A solution is made up of a and a .

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General Chemistry Name:

Unit 7 Note Packet – Solutions Date: Per.:

Recall that matter is classified and categorized:

A solution

A solution is made up of a and a .

 solute- ;

usually appears to change state (for ex., the sugar in water doesn’t stay solid)

 solvent- Types of Solutions-

1. Solid- two solids evenly mixed (ex.: alloys, like gold jewelry-molten (liquid) gold and copper mixed) 2. Gaseous- two or more gases mixed (ex.: air)

3. Liquid- solvent and solution are liquids; solute can be solid liquid or gas (ex.: vinegar –acetic acid in water, soft drinks -CO

2

and sugar with H

2

O)

OUR FOCUS: Aqueous-

The formation of Solutions

The dissolving process for ionic compounds:

Ionic substances (salts) are composed of ions with positive and negative charges. are attracted to the negative ends of nearby water molecules; are attracted to the positive ends of nearby water molecules (this is called an ). If the attraction is strong enough, the ion will be pulled away from the surface of the crystal. The ion in solution will be surrounded by a shell of water molecules. An equation can be written to describe this process:

CaCl

2

 The dissolving process for covalent compounds:

Covalent compounds are molecules. As with ionic compounds dissolving, there are attractions between the solute and solvent particles. Attractions may be classified as

.

If the attraction is strong enough, an outer molecule will be pulled away from the surface of the solid.

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solvation – the interaction between solute and solvent particles

hydration – the interaction between solute and water molecules (when the solvent is water)

 energy is required to break bonds/attractions

 energy is released when bonds/attractions form

If breaking attractions requires more energy than is released in forming attractions, heat will be absorbed in the overall process. 

If breaking attraction requires less energy than is released in forming attractions, heat will be given off in the overall process. 

Concentration of a Solution Concentration=

 Concentrated-

 Dilute-

We usually use the mole to help us express concentration of a solution. There are three ways to express

concentration: 1.) , 2.) , and 3.) ____________________

MOLARITY

Molarity (M) - the number of moles of a substance per liter of solution

moles of a substance = molarity (M) liters of solution

Example- A chemistry teacher needs to make 500. mL of dilute calcium chloride solution. She puts 0.050 moles of CaCl

2

in 500. mL of solution. What is the concentration of the solution?

1) calculate molarity by dividing number of moles by number of liters

Example- A household cleaner contains 10.0 g NaOH in a 0.100 L solution. What is the molarity of the cleaning solution?

1) convert grams to moles

2) calculate molarity by dividing moles by number of liters

Making a solution from the original solute is simple process:

1) determine the mass of solute needed and measure on a balance 2) pour the solute into the proper size volumetric flask

3) add enough distilled water to make the necessary amount of solution

4) cover the volumetric flask and mix completely by inverting

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Example- A student needs to use 0.50 M Pb(NO

3

)

2

for a lab. How many grams of Pb(NO

3

)

2

must be used to make 250. mL of the solution? Briefly describe how the solution would be made.

1) determine the number of moles needed to make the required amount of solution

2) convert moles to grams

3) briefly describe how to make the solution

Dilutions and Solutions

It’s very common to make a more dilute solution by diluting a small amount of more concentrated solution (called the stock solution). Calculations to determine the amount of concentrated solution needed are based on a simple idea: the number of moles of a solute does not change when a solution is diluted. Therefore:

M

1

V

1

= M

2

V

2

where M

1

describes the molarity of the stock (more concentrated) solution

M

2

and V

2

describe the molarity and volume of the more dilute solution you’re making V

1

is unknown;

Making a dilution is also a simple process:

1) determine volume of concentrated solution needed and measure in a graduated cylinder or volumetric pipet 2) pour more concentrated solution into the proper size volumetric flask

3) add enough distilled water to make the necessary amount of solution 4) cover the volumetric flask and mix completely by inverting

Example- 0.10 M HCl is needed for a lab. How would 250. mL of 0.10 M HCl be prepared from 6.0 M HCl stock solution?

1) solve for V

1

2) briefly describe the process of making the solution

Stoichiometry with Solutions

Now that you know about concentration – you have another fact you can use to work with moles.

Because Molarity = moles/liters, this can be used as a conversion factor.

For example, how many moles of NaCl are in 0.25L of a 2.0 M solution? Remember, the Molarity gives you your conversion factor.

Set up the problem using dimensional analysis:

0.25 L x 2.0 mol = 0.50 mol NaCl

1.0 L

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Try another – how many moles of iron (III) chloride are in 0.32 L of a 1.5 M solution?

0.32L x 1.5 mol = 1.0 L

And one more – how many moles of silver nitrate are in 0.50 L of a 0.05 M solution?

Once you have determined the number of moles, you can do any other conversion you already know about moles (mass, mole ratio in a reaction, etc).

That means – we have an even-more-improved mole map!

Remember – to get from volume of a solution to moles, you need the concentration (the Molarity)!

Consider the following reaction: AgNO

3 (aq)

+ HCl

(aq)

 AgCl

(s)

+ HNO

3 (aq)

If 0.1L of a 0.1M solution of AgNO

3

is mixed with unlimited HCl, how many grams of AgCl will be made?

0.1L x 0.1mol AgNO

3

x 1 mol AgCl x 143.32 g AgCl = 1.0 L 1 mol AgNO

3

1 mol AgCl Trace the path taken on the mole map above.

Try another one:

Liters of substance A (if it is a GAS)

Moles of substance A Grams of

substance A

Particles of substance A

Volume of solution of substance A

Moles of substance B

Volume of solution of substance B

Particles of substance B

Grams of substance B Liters of substance B

(if it is a GAS)

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If 0.2 L of a 1.0 M solution of HCl reacts with unlimited silver nitrate, how many grams of AgCl will be made?

0.2 L x 1.0 mol HCl x _ x _ = 1.0 L

And one more:

A solution of 0.2 M iron (III) chloride reacts with aqueous sodium hydroxide to produce a precipitate of iron (III) hydroxide and aqueous sodium chloride.

Write and balance the equation.

If 0.025 L of the iron (III) chloride solution reacts with unlimited NaOH, how many grams of precipitate will be made?

Saturation

Is there a limit to how much sugar you can dissolve in a cup of tea?

 – a solution that contains as much solute as can possibly be dissolved under the existing conditions of temperature and pressure

 - a solution with less than the maximum amount of solute

 – a solution that contains a greater amount of solute than that needed to form a saturated solution (this is an unstable situation; such a solution will not remain supersaturated for long) Supersaturation Demo Notes:

Solubility

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Solubility is .

 determined experimentally

 expressed in grams of solute per 100 grams of solvent (at a specific T and P)

On a graph showing solubility versus temperature, (see Table D in your Reference Tables), a line represents for a particular substance at different temperatures. A point above the line represents while a point below the line indicates an solution.

Find the potassium nitrate (KNO

3

) curve on Table D and use it to answer the following questions.

1. How many grams of KNO

3

will dissolve in 100g of water at 50°C? _________

2. If you only put 70g of KNO

3

in the water at 50 °C, how would you describe the solution?

__ How much more solute could you dissolve?

3. It is HARD to make a supersaturated solution. You would be lucky to get 70g of KNO

3

to dissolve at 40 °C.

Find that spot on the graph. How much would you expect to dissolve at 40°C? ________________

Solubility is affected by:

1.  how temperature affects solubility of a solid depends on if the dissolving process was endothermic or exothermic:

a. endothermic when dissolving – higher temperature results in increased solubility b. exothermic when dissolving – higher temperature results in decreased solubility

 gaseous solutes are less soluble at higher temperatures a.

2.  “like dissolves like” – polar and ionic solutes dissolve in polar solvents but not in nonpolar solvents;

nonpolar solutes dissolve in nonpolar solvents but not in polar solvents (ex.: nonpolar oil molecules stay separated from polar water molecules in salad dressing)

3.  solutions of solids and liquids not affected much by pressure

 gas solubility is strongly affected by pressure

a. increased pressure increases the rate at which gas molecules strike the surface of solute b. Henry’s Law: the solubility of a gas is proportional to the partial pressure of the gas above the

liquid Dissolving rate is affected by:

1. – solvent particles move faster at higher temperatures, so more solvent particles come into contact with the solid solute in a shorter period of time

2. – because solvent particles pull solute particles from the surface of the solute 3. – because dissolved solute builds up next to the solid solute causing the dissolving

process to slow down