MIXTURES - Brim's Science Stuff



1. SC7. Students will characterize the properties that describe solutions…

1. a. Explain the process of dissolving in terms of solute/solvent interactions:

2. • Observe factors that effect the rate at which a solute dissolves in a specific solvent,

3. • Express concentrations as molarities,

4. • Prepare and properly label solutions of specified molar concentration,

5. • Relate molality to colligative properties.

SOLUTIONS

Chapter 14

MIXTURES

There are three basic types of mixtures: solutions, colloids, and suspensions. The particle size determines the type of mixture. See the table below for differences between the three types.

SOLUTIONS COLLOIDS SUSPENSIONS

Particle ions, atoms, small large molecules or large particles or

Type molecules particles aggregates

Particle 0.1 – 1 nm 1 – 1000 nm >1000 nm

Size

Effect of stable, will not stable, will not unstable, will

Gravity settle out settle out separate and settle

out upon standing

Effect of no effect, light scatters light scatters light

Light passes through Tyndall Effect Tyndall Effect

Filtration cannot be filtered cannot be filtered particles can be

filtered out

Examples sugar water, salt whipped cream, oil and water

water, any ionic milk, fog, jello, dirt and water

substance in water blood, aerosol cornstarch and water

(see page 477)

TYNDALL EFFECT - particles dispersed in a mixture are big enough to scatter light.

SOLUTIONS

A solution is a homogeneous mixture of two or more substances in a single physical state. (That means when a solid dissolves in a liquid, the solid is acting like a liquid – hence the bit about the single phase.)

Solutions form when particles of solute are interspersed evenly throughout the solvent.

SOLVENT: the substance doing the dissolving and in the greater amount. The most common solvent is water. Called AQUEOUS.

SOLUTE: the substance being dissolved and in less abundant.

SOLUBLE: when a substance dissolves into another substance

INSOLUBLE: when a substance does not dissolve into another substance.

MISCIBLE: two liquids that are soluble in each other.

IMMISCIBLE: two liquids that are insoluble in each other.

There are nine basic types of solutions based on the phases of the solvent. See page 479 for other examples

Liquid Solutions:

1. solid in liquid (salt water)

2. liquid in liquid (vinegar)

3. gas in liquid (carbonated drink)

Gaseous Solutions:

All gaseous mixtures are solutions

1. solid in gas (soot in air)

2. liquid in gas (humid air)

3. gas in gas (air)

Solid Solutions

Homogeneous mixtures of solids are usually made from liquid solutions that have been mixed and then solidified (frozen)

1. solid in solid (alloys – brass, bronze)

2. liquid in solid (dental fillings)

3. gas in solid (charcoal gas mask)

SOLUTION EQUILIBRIUM

A solution is in dynamic equilibrium when the number of solute particles returning to the crystal surface is equal to the number of solute particles leaving the crystal surface. Saturation is dependent upon temperature and pressure.

A SATURATED solution is a solution with the maximum amount of solute dissolved at a given temperature (Any more added goes to the bottom). It has reached dynamic equilibrium.

An UNSATURATED solution has the ability to dissolve more solute at a given temperature

If a solution is SUPERSATURATED, then a hot solution is saturated and cooled. An unstable condition results because the solution holds more solute that it normally does at a given temperature.

VIDEO: Solutions (Disc chapter 20)

Be able to answer the following:

1. What are ion-dipole attractions? During the process of solvation, how do ion-dipole attractions compare to the ionic bonds inside an ionic crystal?

2. Describe the processes that are balanced when solubility equilibrium has been reached. If more solute is added to a solution in equilibrium, will this always result in more dissolved solute particles in the solution? Explain.

3. How does a solution become supersaturated? After crystals precipitate out of a supersaturated solution, would you still call it supersaturated? Explain.

THE DISSOLVING PROCESS:

How a solution forms

Dissolving occurs when the solute is pulled apart by the solvent. This takes place at the surface of the solute. The solvent surrounds the solute. This process of surrounding the solute is called SOLVATION. When the surrounding is done by water, this is called HYDRATION, a particular type of solvation.

When ionic compounds separate into their ions in a solvent, DISSOCIATION occurs.

Combinations of solute and solvent are based on the properties of each. There are four main types:

1. Polar Solvent – Polar Solute:

The polar solvent is attracted to the polar solute. The solvent gradually surrounds the solute. The particles attach themselves due to polar attraction. Solvation occurs. Like dissolves like.

2. Polar Solvent – Nonpolar Solute:

Polar solvent particles are attracted to each other and not the solute. Solvation does not occur and a solution is unlikely.

3. Nonpolar Solvent – Polar Solute:

Nonpolar solvent particles have little attraction to the polar solute. Solvation does not occur and a solution is unlikely.

4. Nonpolar solvent – nonpolar solute:

Random motion of solute particles causes them to leave the surface of the solute and become evenly dispersed in the nonpolar solvent. Solvation occurs. Like dissolves like.

VIDEO: Polar and Nonpolar Solvents (Disc chapter 30)

Be able to answer the following:

1. Why are some liquids immiscible? Explain in terms of intermolecular forces.

2. How would you predict whether carbon tetrachloride is a polar or nonpolar solvent? What evidence have you observed that supports your prediction?

3. Based on chemical formula alone, can you tell whether iodine, I2, is a polar or nonpolar solute? What evidence have you observed that supports your prediction?

4. Based on the chemical formula alone, can you tell whether copper (II) chloride, CuCl2, is a polar or nonpolar solute? What evidence have you observed that supports your prediction?

SOLUBILITY

Solubility is how much solute can dissolve in a given amount of solvent. It is measured in g/L or mol/L. It is usually the grams of solute per 100g of solvent. A CONCENTRATED solution is said to have a high ratio of solute to solvent. A DILUTE solution is the opposite of this.

For solids: solubility usually increases with increased temperature

For gases: solubility usually decreases with increased temperature

Factors Affecting Solubility

1. Nature of the solvent and solute - “like dissolves like”. This means that polar solvent dissolves a polar solute and nonpolar solvent dissolves nonpolar solute. But polar does not dissolve nonpolar.

2. Temperature - increase the temperature and solubility increases (except gases).

3. Pressure - increase the pressure and you increase solubility (only with gases).

4. How much is already dissolved.

Factors Affecting the RATE of Solubility

(How fast something will dissolve)

1. Agitation (shaking, stirring)

2. Increased temperature (except gases)

3. Smaller Particle size

These three allow more solvent to come into contact with the solute faster.

Heat of Solution

Energy is released and absorbed as substances dissolve. The solute particles must separate and the solvent particles must separate to make room for the solute. Both of these processes take energy – thus this part is endothermic.

When the solute and the solvent mix, the particles are attracted to each other. When this happens, energy is released – thus this part is exothermic. The overall change in energy is the heat of solution. Most solutions are endothermic.

CONCENTRATION

Concentration is how many particles of a solute are dissolved. It is NOT dependent upon the sample size and it can be measured. There are many units associated with concentration. Below are a few:

Grams/100.0 grams measures solubility

Parts per million (ppm) measures small concentrations

Parts per billion (ppb) measures pollutants in small concentrations

Molarity used in lab chemistry

Molality used for special calculation such as freezing point depression

We will learn to calculate molarity, molality, mass percent, and mole fraction

MOLARITY, M

This is the ratio between the moles of dissolved substance and the volume of the solution expressed in liters. It is the most useful measurement of concentration.

Molarity, M = moles of solute

volume of solution in liters

A one-molar (1M) solution of hydrochloric acid contains one mole of hydrochloric acid in one liter of water. (Which means it contains 36.46g of HCl in 1 liter of water.)

Sample Problem:

Sandy dissolves 45.0 g of NaCl in 2.5 liters of solution. What is the concentration in molarity of NaCl?

Mass = 45.0 g 2 steps: mol = 45.0 g NaCl 1 mol NaCl = 0.770 mol

V = 2.5 L 58.44 g NaCl

Molar Mass = 58.44 g/mol

M = mol = 0.770 mol = 0.31M

L 2.5L

1 step: M = 45.0 g NaCl 1 mol NaCl = 0.31M

2.5 L 58.44 g NaCl

PRACTICE:

What is the molarity of 58.5g of NaCl dissolved in 2.0L of solution?

MOLALITY, m

This is concentration expressed in terms of moles of solute per kilogram of solvent. Volume is not a factor.

Molality, m = moles solute

Kg solvent

A 1.0 molal aqueous sugar solution: 1 mole sugar

1 kg water

Sample Problem:

Calculate the molality of 98.0g RbBr in 0.824 Kg water.

m = 98.0 g RbBr 1 mol RbBr = 0.719m

0.824Kg H2O 165.38g RbBr

PRACTICE:

Calculate the molality of 85.2g SnBr2 in 140.0g water.

MASS PERCENT

Scientists frequently express the concentration of solutions in mass percent.

Mass % = g of solute x 100

g of solute + g of solvent

A 5% solution of sodium hydroxide contains 5g NaOH in each 100g of solution (95g solvent and 5g of solute).

VOLUME PERCENT

Consumer products frequently express their concentration of solutions in volume percent.

Volume % = L of solute x 100

L of solute + L of solvent

A 5% solution of sodium hydroxide contains .05L NaOH in each 1.0L of solution (0.95L solvent and 0.05L of solute).

MOLE FRACTION

The concentration of solution can also be expressed in mole fractions.

Mole Fraction = mole of solute

mole of solute + mole of solvent

PREPARING AND DILUTING SOLUTIONS

A 3M solution of hydrochloric acid is not bought but made from 12M stock solutions. In addition, a 1.0M solution of sodium hydroxide is made from a calculated amount of solid sodium hydroxide added to water. It is important to know how to make different concentration of solutions.

Prepare 1.0 liter of a 1.0M aqueous solution of NaOH

M = mol 1.0M = mol mol = (1.0M)(1.0L) = 1.0mol

L 1.0L

1.0mol NaOH 40.00g NaOH = 40.00g NaOH So, you will put 40.00g

1 mol NaOH NaOH in a flask, add 1.0L

water, and mix well.

Prepare 1.0L of a 6.0M aqueous solution of KCl

To dilute a solution, you can form a ratio between molarity and volume.

M1V1 = M2V2

Diluting a 12.0M solution of HCl to 6.0M HCl

What volume would you use to make 0.500L of 6.0M HCl solution?

V1 = M2V2 = (6.0M HCl)(0.500L HCl) = 0.250L HCl

M1 (12.0M)

Thus you would need 0.250L HCl and 0.250L water to make the solution.

Preparing a 0.1M solution of HCl

What volume would you use of 12.0M HCl to make 1.0L of 0.1M HCl?

COLLIGATIVE PROPERTIES

“colligative” – depending upon the collection

These are properties that depend on the concentration (the number of) of the solute particles. They are independent of the type of solute particle.

Three Factors:

1. Vapor pressure reduction

2. Boiling Point Elevation

3. Freezing Point Depression

VIDEO: Colligative Properties (Disc chapter 32)

Think about the following:

1. Support one of the following hypotheses about why it helps to add salt to the water before cooking the pasta: a) The salt brings the water to a boil sooner; or b) The salt brings the water to a boil at a higher temperature.

2. Which antifreeze solution boils at the highest temperature? Which solution would you want in your car on a hot summer day? Explain your reasoning.

3. On cold winter days, unprotected radiator water can freeze and expand, which can ruin an engine by cracking the engine block. Which antifreeze mixture would you want in your car on a cold winter day?

Vapor Pressure Reduction

A Review: Vapor pressure arises because some molecules of pure liquid leave the liquid surface and enter the gas phase. At the same time, condensation occurs. If the rate of condensation and the rate of vaporization are equal, then the system is in equilibrium. The gas pressure resulting from the vapor molecules over the liquid is VAPOR PRESSURE.

The vapor pressure of a solvent containing a non-volatile solute is LOWER than the vapor pressure of the pure solvent. It is NOT dependent upon the type (identity) of the solute.

WHY: The solute takes up space at the surface of the liquid and prevents some solvent molecules from leaving the liquid. Now, more molecules leave the gas phase than enter it and therefore, the vapor pressure is lower.

Boiling Point Elevation

A Review: The boiling point of a substance is the temperature at which the vapor pressure of a liquid is equal to the external pressure on its surface (usually atmospheric pressure).

An addition of solute lowers that vapor pressure, therefore a higher temperature is necessary to get the vapor pressure up to the external (atmospheric) pressure so that the solution will boil. This is the boiling point elevation.

(Tb is the difference between the boiling point of the solution and the boiling point of the pure solvent. It is directly proportional to the number of solute particles per mole of solvent particles.

Dissociation factor molality

(Tb = (Kb)(df)(m)

constant (molal BP elevation constant)

The “df” (dissociation factor) is how many particles the solute breaks up into.

For a nonelectrolyte, the df is 1)

Water’s “Kb” value is 0.515(C/m

Freezing Point Depression

A Review: The freezing point of a substance is the temperature at which the vapor pressure of the solid and liquid phases are the same.

When a dissolved solute lowers the freezing point of its solution, you have freezing point depression. Tf is directly proportional to the molality of the solution.

Tfp = (Kfp)(df) (m)

Water’s “Kfp” value is 1.853(C/m

CALCULATING FOR A NON-IONIC SOLUTE

Determine the freezing and boiling point for 85.0g of sugar (C12H22011) in 392g water.

Need these things: 1. Molality

2. 2. number of particles (should be one)

3. 3. Change in temp

392 g H2O 1 kg = 0.392Kg df = 1 (non-ionic solute)

1000 g

m = 85.0 g C12H22011 1 mol C12H22011 = 0.633m

0.392 Kg H2O 342.34 g C12H22011

(Tfp = 1.853(C 1 0.633 m = 1.17(C (Tbp = 0.515(C 1 0.633 m = 0.326(C

m m

0.00(C – 1.17(C = -1.17(C 100.000(C + 0.326(C = 100.326(C

PRACTICE:

What is the freezing point for a solution of 210.0g of glycerol dissolved in 350.0g of water? (The molecular mass of glycerol is 92.11 g/mol)

CALCULATING FOR AN IONIC SOLUTE

Determine the freezing and boiling point for 21.6 g NiSO4 in 100.0 g water.

Need these things: 1. molality

2. number of ions

3. change in temp

100.0g H2O 1 kg = 0.1000 Kg

1000 g

m = 21.6 g NiSO4 1 mol NiSO4 = 1.396 = 1.40 m

0.1000 Kg H2O 154.76 g NiSO4

There are two ions, Ni+2 and SO4-2, so the df is 2

(Tfp = 1.853(C 2 1.40 m = 5.19(C (Tbp = 0.515(C 2 1.40 m = 1.44(C

m m

0.00(C – 5.19(C = -5.19(C 100.00(C + 1.44(C = 101.44(C

PRACTICE:

What is the boiling point of a solution containing 34.3 g of Mg(NO3)2 dissolved in 0.107 kg of water? (The formula mass of magnesium nitrate is 148.32 g/mol)

DETERMINING MOLAR MASS

Colligative properties provide a useful means to experimentally determine the molar mass (molecular mass in one mole) of an unknown substance.

Steps: 1. Solve for molality.

2. Solve for moles of solute.

3. Solve for molar mass

EXAMPLE:

A 10.0 g sample of an unknown organic compound is dissolved in 0.100 kg water. The boiling point of the solution is elevated to 0.433oC above the normal boiling point of water. What is the molar mass?

Step One: (Tbp = Kbp df m m = (Tbp = 0.433oC = 0.841 m (mol/kg)

Assume df = 1 Kbp 0.515oC/m

Step Two: m = mol solute mol solute = m x kg solvent

Kg solvent = 0.841 mol/kg x 0.100 kg

= 0.0841 mol

Step Three: mol solute = mass solute

molar mass solute

molar mass = mass solute = 10.0 g = 118.90606 = 119 g/mol

mol solute 0.0841 mol

PRACTICE:

A solution of 3.39 g of an unknown in 10.00 g of water has a freezing point of –7.31oC. The solution is a nonelectrolyte. What is the molar mass of the compound? (The molal freezing point constant is 1.853(C/m.)

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