SOLUTIONS. TOPICS Solution Formation Solubility Solution concentration.
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Transcript of SOLUTIONS. TOPICS Solution Formation Solubility Solution concentration.
STANDARDS ADDRESSED 6. Solutions are homogeneous mixtures of two or more
substances. As a basis for understanding this concept:
a. Students know the definition of solute and solvent
b. Students know how to describe the dissolving process at the
molecular level by using the concept of random molecular
motion
c. Students know temperature, pressure, and surface area
affect the dissolving process
d. Students know how to calculate the concentration of a solute
in terms of grams per liter, molarity, parts per million, and
percent composition.
OBJECTIVE
By the end of the lesson, the student will: Understand the dissolving process at the molecular level. Understand how the structure of the water molecule
allows it to dissolve many substances Know the factors which influence:
solubility and why dissolving rate and why
Know how to calculate the concentration of a solution
SOLUTION FORMATION
Solution: a homogeneous mixture of two or more substances Solvent: a substance which dissolves another substance
Solute: the dissolved particles; whatever is being dissolved
Solution Formation Vocab:
Soluble: a substance that dissolves in a solvent
Insoluble: a substance that does not dissolve in a solvent
Examples of Solutions
Gas in Liquid: Carbonic Acid (carbonation): CO2 in H2O
Liquid in Liquid Vinegar: Acetic Acid in water
Solid in Solid Brass: copper and zinc
Animation: Dissolving Process
http://www.youtube.com/watch?v=EBfGcTAJF4o (53 sec)
http://www.youtube.com/watch?v=CLHP4r0E7hg&feature=related (42 sec)
http://www.youtube.com/watch?v=xdedxfhcpWo&feature=related (1:35)
The structure of the water molecule creates a polar molecule due to the differences in electronegativities of the oxygen and hydrogen atoms.
Partial negative charge on the oxygen attracts the partial positive charge on the neighboring molecule. Molecules arrange themselves so as to have opposite charges aligned, pulling it away from its ‘structure’, “one ion or molecule at a time”
DISSOLVING PROCESS:
Once released, the ions are surrounded by water molecules.
The dissolving process is reversible
The dissolved solute moves around in the solution and when it
comes into contact with un-dissolved solute particles it re-
crystallizes, meaning it returns to the solid state. And the
process repeats itself.
NaCl Na+(aq) + Cl- (aq)
Saturation
When the rates of the dissolving and
recrystallization become the same, the
solution is saturated at that temperature. A
dynamic equilibrium is reached,
meaning the rate of the forward reaction is the
same as the reverse reaction.
FACTORS AFFECTING DISSOLVING RATE
Where does dissolving occur?
On the surface of a substance To increase dissolving rate, contact between
solvent and solute must increase 3 common ways to increase the collisions between solute and
solvent Temperature Amount of Surface Area Exposed Agitation
Temperature
Temperature: raising the temperature increases the kinetic energy of the particles, resulting in more frequent and forceful collisions.
Amount of Surface Area Exposed
Amount of Surface Area Exposed: breaking the solute into smaller pieces increases its surface area. A greater surface area allows more collisions to occur and therefore, faster dissolving
Agitation (mixing, stirring, etc.)
Agitation: stirring moves dissolved solute particles away from the contact surfaces more quickly and thereby allows new collisions between solute and solvent particles to occur. Without stirring, solvated particles move away from the contact areas slowly.
SOLUBILITY the maximum amount of solute that will dissolve in a
given amount of solvent at a specified temperature and pressure
Saturated solution: contains the maximum amount of solute. You cannot dissolve any more solute in the solvent.
Unsaturated solution: contains less than the maximum amount of solute. You can dissolve more solute in the solvent.
Supersaturated solution
contains more than the normal maximum amount of solute. This is usually achieved by heating the solution in order to dissolve more solute, then the solution is cooled. This makes a supersaturated solution. (ie.Rock candy; hot mineral springs)
3 Factors Affecting Solubility 1. Temperature:
many substances are more soluble at high temperatures than low temperatures. Exception: Gases
Solubility of Gases with Increased Temperature
As Temperature increases, Solubility of gases decreases
Pressure affect on gases
2. Pressure: affects the solubility of
gaseous solutes. The solubility of a gas in any solvent increases as the pressure above the solution increases
3 Factors Affecting Solubility
2. Pressure: affects the solubility of gaseous solutes. The solubility of a gas in any solvent increases as the pressure above the solution increases
3 Factors Affecting Solubility 2. Pressure: affects
the solubility of gaseous solutes. The solubility of a gas in any solvent increases as the pressure above the solution increases
3 Factors Affecting Solubility 3. Intermolecular forces: “like dissolves like;”
meaning polar substances dissolve polar substances and non-polar substances dissolve non-polar substances.
Miscibility: the ability to mix without separating into two phases
Miscible: substances that mix togetherie. Vinegar and water
Immiscible: substances that do not mix together
ie. Oil and water
SOLUTION CONCENTRATION: the amount of solute in a solution
Dilute solution: contains small amounts of solute (dissolved particles)
Concentrated solution: contains large amounts of solute
MASS PERCENT:the number of grams of solute per 100 grams of solution
Mass Percent = mass of solute x 100%
mass of solution
EXAMPLE 1
Calculate the mass percent of NaCl in a solution containing 15.3 g of NaCl and 155.0 g water. (8.98%NaCl)
EXAMPLE 2
Calculate the mass percent of a solution containing 27.5 g of ethanol and 175 mL of water. The density of water is 1.0 g/mL. 13.6%)
Example 3
You have 1,500 g of a bleach solution. The percent by mass of the solute sodium hypochlorite, NaOCl, is 3.62%. How many grams of NaOCl are in the solution? (54.3 g)
MOLARITY (M): the number of moles of solute dissolved per liter of solution.
Note: Here, volume refers to the total volume of the solution,not the volume of the solvent.
Molarity (M) = moles of solute
liters of solution
M is pronounced “molar”
EXAMPLE 1
Calculate the molarity if water is added to 2.0 mol of
glucose to give 5.0 L of solution. (0.40 M)
EXAMPLE 2 A solution contains 0.900 g of NaCl per 100 mL of solution.
What is the molarity? (0.154 M)
SOLUTION DILUTION:
adding solvent to a concentrated (stock) solution to make a less concentrated (dilute)solution
Moles Before= Moles After
Number moles before dilution = number moles after
dilution
Use the dilution equation
M1 x V1 = M2 x V2
QUESTION 1
How many milliliters of 12.0 M HCl are needed
to prepare 750. ml of 0.250 M HCl? (15.6 mL)
Example 3 What is the molarity of the resulting solution when
the following mixture is prepared? 150.0 mL of water is added to 55.0 mL of 6.50 M NaOH. (1.74 M)
Example 4 How much water must be added to 125 mL of a
4.50 M NaCl solution to produce a 2.75 M solution? (80. mL)
SUMMARY OF STEPS
1. Construct a chart of values
2. Rearrange equation so that unknown is
isolated
3. Convert grams to moles, if necessary
4. Calculate and report to correct # of sig. figs!
SOLUTION STOICHIOMETRY
There are several different ways to solve solution stoichiometry problems. Here are a few guidelines/suggestions to get you started:
Write a balanced chemical equation use Molarity as mol…
L
***allows conversion between moles and volume or concentration and volume
use mol-mol ratio if necessary there may be limiting reagent problems…in this
case determine how many moles there are for each substance
sample mini road map:
mL→L→mol→mol→g
L
…from moles you can go to L (volume) or grams (mass)…or even particles via Avogadro’s #!
Example 1 1) Calculate the mass of solid NaCl that must be added
to 1.50 L of a 0.100 M AgNO3 solution to precipitate all the Ag+ ions in the form of AgCl
Example 2 2) How many milliliters of 0.0500M Pb(NO3)2 are
needed to react with 2.00L of a 0.0250M Na2SO4 solution in order to produce a precipitate? How many grams of precipitate are formed?