CHAPTER 13

SOLUTIONS

BASIC DEFINITIONS

Solution – a homogeneous mixture of 2 or more substances in a single phase

Solute – The dissolved substance usually in smaller amount

Solvent – the dissolving medium usually in larger amount

Visual Concepts

Solutes, Solvents, and Solutions

Chapter 13

Solvent Solvent - present in - present in greater amountgreater amount

Solute Solute - substance - substance being dissolvedbeing dissolved

Visual Concepts

Solutions

Chapter 13

MORE DEFINITIONS

Saturated – The maximum amount of solute is dissolved in the solventUnsaturated – Less than the maximum amount of solute is dissolved in the solventDilute – The amount of solvent is much greater than the amount of solute dissolved in it.

MORE DEFINITIONS

Electrolyte – A substance that dissolves in water to give a solution that will conduct electricity

Nonelectrolyte – A substance that dissolves in water to give a solution that will not conduct electricity

TYPES OF SOLUTIONS

Gases (Air is a mixture of O2, N2, CO2,

H2O, etc.)

Solids (White gold is a mixture of gold & palladium)

Liquids (salt dissolved in water; carbon dioxide in soda)

Particle Model for a Suspension

Chapter 13 Section 1 What Is a Solution?

SUSPENSION

A heterogeneous mixture in which particles in a solvent are so large, they will settle out unless it is constantly stirred

Particles can be filtered out due to their size (>1000 nm in diameter)

Visual Concepts

Suspensions

Chapter 13

COLLOIDS

A heterogeneous mixture in which particles in a solvent remain suspended by the movement of surrounding molecules Think of fruit suspended in jelloParticles are not easily filtered due to their size (<1000 nm in diameter)

Visual Concepts

Factors Affecting the Rate of Dissolution

Chapter 13

FACTORS AFFECTING SOLUBILITY

1. Surface Area – increasing the surface area gives molecules more places to interact causing the solute to dissolve faster

2. Agitation/Stirring – stirring gives molecules more opportunities to interact – faster dissolving

3. Temperature – Adding heat adds energy to molecules = faster dissolving

4. “Like dissolves like” – polar solutes dissolve in polar solvents or Nonpolar solutes dissolve in Nonpolar solvents

Visual Concepts

Like Dissolves Like

Chapter 13

CONCENTRATION CALCULATIONS

Visual Concepts

Concentration

Chapter 13

Concentration, continuedCalculating Concentration, continued

• Concentrations can be expressed in many forms.

Section 2 Concentration and MolarityChapter 13

MOLARITY (M)

MOLARITY (m)

Moles of solute per Liter of solution

molarity (M)

amount of solute (mol)

volume of solution (L)

SAMPLE PROBLEM A

You have 3.50 L of solution that contains 90.0 g of sodium chloride, NaCl. What is the molarity of that solution?

SAMPLE PROBLEM A

You have 3.50 L of solution that contains 90.0 g of sodium chloride, NaCl. What is the molarity of that solution?

Step 1: Outline what you know.

M = ? Mol / L

mol = 90.0 g

L = 3.50 L

SAMPLE PROBLEM A

You have 3.50 L of solution that contains 90.0 g of sodium chloride, NaCl. What is the molarity of that solution?

Step 2: Convert any units necessary.

mol = 90.0 g NaCl

90.0 g NaCl x1 mol NaCl58 g NaCl

= 1.55 mol NaCl

SAMPLE PROBLEM A

You have 3.50 L of solution that contains 90.0 g of sodium chloride, NaCl. What is the molarity of that solution?

Step 3: Plug into the equation and solve.M = ? Mol / Lmol = 1.55 molL = 3.50 L

molarity (M)

amount of solute (mol)

volume of solution (L)

SAMPLE PROBLEM A

You have 3.50 L of solution that contains 90.0 g of sodium chloride, NaCl. What is the molarity of that solution?

Step 3: Plug into the equation and solve.

molarity (M)

amount of solute (mol)

volume of solution (L)

M=

1.55 mol3.50 L

= 0.44 mol / L

SAMPLE PROBLEM B

You have 0.8 L of a 0.5 M HCl solution. How many moles of HCl does this solution contain?

SAMPLE PROBLEM B

You have 0.8 L of a 0.5 M HCl solution. How many moles of HCl does this solution contain?

Step 1: Outline what you know.

M = 0.5 Mol / L

mol = ? mol

L = 0.8 L

SAMPLE PROBLEM B

You have 0.8 L of a 0.5 M HCl solution. How many moles of HCl does this solution contain?

Step 2: Convert any units necessary.

All units correct! Moving on…

SAMPLE PROBLEM B

You have 0.8 L of a 0.5 M HCl solution. How many moles of HCl does this solution contain?

Step 3: Plug into the equation and solve.M = 0.5 Mol / Lmol = ? molL = 0.8 L

molarity (M)

amount of solute (mol)

volume of solution (L)

SAMPLE PROBLEM B

You have 0.8 L of a 0.5 M HCl solution. How many moles of HCl does this solution contain?

Step 3: Plug into the equation and solve.

molarity (M)

amount of solute (mol)

volume of solution (L)

0.5 M=

mol

0.8 L

(0.8 L)(0.8 L)

SAMPLE PROBLEM B

You have 0.8 L of a 0.5 M HCl solution. How many moles of HCl does this solution contain?

Step 3: Plug into the equation and solve.

mol = 0.4

MOLALITY (m)

MOLALITY (m)

Moles of solute per kilogram of solvent

molality (m)

amount of solute (mol)

mass of solvent (kg)

SAMPLE PROBLEM C

A solution was prepared by dissolving 17.1 g of sucrose (table sugar, C12H22O11) in 125 g of water. Find the molal concentration of this solution.

SAMPLE PROBLEM C

A solution was prepared by dissolving 17.1 g of sucrose (table sugar, C12H22O11) in 125 g of water. Find the molal concentration of this solution.

Step 1: Outline what you know.

m = ? Mol / kg

mol = 17.1 g

Kg = 125 g

342.34 g C12H22O11

SAMPLE PROBLEM C

A solution was prepared by dissolving 17.1 g of sucrose (table sugar, C12H22O11) in 125 g of water. Find the molal concentration of this solution.

Step 2a: Convert any units necessary.

mol = 17.1 g C12H22O11

17.1 g C12H22O11 x1 mol C12H22O11

= 0.05 mol C12H22O11

1000 g

SAMPLE PROBLEM C

A solution was prepared by dissolving 17.1 g of sucrose (table sugar, C12H22O11) in 125 g of water. Find the molal concentration of this solution.

Step 2b: Convert any units necessary.

kg = 125 g H2O

125 g H2O x1 kg

= 0.125 kg H2O

SAMPLE PROBLEM C

A solution was prepared by dissolving 17.1 g of sucrose (table sugar, C12H22O11) in 125 g of water. Find the molal concentration of this solution.

Step 3: Plug into the equation and solve.m = ? Mol / kgmol = 0.05 molkg = 0.125 kg

molality (m)

amount of solute (mol)

mass of solvent (kg)

SAMPLE PROBLEM C

A solution was prepared by dissolving 17.1 g of sucrose (table sugar, C12H22O11) in 125 g of water. Find the molal concentration of this solution.

Step 3: Plug into the equation and solve.

m=

0.05 mol

0.125 kg= 0.400 mol /

kg

molality (m)

amount of solute (mol)

mass of solvent (kg)

SAMPLE PROBLEM D

How much iodine must be added to prepare a 0.480 m solution of iodine in CCl4 if 100.0 g of CCl4 is used?

SAMPLE PROBLEM D

How much iodine must be added to prepare a 0.480 m solution of iodine in CCl4 if 100.0 g of CCl4 is used?

Step 1: Outline what you know.

m = 0.480 m

mol = ? mol

kg = 100.0 g

SAMPLE PROBLEM D

How much iodine must be added to prepare a 0.480 m solution of iodine in CCl4 if 100.0 g of CCl4 is used?

Step 2: Convert any units necessary.

kg = 100.0 g

Kg = 0.100 kg

SAMPLE PROBLEM D

How much iodine must be added to prepare a 0.480 m solution of iodine in CCl4 if 100.0 g of CCl4 is used?

Step 3: Plug into the equation and solve.m = 0.480 Mol / kgmol = ? molkg = 0.100 kg

molality (m)

amount of solute (mol)

mass of solvent (kg)

SAMPLE PROBLEM D

How much iodine must be added to prepare a 0.480 m solution of iodine in CCl4 if 100.0 g of CCl4 is used?

Step 3: Plug into the equation and solve.

0.480 m =

mol

0.100 kg

(0.100 kg)(0.100

kg)

molality (m)

amount of solute (mol)

mass of solvent (kg)

SAMPLE PROBLEM D

How much iodine must be added to prepare a 0.480 m solution of iodine in CCl4 if 100.0 g of CCl4 is used?

Step 3: Plug into the equation and solve.

mol = 0.048

PARTS PER MILLION (ppm)

ppm = grams of soluteLiters of solution

X 1000

Grams of solute in 1 million grams of solution.

Useful in very dilute samples, but for the sake of simplicity, your worksheet will not work with very small exponents.