containing the solution to be titrated,
which includes an indicator. The
purpose of the indicator is to indicate
the point of neutralization by a color
change.
Key Titration Terms
• A titration is a volumetric method used to
determine the concentration of an unknown
solution by reacting it with a standard
solution.
• A standard solution is a solution of known
concentration.
• The equivalence point in a titration is
reached when enough standard solution has
been added to completely react with the
unknown solution.
• The end point in a titration is reached when
the indicator changes color.
Titration
The picture on the left shows the tip
of a buret, with air bubble, which is
not good, and also shows the stopcock. Note the position of the stopcock is in the “off” position. This
picture shows the color of the
phenolphthalein indicator at the endpoint. In this experiment a 23.00 mL
aliquot of 0.1000 M NaOH titrant is
NaOH + HCl  NaCl + HOH added to 5.00 mL of an unknown HCL
solution. The acid solution in the
beaker starts out clear and becomes
pink when all of the HCL has been
consumed.
Titration
How can we calculate the concentration of acid in
the beaker?
Titration
How can we calculate the concentration of acid in
the beaker?
Normal procedure, yes, a conversion. Steps 1-4, again!
Titration
How can we calculate the concentration of acid in the
beaker?
Normal procedure, yes, a conversion. Steps 1-4, again!
0.100 mole NaOH
L NaOH solution
Titration
How can we calculate the concentration of acid in the
beaker?
Normal procedure, yes, a conversion. Steps 1-4, again!
0.100 mole NaOH 10-3 L solution 23.00 mL soln mole HCl
mole NaOH 0.00500L
L NaOH solution mL solution
0.460 M HCl
Indicators
Indicators are weak organic (carbon containing) acids of
various colors depending on the formula of the acid.
Below is a generic acid.
HA  H+ + Acolorless
1.
pink
Describe the color change when a strong acid is added?
Indicators
Indicators are weak organic (carbon containing) acids of
various colors depending on the formula of the acid.
Below is a generic acid.
HA  H+ + Acolorless
1.
Less pink
pink
Describe the color change when a strong acid is added?
Indicators
Indicators are weak organic (carbon containing) acids of
various colors depending on the formula of the acid.
Below is a generic acid.
HA  H+ + Acolorless
pink
1.
Describe the color change when a strong acid is added? Less pink
2.
Describe the color change when a strong base is added?
Indicators
Indicators are weak organic (carbon containing) acids of
various colors depending on the formula of the acid. Below is
a generic acid.
HA  H+ + Acolorless
pink
1.
Describe the color change when a strong acid is added? Less pink
2.
Describe the color change when a strong base is added? Darker pink
Indicators
Indicators are weak organic (carbon containing) acids of
various colors depending on the formula of the acid.
Below is a generic acid.
HA  H+ + A-
colorless
pink
1.
Describe the color change when a strong acid is added? Less pink
2.
Describe the color change when a strong base is added? Darker pink
3.
Describe the color change when the pH is lowered?
Indicators
Indicators are weak organic (carbon containing) acids of
various colors depending on the formula of the acid.
Below is a generic acid.
HA  H+ + Acolorless
pink
1.
Describe the color change when a strong acid is added? Less pink
2.
Describe the color change when a strong base is added? Darker pink
3.
Describe the color change when the pH is lowered? Less pink
Indicators
Indicators are weak organic (carbon containing) acids of
various colors depending on the formula of the acid.
Below is a generic acid.
HA  H+ + Acolorless
pink
1.
Describe the color change when a strong acid is added? Less pink
2.
Describe the color change when a strong base is added? Darker pink
3.
Describe the color change when the pH is lowered? Less pink
4.
Describe the color change when the pH is raised?
Indicators
Indicators are weak organic (carbon containing) acids of
various colors depending on the formula of the acid.
Below is a generic acid.
HA  H+ + Acolorless
pink
1.
Describe the color change when a strong acid is added? Less pink
2.
Describe the color change when a strong base is added? Darker pink
3.
Describe the color change when the pH is lowered? Less pink
4.
Describe the color change when the pH is raised? Darker pink
Color versus pH of Many Different indicators
How can we make an indicator?
How can we make an indicator?
Step One
Red Cabbage
Step Two
Cook the Cabbage
Step Three
Filter the Juice
What color is the juice after filtering?
What color is the juice after filtering? The color of pH 6, 7, or
8
Colors of cabbage juice at various pH values
CH#4
REVIEW
To the left is a plot that shows the pH
of an HCl solution as a function of the
added volume of 0.011 M NaOH.
Which of the following plots would
correspond to the same titration but
using 0.022 M NaOH?
A)
Titraction of HCl and NaHO
B)
C)
Consider the following arguments for each answer
and vote again:
A. The shape of both titration curves is the same, but the
pH for titration with the stronger base should be
higher at every point on the curve.
B. The titration curve stays the same up to the
equivalence point, but the pH will be higher when
there is excess strong base.
C. The curve will be shifted to the left because only half
the volume of 0.022 M NaOH will be required to
reach the equivalence point.
Titraction of HCl and NaHO
An acetic acid (HAc) solution of
pH 3.0 is diluted by a factor of 10
with water. What is the new pH of
the solution?
A) < 4.0
Dilution of Acidic Acid
B) 4.0
C) > 4.0
Consider the following arguments for each answer and
vote again:
A. Dilution of an acetic acid solution will drive the
HAc/Ac- equilibrium toward further ionization of
HAc. Therefore, the final pH will be between 3 and 4.
B. Dilution of an acidic solution with pH 3.0 by a factor
of 10 will result in a decrease in the H3O+
concentration from 0.001 M to 0.0001 M, giving a pH
of 4.0.
C. Dilution of a weak acid solution with pH 3.0 will give
a less acidic solution than the dilution of a strong acid
solution with the same pH.
Dilution of Acidic Acid
Which of the following three
solutions would have the highest
pH?
A) 10-3 M NaOH
B) 10-6 M H2SO4
pH of HCl,H SO ,and NaOH Solutions
C) 10-12 M HCl
Consider the following arguments for each answer and
vote again:
A. Of the three solutions, only the NaOH solution is basic,
so its pH must be the highest.
B. Although H2SO4 is a strong acid, it dissociates in water
to form the base SO42-, making its pH higher than that
of the other two solutions.
C. Since HCl dissociates completely in water, the
concentration of H3O+ is 10-12 M for this solution.
Therefore, the pH of the solution is 12.
pH of HCl,H SO , and NaOH Solutions
THE END
Download
Related flashcards

Household chemicals

39 cards

Pigments

50 cards

Analytical chemistry

34 cards

Molecular physics

14 cards

Oxohalides

46 cards

Create Flashcards