Solutions

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Solutions Chapters 4.1, 4.5, and 13

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Solutions. Chapters 4.1, 4.5, and 13. Solutions. Solutions are homogeneous mixtures. Solutions are substances composed of dissolved particles (solute) in a dissolving medium (solvent). Electrolytic Properties. - PowerPoint PPT Presentation

Transcript of Solutions

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Solutions

Chapters 4.1, 4.5, and 13

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Solutions are substances composed of dissolved particles (solute) in a dissolving medium (solvent).

Solutions are homogeneous mixtures.

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Electrolytic Properties

Electrolytes – when ions are dissolved into an aqueous solution they can conduct an electric current; the more ions it breaks down into, the stronger electrolyte it is.

Nonelectrolytes – a substance that doesn’t form ions in solution, thus it doesn’t conduct an electric current.

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Strong Electrolytes = completely ionized when dissolved in water.

Strong Acids and Strong Bases are Strong Electrolytes.

Weak Electrolytes = slightly ionized solution when dissolved in water

Weak acids and bases are weak electrolytes.

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Ionic vs Molecular Compounds in Water

Ionic compounds break down into their component ions in water. Na2SO4

becomes Na+ ions and SO4-2 ions

dispursed through the water. Molecular compounds dissolve in

water, but the entire molecule stays intact. C6H12O6 dissolves and is dispersed through water, but the entire C6H12O6 molecule remains intact.

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Concentrations of Solutions

Molarity = Moles of solute / Liters of Sol’n Calculate the molarity of a solution made

by dissolving 23.4 g of sodium sulfate in enough water to form 125 mL of solution.

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Expressing Conc. Of an Electrolyte

When ionic compounds dissolve, the relative concentrations depend on the chemical formula.

1.0 M of NaCl is 1.0 M of Na+ and 1.0 M of Cl-

If I had 1.0 M of Na2SO4, what is the concentration of Na+ and SO4

-2?

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Preparation of Standard Solutions

Standard solutions = solution whose concentration is accurately known.

How much x How strong x What does it weigh?

L x mol / L x g / mol = grams required to prepare the standard

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Dilutions

You can make substances less concentrated by adding water to the solution.

Moles of solute before dilution = moles of solute after dilution

M1V1 = M2V2

How many mL of 3.0 M H2SO4 are needed

to make 450 mL of 0.10 M H2SO4?

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NaCl dissolving in water.

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The Solution Process

As individual solute ions break away from the crystal, the negatively and positively charged ions become surrounded by solvent molecules and the ionic crystal dissolves.

This is caused by Random Molecular Motion.

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“Like dissolves Like”

Polar molecules and ionic compounds tend to dissolve in polar solvents.

Nonpolar molecules dissolve in nonpolar solvents.

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Energy Changes and Solution Formation

Heat of solution ( Hsoln ) is the energy change for making a solution.

Most easily understood if broken into steps:

1. Break apart solute (endothermic)

2. Break apart solvent (endothermic)

3. Mixing solvent and solute (often exothermic)

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1. Break apart the Solute Have to overcome IMFs to break the solute into

individual components (expanding the solute). H1 >02. Break apart Solvent. Have to overcome IMFs in the solvent to make

room for the solute (expanding the solvent). H2 >0

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3. Mixing solvent and solute

Allowing the solute and solvent to interact to form the solution (often exothermic)

H3 depends on what you are mixing.

Molecules can attract each other means H3

is large and negative. Molecules can’t attract means H3 is small

and positive. This explains the rule “Like dissolves Like”

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Energy

Reactants

Solution

H1

H2

H3

Solvent +

Sep. Solute

Sep. Solute + Sep. Solvent

Size of H3 determines whether a solution will form

Hsoln

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Enthalpy (Heat) of Solution

Hsol’n = H1 + H2 + H3

Hsol’n may be positive or negative

Enthalpy (heat) of hydration, Hhyd includes steps 2 and 3.

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Types of Solvent and solutes

If Hsoln is small and positive, a solution will still form because of entropy.

Entropy = measure of disorder There are many more ways for them to

become mixed than there is for them to stay separate.

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Solution Formation, Spontaneity, and Disorder

Processes in which the energy content of the system decreases tend to occur spontaneously. (tend to be exothermic)

You let go of something, it falls to the floor Your room gets messy

Change tends to occur in the direction that leads to lower energy or enthalpy for the system.

The amount of disorder in a system a called entropy… processes in which the disorder (entropy) of the system increases tend to occur spontaneously.

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The solution process involves 2 factors:

1. Change in enthalpy

2. Change in entropy Formation of solutions is favored by the

increase in entropy that accompanies mixing… therefore a solution will form unless solute-solute or solvent-solvent IMFs are too strong relative to the solute-solvent interactions.

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Overview of factors favoring the solution process

1. Negative Hsol’n

2. Increase in entropy

3. For positive values of Hsol’n, it is the increase in entropy that outweighs the increase in energy and causes the solution process to occur.

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What is the difference between the IMFs at work in pure substances (chapter 11) and the IMFs in solutions?

Not Much! IMFs in solutions are between unlike particles (solute-solvent interactions) while IMFs in pure substances are among like particles (solute-solute or solvent-solvent)

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Saturated solutions and Solubility

The process opposite of the solution process is called crystallization.

Solubility = the amount of solute that will dissolve in a solvent at a given temperature.

What is meant by the terms: Saturated, Supersaturated, and Unsaturated

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Factors Affecting Solubility

1. Structure (Solute-Solvent Interactions)

2. Pressure

3. Temperature

4. Agitation

5. Particle Size

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Structure (Solute-Solvent Interactions)

“Like dissolves Like” Polar (hydrophillic) dissolves in polar Nonpolar (hydrophobic) dissolves in

nonpolar Miscible – two “like” liquids dissolve in

one another (Alcohol and Water) Immiscible – two “un-like” liquids do not

dissolve in one another (oil and water).

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Pressure Effects

Changing the pressure doesn’t effect the amount of solid or liquid that dissolves because they are incompressible.

It does effect gases. How would an increase in pressure

effect gas solubility? A decrease?

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Dissolving Gases

Pressure effects the amount of gas that can dissolve in a liquid.

The dissolved gas is at equilibrium with the gas above the liquid.

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The gas is at equilibrium with the dissolved gas in this solution.

The equilibrium is dynamic.

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If you increase the pressure the gas molecules dissolve faster.

The equilibrium is disturbed.

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The system reaches a new equilibrium with more gas dissolved.

This process is known as Henry’s Law.

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Henry’s Law

the amount of a gas dissolved in a solution is directly proportional to the pressure of the gas above the solution

Sg=kPg

Pg = partial pressure of the gaseous solute above the solution

Sg = concentration of the dissolved gask = constant characteristic of a particular

solution (dif. for every solution)

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Sample Henry’s Law problem

Calculate the concentration of CO2 in a soft drink that is bottled with a partial pressure of CO2 of 4.0 atm over the liquid at 25oC. The Henry’s Law constant for CO2 in water at this temperature is 3.1 x 10-2 mol/L-atm.

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Temperature Effects Increased temperature increases the rate

at which a solid dissolves. We can’t predict whether it will increase

the amount of solid that dissolves. We must read it from a graph of

experimental data.

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20 40 60 80 100

Temperature

Solubility

Where is:Saturated?

Unsaturated?Supersaturated?

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Temperature Effects

Solubility of gases always decreases with increasing temperature.

Why do you think that is?

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Small particles dissolve faster than larger particles because there is

more surface area exposed.

Aids in dispersion of the particles and makes the solute dissolve

faster.

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Ways of expressing Concentration

Molarity = moles of solute Liters of solution

% mass = Mass of solute x 100 Mass of solution

Mole fraction of component A

A = nA

nA + nB + … (Moles of component A divided by total moles

of all components)

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Molality = moles of solute Kilograms of solvent

Molality is abbreviated m Parts per million (ppm) =

mass of component in sol’n x 106 total mass of sol’n

Ways of expressing Concentration

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Sample problems

A)A solution is made by dissolving 13.5g of glucose (C6H12O6)in 0.100 kg of water.

What is the mass percentage of solute in this solution? B) A 2.5g sample of groundwater was found to contain 5.4 g of Zn+2. What is the concentration of Zn+2 in ppm?

A solution is made by dissolving 4.35g glucose in 25.0 mL of water. Calculate the molality of glucose in the solution.

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A solution of hydrochloric acid contains 36% HCl by mass. A) Calculate the mole fraction of HCl in the solution. B) calculate the molality of HCl in the solution.

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Gases are predictable

As temperature increases, solubility decreases.

Gas molecules can move fast enough to escape.

Thermal pollution.

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Vapor Pressure of Solutions

A nonvolatile solvent lowers the vapor pressure of the solution.

The molecules of the solventmust overcome the force of both the other solvent molecules and the solute molecules.

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Raoult’s Law:

Psoln = solvent x Psolvent Vapor pressure of the solution =

mole fraction of solvent x vapor pressure of the pure solvent

Applies only to an ideal solution where the solute doesn’t contribute to the vapor pressure.

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Aqueous Solution

Pure water

Water has a higher vapor pressure than a solution

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Aqueous Solution

Pure water

Water evaporates faster from for water than solution

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The water condenses faster in the solution so it should all end up there.

Aqueous Solution

Pure water

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Sample problem

Glycerin (C3H8O3) is a nonvolatile nonelectrolyte with a density of 1.26 g/mL at 25oC. Calculate the vapor pressure at 25oC of a solution made by adding 50.0mL of glycerin to 500.0 mL of water. The vapor pressure of pure water at 25oC is 23.8 torr.

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What is the composition of a pentane-hexane solution that has a vapor pressure of 350 torr at 25ºC ?

The vapor pressures at 25ºC are pentane 511 torr hexane 150 torr.

What is the composition of the vapor?

Review Question

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Colligative Properties

Because dissolved particles affect vapor pressure, they also affect phase changes.

Colligative Properties are dependent on the number of solute particles, but not their identity.Boiling-Point elevationFreezing-Point depressionOsmotic Pressure

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Boiling-Point Elevation

Since non-volatile solutes lower the vapor pressure, the Boiling Point will rise. Why?

The equation is: T = Kbmsolute

T is the change in the boiling point Kb is a constant determined by the

solvent.msolute is the molality of the solute

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Freezing-Point Depression

Because a non-volatile solute lowers the vapor pressure of the solution it lowers the freezing point. Why?

The equation is: T = Kfmsolute

T is the change in the freezing point Kf is a constant determined by the

solventmsolute is the molality of the solute

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1 atm

Vapor Pressure of solution

Vapor Pressure of pure water

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1 atm

Freezing and boiling points of water

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1 atm

Freezing and boiling points of solution

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1 atm

TfTb

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Electrolytes in solution Since colligative properties only depend

on the number of molecules. Ionic compounds should have a bigger

effect. When they dissolve they dissociate. Individual Na and Cl ions fall apart. 1 mole of NaCl makes 2 moles of ions. 1mole Al(NO3)3 makes 4 moles ions.

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Electrolytes have a bigger impact on on melting and freezing points per mole because they make more pieces.

Relationship is expressed using the van’t Hoff factor i

i = Moles of particles in solution

Moles of solute dissolved The expected value can be determined

from the formula.

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The actual value is usually less because

At any given instant some of the ions in solution will be paired.

Ion pairing increases with concentration.

i decreases with in concentration. We can change our formulas to

H = iKm

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Determination of Molar Mass

A solution of an unknown nonvolatile nonelectrolyte was prepared by dissolving 0.250g of the substance in 40.0g of CCl4. The boiling point of the resultant solution was 0.347oC higher than that of the pure solvent. Kb is 5.02 oC/m. Calculate the molar mass of the solute.

88.0 g/mol

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Another type of problem…

Determine the formula and molecular mass for an organic compound from the following info: 5.00 g of the solid was dissolved in 100.0g of benzene. The boiling temp. of the solution was 82.3oC. The solute is 46.7%N, 6.67%H, 26.7%O, and the remainder is carbon. The boiling temp. of pure benzene is 80.2oC; Kb=2.53oC kg/mol

A) Determine the molecular mass and molecular formula of the solid. B) Determine the mole fraction of the solid C) If the density of the solution is 0.8989 g/mL, what is the molarity of the solution?

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Colloidal Dispersions (Colloids)

Tiny particles suspended in some medium

Subject to coagulation or clumping Ex: paint, glue, gelatin, smoke…

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Tyndall Effect

The scattering of visible light by colloidal particles or suspensions.

Solutions do not exhibit the Tyndall Effect.

Why is it usually recommended that drivers use low beams when driving under foggy conditions at night?