# SOLUTIONS

### TOPICS

Solution Formation

Solubility

Solution concentration

 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): CO 2 in H 2 O  Liquid in Liquid  Vinegar: Acetic Acid in water  Solid in Solid  Brass: copper and zinc

### Animation: Dissolving Process

The dissolving process at the molecular level:

## Let’s look at the structure of the water molecule

 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

### 3 Types of Solutions

Saturated solution

Unsaturated solution

Supersaturated solution

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)

### Supersaturated Solutions

STALAGMITES ROCK CANDY

### 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 together ie. 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)

### Example 3

 How many moles of Na 2 SO 4 are needed to prepare 1.5L of 0.20 M Na 2 SO 4 ? (0.30 moles solute)

### Example 4

 How many grams of solute are contained in 250 mL of 3.0 M NaOH? (30. g NaOH)

### Example 5

 How would you prepare 250 mL of a 0.500 M NaCl solution? (7.31 g NaCl)

### 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  M 1 x V 1 = M 2 x V 2

### QUESTION 1

 How many milliliters of 12.0 M HCl are needed to prepare 750. ml of 0.250 M HCl? (15.6 mL)

### QUESTION 2

 How would you prepare 5.00 L of a 1.50 M KCl from a 12.0 M stock solution?

(0.625 L)

### 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 AgNO 3 the Ag + solution to precipitate all ions in the form of AgCl

### Example 2

 2) How many milliliters of 0.0500M Pb(NO 3 ) 2 are needed to react with 2.00L of a 0.0250M Na 2 SO 4 solution in order to produce a precipitate? How many grams of precipitate are formed?