Solutions

Chapters 20 and 21

Properties of Solutions, Suspensions, and

Colloids

Solutions

Solutions: a homogeneous mixture of two or more substances in a single phase of matter.

In a simple solution where, for example, salt is dissolved in water, the particles of one substance are randomly mixed with the particles of another substance.

The dissolving medium (water) in a solution is called the solvent.

The substance dissolved (salt) in a solution is called the solute.

The solute is generally designated as that component of a solution that is of lesser quantity.

If we had a mixture of 25 mL of ethanol and 75 mL of water, the ethanol would be the solute and water would be the solvent.

If we had a 50% to 50% ratio, it would be unnecessary to designate solvent or solution.

4 simple solution situations can be considered.

When deciding what type of solvent to use with a given solute it is important to identify what types of substances you have.

1. Ionic or Polar (partial + or – charges)

2. Nonpolar (equal sharing of e-) Remember the rule:

Like Dissolves Like

Solvent-Solute Combinations

Remember:

LIKE DISSOLVES LIKE

Solvent Type Solute Type Is solution likely?

Polar Polar Yes

Polar Nonpolar No

Nonpolar Polar No

Nonpolar Nonpolar Yes

The Dissolving Process

Water is a polar solvent and is attracted to polar solutes.

Salt is polar (ionic). Water molecules

surround and isolate the surface ions. The ions become hydrated and move away from each other in a process called dissociation.

Suspensions

If the particles in a solvent are so large that they settle out unless the mixture is constantly stirred or agitated, the mixture is called a suspension.

Think of a sedimentator (models a river bed).

If left undisturbed, the larger, denser particles sink due to gravity. Particles over 1000nm in diameter form suspensions.

These particles can be filtered out of the heterogeneous mixtures.

Remember our snow globe lab?

Colloids Particles that are intermediate in

size between those in solutions and suspensions form mixtures known as colloidal dispersions.

Particles between 1nm and 1000 nm in diameter may form colloids. After the larger particles settle out (suspensions), the water may still be cloudy because colloidal particles remain dispersed in the water.

Milk is an example of a colloid.

Suspensions Colloids Solutions

The Tyndall Effect

Many colloids appear homogeneous because the individual particles cannot be seen. The particles are, however, large enough to scatter light.

Tyndall effect is a property that can be used to distinguish between a solution and a colloid.

When a laser is passed through a solution and a colloid at the same

time, it is evident which glass contains the colloid.

Solutions Colloids SuspensionsHomogeneous Heterogeneous Heterogeneous

Particle size:

0.01-1 nm; can be atoms, ions, molecules

Particle size:

1-1000 nm, dispersed; can be large molecules

Particle size:

Over 1000 nm, suspended; can be large particles

Do not separate on standing

Do not separate on standing

Particles settle out

Cannot be separated by filtration

Cannot be separated by filtration

Can be separated by filtration

Do not scatter light

Scatter light

(Tyndall effect)

Not transparent

May scatter lite

Solution Concentration

Molarity is simply a measure of the "strength" of a solution.  A solution that we would call "strong" would have a higher molarity than one that we would call "weak".

If you ever made or drank a liquid made from a powdered mix, such as Kool-Aid or hot cocoa, you probably are familiar with the difference between what is called a "weak" solution or a "strong" solution. 

To make Kool-Aid of "normal" strength = 4 scoops of

powder-----------------------

                                              2 quarts of water                                                                        To make Kool-Aid twice the "normal" strength =

8 scoops of powder  4 scoops of powder -------------------- or --------------------  

2 quarts of water   1 quart water                                                               

Solution Concentration

Molarity:*One-molar (M) = 1 mole solute

1Liter solvent One mole of NaCl (22.99 + 35.45

= 58.44 grams) is dissolved in enough water (1 Liter) to make a 1M solution.

Molarity Calculations Calculate the molarity of:

*35.2 grams of CO2 in 500. mL.

Step 1: convert 35.2 g of CO2 into moles

35.2g (1 mole ) = 0.800 mol 1 (44.01 g)Step 2: divide moles by volume in liters

0.800 mol CO2 = 1.60 M CO2

0.500 L

Molarity Calculations

One of the most common molarity calculations is to find the grams of solute necessary to produce a known quantity of a desired molarity.

Your teachers do this for most solutions used in lab because we purchase our solutes in salt form. It is more economical to do a little math and make the solutions ourselves.

Calculate grams of NaOH needed to make 0.500 L of a 1.50 M NaOH solution.

Since M = moles of solute : liters of solution

1.50molNaOH( 0.500L)(40.00g) 1L ( 1 ) ( 1mol ) =

30.0 gNaOHneeded

Not as common but testable is to calculate volume if grams or moles of solute is known for a desired molarity.

Calculate volume of solution if 20.0 g of NaOH is used to produce a 0.750 M NaOH solution.

20.0 g NaOH ( 1mol ) ( 1L ) 1 (40.00g) (0.750mol)

=0.667 Lsolution

Making Diluted Solutions

Often in chemistry, you will have to make a dilute solution using a more concentrated solution.

Use the following equation:

M1V1 = M2V2

M1 and M2 are the initial and final molar solutions.(M1=highest molarity)

V1 and V2 are the initial and final volumes of solutions.

Initial M and V are based on the most concentrated substance.

Final M and V are based on the diluted substance desired.

When calculating sulfuric acid, H2SO4, you must adjust M1 and/or M2 because sulfuric acid dissociates into two moles of H+ ions. Multiply molarities by 2.

EX: (12M x 2) = 24M

If you needed 200. mL of a 0.10 M solution, how many mL would you use of the following?

1. 1.0 M HC2H3O2

EX: M1V1 = M2V2

M1 = 1.0 M HC2H3O2 (more

concentrated)

V1 = ?

M2 = 0.10 M HC2H3O2 (diluted)

V2 = 200. mL

Isolate unknown (V1) from M1V1 = M2V2

because you are finding the volume of the higher molarity solution you will need to make the diluted solution.

V1 = M2V2

M1

0.10M HC2H3O2 x 200. mL =

1.0MHC2H3O2

20. mL 1.0 M HC2H3O2

180 mL of H2O

In order to obtain a 200. mL volume of the diluted solution you would:*Pipet 20 mL of 1.0 M HC2H3O2 into a 200 mL to 250 mL flask that contains 180. mL of water. (final volume is 200. mL)

*Always add acid to water.

Solubility Solubility is defined as the amount

of a substance that can be dissolved in a given quantity of solvent.

Any substance whose solubility is less than 0.01 mol/L will be referred to as insoluble.

We can predict whether a precipitate (insoluble substance) will form when solutions are mixed if we know the solubilities of different substances.

Experimental observations have led to the development of a set of empirical solubility rules for ionic compounds.

EX: Experiments demonstrate that all ionic compounds that contain the nitrate anion, NO3

-, are soluble in water.

Solubility Rules

1. All common compounds of Group I and ammonium ions are soluble.

2. All nitrates, acetates, and chlorates are soluble.

3. All binary compounds of the halogens (other than F) with metals are soluble, except those of Ag, Hg(I), and Pb. (Pb halides are soluble in hot water.)

4. All sulfates are soluble, except those of barium, strontium, calcium, lead, silver, and mercury (I). The latter three are slightly soluble.

5. Except for rule 1, carbonates, hydroxides, oxides, silicates, and phosphates are insoluble.

6. Sulfides are insoluble except for calcium, barium, strontium, magnesium, sodium, potassium, and ammonium.

Ionic Equations: An ionic equation is a chemical

equation in which electrolytes (soluble ions that conduct electricity) are written as dissociated ions. Ionic equations are used for single and double replacement reactions which occur in aqueous solutions.

In an aqueous reaction ions that are found as both reactants and products are not part of a reaction. They are termed spectator ions and essentially cancel out of the ionic equation.

To write net ionic equations, follow these simple rules:

1. Write a balanced equation.2. Repeat the equation with

reactant(s) and product(s) dissociated where appropriate.

– Salts: written in ionic form if soluble, and in undissociated form if insoluble. *Know the solubility rules.

EX: KCl K+ + Cl-

– Strong Acids/Bases: written in dissociated form. HCl, HBr, HI, HClO3, HClO4, H2SO4, HNO3;

*Bases (OH-) -Follow solubility rules

– Weak Acids/Bases: written in undissociated form. EX: HF, H3PO4, Mg(OH)2(s)

– Oxides: Oxides are always written in molecular or undissociated form. EX: MgO(s), H2O(l)

– Exception: Group 1 metal oxides– Gases: Gases are always written

in molecular form.EX: SO2(g), NH3, H2, O2

3. Cancel all spectator ions and rewrite the remaining net ionic equation.

Step 1:AgNO3 + NaCl AgCl + NaNO3

Step 2:Ag+ + NO3

- + Na++ Cl- AgCl(s) + Na+ + NO3

-

Step 3:Ag+ + Cl- AgCl(s)

What Does This Represent?

NaCl(aq) NaCl(aq)

CCCCCCC

Saline

Saline

Over

The 7 Seas

Colligative Properties

Colligative comes from the Greek word kolligativ meaning glue together.

We use this term for the properties of substances (solutes and solvents) together.

Colligative properties of solutions is used to describe the effects of antifreeze/summer coolant.

Saturated Solutions A solution at equilibrium with undissolved

solute is said to be saturated. Additional solute will not dissolve if added

to this solution. It is possible to dissolve less solute than

needed to form a saturated solution. These solutions are unsaturated.

A supersaturated solution can be made by dissolving the solute under high temps and then carefully cooling them. These are unstable solutions.

Saturated Solution

Supersaturated Solution

Factors Affecting Solubility

Solubility depends on the nature of both the solvents and solutes, temperature, and for gases, on pressure.

The solubility of most solid solutes in water increases as the temp of the solution increases.

This means that more sucrose C12H22O11 can be dissolved in hot water than cold, the basis for making “rock candy”.

The graph

represents the

solubility of substances, including NaCl,

NaNO3, and KNO3 at different temps.

Notice that when

temp increases,

solubility increases for most substances.

Solubility of Gases

Solubility of gases increases if the pressure is increased and/or the temperature decreases.

Think about the carbonation in a warm soda vs a cold soda. The cold soda stays carbonated longer.

Based on the solubility curve, the solubility of NH3 and SO2 (both gases) decreases as temperature increases.

Molality

Recall the units for Molarity:moles

LMolality is the measure of the number of

moles of a solute per 1000g of solvent.moles solute

1000 g ( 1kg) solvent Molality is best used to describe colligitive

properties and is represented by m.

Boiling Point and Freezing Point

Review the phase diagram of a pure substance.

How will the phase diagram of a solution (freezing and boiling points) differ from those of a pure solvent?

The addition of a nonvolatile solute will require a higher temperature in which to reach boiling point, thus:

Boiling point elevation The addition of a nonvolatile solute will

require a lower temperature in which to reach freezing point, thus

Freezing point depression

Pure water Water with NaCl

The water with the solute of NaCl has fewer liquid molecules becoming gases.

This will increase the temp needed to change the state from (l) (g)

Calculating Freezing and Boiling Points

The following table contains the molal (K) Boiling Point Elevations, Kb, and Freezing Point Depressions, Kf.

Solvent Normal boiling pt (°C)

Kb

(°C/m)

Normal freezing pt (°C)

Kf

(°C/m)

Water 100.0 0.52 0.0 1.86

Benzene 80.1 2.53 5.5 5.12

Ethanol 78.4 1.22 -114.6 1.99

CCl4 76.8 5.02 -22.3 29.8

Chloroform 61.2 3.63 -63.5 4.68

The data for the table was found by doing experiments.

It has been found that 1 mole of a nonvolatile solute particles will raise the boiling temperatures of 1 kg of water by 0.52 C°.

The same concentration of solute will lower the freezing point of 1 kg of water by 1.86 C°.

These two figures are the molal boiling point constant (Kb) and the molal freezing point constant (Kf).

A 1m solution of sugar in water contains 1 mol of solute particles per 1 kg of solvent.

A 1m solution of NaCl in water contains 2 mol of solute (because NaCl is an ion, it will dissociate in water into Na+ and Cl- ions) per 1 kg of solvent.

How many mol of solute would a 1m calcium nitrate have per 1 kg of solvent?

That’s right!

3 mol because of the Ca+2 and the two NO3

- ions.

Calculating Changes in Kb and Kf

Boiling point elevation is:

ΔTb = Kbm (molality) (change in boiling point) (boiling point constant) Freezing point depression:

ΔTf = Kfm (molality) (change in freezing point) (freezing point constant)

If 55.0 grams of glucose (C6H12O6) are dissolved in 525 g of water, what will be the change in boiling and freezing points of the resulting solution?

Step 1: Convert g of glucose to moles

55.0 g (1 mol)

1 (180.18 g) = 0.305 mol

Step 2: Convert g of water to kg

525g 0.525kg

Step 3: Calculate m

0.305mol/0.525kg = 0.581 m

Step 4: Obtain molal Kb and Kf from table.

Step 5: Place values into equation

ΔTb = Kbm

ΔTb = (0.52°C/m)(0.581m) = 0.302 °C

This means that the boiling point will be elevated by 0.302 °C.

(100 °C + 0.302 °C) This solution will reach boiling point at

100.302 °C. Now calculate the change in freezing.

Calculate the change in freezing point of 24.5g nickel(II) bromide dissolved in 445 g of water. (assume 100% dissociation)

Step 1: Convert g of NiBr2 into moles

24.5g ( 1 mol)

1 (218.49 g) = 0.112mol

Step 2: Convert solvent to kg

445 g 0.445kg

Step 3: Calculate m

0.112 mol/0.445kg = 0.252m

• We now have to take 0.252 and multiply by 3 because the dissociation of the ionic compound makes 3 moles of ions (solute) per kg of solvent:

NiBr2 Ni+2 + 2Br –

0.252 x 3 = 0.756m• Step 4: Obtain molal Kf from table.• Step 5: Place values into equation

ΔTf = (1.86°C/m)(0.756m) = 1.41 °C(0 °C - 1.41 °C)

*Freezing point has been depressed to -1.41 °C.

Coolant is used because it takes higher temperatures to reach boiling point.

Antifreeze needs lower temperatures in order to freeze.

This also why salt is used on frozen roads and walkways. The salt dissolves in the water and lowers the freezing point of water. It now takes colder temps to turn the water into ice.

A 10-percent salt solution freezes at 20 F (-6 C), and a 20-percent solution freezes at 2 F (-16 C).

Practice problems: Compute both boiling and freezing points of these solutions: (assume 100% dissociation of all ionic compounds)

1. 27.6 g Mg(ClO4)2 in 100.g of water.

2. 100.0 g of C10H8 (naphthalene) in 250. g of C6H6 (benzene).

3. 25.9 g of C7H14BrNO4 (3-bromo-2-nitrobenzoic acid) in 150. g of benzene.

4. 55.6 g of Al2(SO4)3 in 500. g of water.

5. 1500.g of NaCl in 4500. g of water.

Brief Summary Heterogeneous liquid mixtures are

classified as suspensions (large particles that settle out), or colloids (small particles that stay dispersed).

Homogeneous mixtures are solutions made of a solute dissolved in a solvent.

Solutes and solvents must be alike in polarity in order to produce a solution.

The concentration of a solution is molarity (molar) and has the unit M, which includes moles of solute per unit volume of solvent.

When preparing a dilute solution from a concentrated solution, use the formula:

M1V1 = M2V2

Where initial volume and molarity of concentrated solution (EX: 12M HCl) is compared to final volume and molarity of diluted solution (EX: 6M HCl).

Solubility of solutes can be reflected in a solubility graph.

Solubility of solid substances generally increases as temperature increases.

Solubility of gases decreases with increased temperature.

Ionic equations can be written to express the net reaction occurring in a system after the spectator ions have been removed.

Solubility rules for substances have been experimentally determined. They indicate what substances are or are not water soluble.

Colligative properties demonstrate the properties of the solution rather than solute and solvent independently.

A solution with undissolved solute is termed unsaturated.

A solution with undissolved solute is termed saturated.

A solution that has more dissolved solute at a particular temp due to being dissolved at a higher temp is termed supersaturated.

Boiling points and Freezing points of solutions can be calculated using molality.

Molality (molal) is described by unit m and expresses moles of solute per kg of solvent.

When calculating BP and FP differences use equation: Kfp or bp = Kb or f x m

Kb or f is a standard and must be given

m must be calculated and adjusted to express moles contributed.

-molecules contribute only 1 mol.

-ionic compounds contribute the number of moles they dissociate into.

EX: K2SO4 2 mole K+ + 1 mole SO4-2

After calculating difference, refer to normal BP and FP of solvents and:

Boiling Point Elevation add difference to normal BP.

Freezing Point Depression subtract difference from normal FP.