classification of matter: 5.1- The Basics

All Matter

Mixtures Pure Substances

Heterogeneous(Mechanical Mixtures and Colloids)

Homogeneous(Solutions)

Elements

Compounds

Classifying Solutions Chapter 5: Solutions

is any that has

are a type of matter that has

eg)

are combinations of matter that can be

matter solid, liquid or gas mass

pure substances

definite fixed composition

elements and compounds

mixtures separated by physical means

volume

do NOT have definite proportions

and

and

have and the different components are

heterogeneous mixtures (mechanical mixtures) varying composition

usually visible

are which havesolutions homogeneous mixtures

uniform composition

composed of at least one substance dissolved in another

and the different components arenot visible

is the and is usually the substance present in the

is what is

solvent dissolver, largest quantity (mass, volume, amount)

eg)

solute dissolved in the solvent

eg)

water

salt

Types of Solutions both solvents and solutes can be

you can have various combinations of solute and solvent phases

solids, liquids or gases

eg) liquid in liquid –

mercury amalgam fillingsliquid in solid –alloyssolid in solid –

carbonated beveragesgas in liquid –Kool Aidsolid in liquid –

ethylene glycol (antifreeze)

aqueous an solution is any solution in which is thewater solvent

water is called the therefore we will be concerned mainly with

“universal solvent” aqueous solutions

water dissolves a lot of different solutes because of it’s …this is good because

…also bad because dissolve in it easily

unique properties75%

toxinsof the earth (and our bodies) is water

Solutions in our Societysolutions are all around us and affect our

lives in many ways

we have developed many to meet the and needs of humans

eg)tests to monitor drinking water

technologiespersonal industrial

hair products,breathalyzer test,

in using solutions in our lives, care must be taken to ensure responsible use

eg) – set up in Canada in 1985 to ensure that companies that deal with chemicals are using them in a safe and appropriate manner

Responsible Care Program

using solutions can have intended and unintended consequences for

released into the environment are taken in by organisms

this can have toxic effects

eg) H2S(g)…sour gas will cause a loss of consciousness at 700 ppm

humans and the environment

toxins

immediate

sometimes the of the food chain are by the toxins since the levels are not that high…but the toxins are then passed along to the of the food chain

organisms at the of the food chains (like humans!) can end up with of these chemicals

eg)

lower levelsunaffected

upper levels

toptoxic levels

mercury, lead, arsenic, PCB’s, DDT etc.

is the increase in the concentration of toxins as you

biomagnification (bioaccumulation) move up the food chain

we must assess the and of using technologies that contain certain substances

eg) heavy metals released from mineral processing, power plants –

risks benefits

mercury, arsenic

many chemical reactions (either in our bodies or in the lab) are if the reactants are not dissolved in

very slow

dissolving in water allows the solute particles toseparate, disperse and collide

particle collisions are necessary for to occur

reactions

water

with other solute particles

Electrolytes vs Non-Electrolytes

are aqueous solutions that

electrolytes (weak and strong)conduct electricity

are aqueous solutions thatnon-electrolytes

do not conduct electricity

eg)

eg) molecular compounds in solution

all soluble ionic compounds, very polar molecular compounds (like acids, ammonia)

5.2: Explaining Solutions is adissolving physical change

the of a solid solute are held together by

molecules or ionsbonds

when dissolving occurs, these bonds and the of the solute become

breakions or molecules

attracted to the solvent particles

NaCl

H2O

the 3 processes involved in dissolving are:

1. broken endothermic

(requires energy)

2. brokenendothermic

3. form

always exothermic

bonds between molecules or ions of solute –always

bonds between molecules of solvent –always

bonds between molecules or ions of solute and solvent –

overall energy change in dissolving is equal to the of the three steps

if, the overall dissolving process is

more energy is released than is requiredexothermic

if, the overall dissolving process is

less energy is released than is requiredendothermic

sum(law of conservation of energy)

When we dissolve compounds in water, water is not included as a reactant.

Water is involved in the ion interaction but it is not consumed so its not a reactant

We symbolize compounds dissolving in water by write an (aq) as the ‘state’ of the compound

occurs when when they are dissolved in an

dissociation ionic compounds

aqueous solution

are used to show what happens to a substance when it is put into

water

you can have 4 situations:

1. insoluble ionic or molecular compounds they to any great extent use the for ionic

compounds

eg)

C25H52(s) C25H52(s)

AgCl(s)AgCl(s)

break apart into their ions

dissociation equations

do not dissolve

solubility table

2. soluble ionic compounds

dissolve to a great extent to form

are broken

eg)

use solubility table

the number of ions

Cl(aq)+Na+(aq)NaCl(s)

includes which turn litmus paper

OH(aq)+K+(aq)KOH(s)

ions in solution

ionic bonds

bases

balance

blue

3. soluble molecular compounds

dissolve to form

are broken

eg) C12H22O11(aq)

C12H22O11(s)

molecules in solution

intermolecular forces (LD, DD, HB)

Molecular compounds do not have a ‘solubility chart’ but there are a few patterns that can help us.

Non-polar molecules- generally don’t dissolve in water

Polar molecules may be slightly soluble in water

Polar compounds with hydrogen bonding are the most likely to be very soluble in water.

4. acids

they are but are

dissolve to form in solution

eg)

the number of ions

SO42(aq)+2H+(aq)H2SO4(s)

molecular compoundsvery polar

ions (ionize)

balance

turns litmus paperred

Ionization

Ionization: the process by which a neutral atom or molecule is converted to an ion.

Arrhenius proposed that acids ionize by interacting with water solvents

Idea is that the H+ causes the litmus paper to turn color

Examples

Write the equations to show what happens to each of the following in water:

1. potassium chloride

2. carbon dioxide

KCl(s)

K+

(aq) + Cl-

(aq)

CO2(g)CO2(g)

5. sodium phosphate decahydrate

Na3PO4 10H2O(s)

Na+

(aq) + PO4

3- (aq)

4. aluminum sulphate

Al2(SO4)3 (s) Al3+

(aq) + SO4

2-

(aq) 32

3

3. solid hydrogen nitrate

HNO3(s) NO3(aq)+H+(aq)

7. barium sulphate

BaSO4(s) BaSO4(s)

6. gasoline

C8H18(l) C8H18(l)

many chemical reactions (either in our bodies or in the lab) are if the reactants are not dissolved in

very slow

dissolving in water allows the solute particles toseparate, disperse and collide

particle collisions are necessary for to occur

reactions

water

with other solute particles

concentration is

the amount of solute relative to the amount of solvent

dilute = solute, solvent

concentrated = solute, solvent

5.3: Solution Concentration

concentration can be expressed in a variety of ways:

low high

high low

ppm, % mass, % volume,mol/L, mmol/L, mg/dL, %

these expressions can be seen on many products that we use in our daily lives and in industry

eg) vinegar – toothpaste – blood cholesterol levels – breathalyzer - % (g of alcohol per 100 mL of

blood) eg.

5% acetic acid by volume0.243% by mass NaF

250 mg/dL

0.08

molarity = number of moles of solute per litre of solvent

c = n v

n = m M

where: c = concentration in mol/L n = number of moles in mol v = volume in L m = mass in g M = molar mass in g/mol

Molar Concentration

Example 1

A sample of 0.900 mol of NaCl is dissolved to give 0.500 L of solution. What is the molar concentration of the solution?

n = 0.900 molv = 0.500 L

c = n v = 0.900 mol 0.500 L = 1.80 mol/L

Example 2

Calculate the concentration of 100 mL of a solution containing 0.300 mol of sulphuric acid.

n = 0.300 molv = 0.100 L

c = n v = 0.300 mol 0.100 L = 3.00 mol/L

Example 3

Calculate the molar concentration of a 250 mL solution that has 3.2 g of NaCl dissolved in it.

m = 3.2 gv = 0.250 LM = 58.44 g/mol

c = n v = 0.0547… mol 0.250 L = 0.22 mol/L

n = m M = 3.2 g 58.44 g/mol = 0.0547…mol

Example 4

Calculate the number of moles of Pb(NO3)2 needed to make 500 mL of a 1.25 mol/L solution. c = 1.25 mol/Lv= 0.500 L

n = cv = (1.25 mol/L)(0.500 L) = 0.625 mol

Example 5

How many litres of 4.22 mol/L solution would contain 3.69 mol of BaCl2?

c = 4.22 mol/Ln = 3.69 mol

v = n c = 3.69 mol 4.22 mol/L = 0.874 L

Example 6

Calculate the mass of the salt required to prepare 1.50 L of a 0.565 mol/L solution of K3PO4.c = 0.565 mol/Lv = 1.50 LM = 212.27 g/mol

n = cv = (0.565 mol/L)(1.50 L) = 0.8475 molm = nM = (0.8475 mol)(212.27 g/mol) = 180 g

Assignment

Practice questions

Molar concentration of ions

once you have your dissociation equation, you can calculate the concentration of ions in solution using the

5.3 part 2: Molar Concentration of Ions

mole ratiowantedgiven

you may have to calculate the concentration of in solution

ions

areanions negative ions

arecations positive ions

Example 1Calculate the ion concentrations in a 0.050 mol/L solution of KCl.

KCl(s) K+(aq) + Cl-

(aq) 1 1 1

C = 0.050 mol/L C = 0.050 mol/L

1 1

= 0.050 mol/L

C = 0.050 mol/L

1 1

= 0.050 mol/L

g w w

Example 2Calculate the ion concentrations in a 0.050 mol/L solution of Al2(SO4)3.

C = 0.050 mol/L C = 0.050 mol/L 2 1

= 0.10 mol/L

C = 0.050 mol/L 3 1

= 0.15 mol/L

Al2(SO4)3(s) Al3+

(aq) + SO4

2-(aq) 321

w wg

Example 3Calculate the concentrations of dissolved Na3PO4(s) 10H2O that gives a 0.30 mol/L concentration of Na+

(aq) ions.

C = 0.30 mol/L C = 0.30 mol/L

1 3

= 0.10 mol/L

Na3PO4(s) 10H2O

Na+

(aq) + PO4

3-(aq)31 1

w g

Example 4Calculate the concentration of sodium ions in a NaCl(aq) solution made by dissolving 6.33 g of NaCl(s) in 150 mL of water.

NaCl(

s) Cl-

(aq) + Na+(aq)

1 1 1

n = m M = 6.33 g 58.44 g/mol = 0.108…mol

1 1

c = .722 mol/L

c = n v = 0.108…mol 0.150 L = 0.722 mol/L

g w

m = 6.33 g

M = 58.44 g/mol = 0.722 mol/L

Assignment

Practice questions

a solution of is called a

known concentrationstandard solution

there are two ways to make a solution:

1. a measured amount of in a certain volume of

2. a standard solution

5.4 Part 1: Preparing a Standard Solution

dissolve pure solutesolvent

dilute

Calculate the of the required to achieve a specific concentration and volume.

to prepare a solution of known concentration from a : solid solute

Steps

n = cv

m = nM

mass solute

1. ______ g (mass) of _______ (solute).

Measure

2. the solute in _______ mL of water (half of the volume). Dissolve

3. solution to a ______ mLTransfervolumetric flask.

4. flask to _______ mL (final volume) and mix by inverting. Fill

Example

Describe how to prepare 100 mL of a 0.0800 mol/L solution of KMnO4(aq).

v = 0.100 Lc = 0.0800 mol/LM = 158.04 g/mol

n = cv = (0.0800 mol/L)(0.100L) = 0.00800 mol

m = nM = (0.00800 mol)(158.04 g/mol) = 1.26 g

1. Measure out of .2. Dissolve the KMnO4(s) in of distilled H2O(l). 3. Transfer the solution to a volumetric flask.4. Fill to and invert to mix.

1.26 g KMnO4(s)

50 mL100 mL

100 mL

DemoDescribe how to prepare 100 mL of a 0.26 mol/L solution of CuSO4 ° 5 H2O (aq).

v = 0.100 Lc = 0.26mol/LM = 249.72 g/mol

m = nM = (0.026mol)(249.72 g/mol) = 6.5 g

n = cv = (0.26 mol/L)(0.100L) = 0.026 mol

Assignment

Practice questions

not all solutions are available in the concentrations we need

= decreasing the of a solution by adding more

5.4 Part 2 : Dilutions

you can make a out of a solution of known concentration by

less concentrated solutiondiluting it

dilution concentrationsolvent (water)

vici = vfcf

where: vi = initial volume in L vf = final volume in L ci = initial concentration in mol/L cf = final concentration in mol/L

the stays constant!!!

number of moles

ni = nf and n = CV

Note

- ci is always bigger than cf

- vi is always smaller than vf

Example 1

What volume of 1.0 mol/L NaCl solution do you need to make 250 mL of a 0.20 mol/L NaCl solution? Vf = 250 mL = 0.250 LCi = 1.0 mol/LCf = 0.20 mol/L

ViCi = VfCf

Vi(1.0 mol/L) = (0.250L)(0.20 mol/L) Vi = 0.050 L

Example 2

What is the concentration of a 1.50 L solution if it is made by mixing 500 mL of 14.8 mol/L H2SO4(aq) with 1.00 L of water?

Vi = 0.500 LVf = 1.50 LCi = 14.8 mol/L

ViCi = VfCf

(0.500L)(14.8 mol/L) = (1.50L) Cf

Cf = 4.93 mol/L

Assignment

Practice questions prelab

Net Ionic Equations

there are three types of equations:

1. Non-Ionic Equations

this type of equation shows as if they didn’t in solution

all reactants and products

Pb(NO3)2(aq) + 2 KI(aq) 2 KNO3(aq)+PbI2(s)

dissociate

2. Total Ionic Equations

this type of equation is and shows all soluble reactants and products dissociated into ions

more accurate

Pb2+

(aq)

2NO3(aq)+2K +(aq)+PbI2(s)

2I(aq)+2K+ (aq)+2NO3(aq)+

3. Net Ionic Equations

this type of equation shows

Pb2+(aq) 2NO3(aq)+2K +(aq)+PbI2(s) 2I(aq)+2K+(aq)+2NO3

(aq)+

only the chemical change that occurs

arespectator ions (ions that don’t change in the reaction)cancelled

Pb2+

(aq)PbI2(s) 2I(aq)+Net Ionic:

Examples

1. aluminum sulphate and sodium sulphide

Al2(SO4)3(a

q) + Na2S(aq) Al2S3(s) + Na2SO4(aq) 3 3

2Al3+(aq) +++ 3SO4

2-

(aq) 6Na+

(aq) 3S2-(aq) Al2S3(s) + 6Na+

(aq) + 3SO42-

(aq)

Net ionic 2Al3+(aq) + 3S2-

(aq) Al2S3(s)

2. barium hydroxide and aluminum nitrate

Ba(OH)2(aq

)

+ Al(NO3)3(aq) Al(OH)3(s) + Ba(NO3)2(aq

) 2 3

3Ba2+(aq) +++6OH-

(aq) 2Al3+(aq) 6NO3

-(aq) 2

Al(OH)3(s) + 3Ba2+

(aq) + 6NO3

-(aq)

Net ionic 2Al3+(aq) + 6OH-

(aq) 2 Al(OH)3(s)

3 2

Al3+(aq) + 3 OH-

(aq) Al(OH)3(s)

3. fluorine gas and sodium chloride

F2(g) + NaCl(aq) Cl2(g) + NaF(aq) 2 2

F2(g) ++2 Na+

(aq)2 Cl(aq) Cl2(g)+2 Na+

(aq)+2 F(aq)

Net ionic F2(g) +2 Cl(aq) Cl2(g)+ 2 F(aq)

Solubilitythe of a solute is thesolubilit

y amount of a solute that dissolves in a given quantity of solvent

eg) has a solubility ofNaCl(s) 36 g/100mL at 20C

at a given temperature

an is a solution that does have the maximum amount of solute dissolved in it

unsaturated solutionnot

a is a solution that contains thesaturated solution

maximum amount of a dissolved soluteat a given temperature

some will be present

undissolved solute

the solution may still be able to dissolve other solutes

a containssupersaturated solution more dissolved solute than its solubility at a given temperature

http://www.youtube.com/watch?v=AedL_NCv1Pw

the degree to which a solute is soluble depends on thestrength of attraction

1. thesolute particles

2. the solute particles and the solvent particles

Range of Solubility

between:

solutes can be:

1. insoluble –

2. slightly soluble –

less than 0.1 g/100mL,

0.1 g/100 mL

although there is still a tiny bit of dissolving

betweenand1 g/100 mL

3. soluble – 1 g/100 mLgreater than

note that these general rules for solubility do not apply to …the numbers for gases are much and still considered

eg) has a solubility of yet this is considered

O2(g) 0.009 g/100 mL at 20C soluble

gaseslower soluble

A Closer Lookonce a solute , it appears that

the solute-solvent bonds

over time,

when you study a saturated solution, the

dissolvesdon’t break

amount (mass or moles)

undissolved particles become dissolved and dissolved particles crystallize

a saturated solution is said to be in a state ofequilibrium

of undissolved solute at the bottom

remains unchanged

equilibrium occurs when a process and the reverse process take place at the

(dissolving)

crystallization

dissolving

same rate(crystallization)

CuSO4(s) Cu2+(aq) + SO42(aq)

is

Cu2+(aq) + SO42(aq) CuSO4(s)

is

What we do is use a double arrow to show equilibrium:

CuSO4(s) Cu2+(aq) + SO42(aq)

dissolving

crystallization

⇌

eg)

Temperature and Solubility

the for a substance must be provided with a certain and it depends on the of the solute

at , the particles of the solute and solvent have more

when a dissolves in a , the holding the solid together must be

solubility valuetemperature

Solids

solid liquid bondsbroken

higher temperaturesenergy

as temperature the solubility of a solid

increases,

state

increases

when a dissolves in a , additional is needed

the particles of a liquid are held together as as the particles in a solid

Liquids

the solubility of most liquids is

liquid liquidenergy

not affected

notstrongly

not

by temperature

when a dissolves in a , the particles must

gas particles move and have a great deal of

Gases

the temperature would the gas particles and make it for them to dissolve

quicklyenergy

gas liquidlose energy

as temperature the solubility of a gas

increasing speed upharder

increases,decreases

Pressure and Solubility

a change in has apressure negligible effect

the solubility of are greatly affected by

gasespressure

as pressure the solubility of a gas

eg) pop bottles fizzing when opened

on the solubility of solids or liquids

increases,increases

is a process used to the presence of specific substances in solution

qualitative analysis

identifies the of particular substances

quantitative analysis

eg) breathalyzer, testing drinking water for dissolved solutes

5.5 Solubilityidentify

amount

a precipitation reaction is a reaction in which a is formed (the )

Precipitation Reactions

you can add a solution to an solution to see if a precipitate forms

precipitatesolid product

double replacement

unknownknown

this can help you distinguish between possible ions in the unknown solution

ExampleYou have a solution that may contain either aluminum ions or potassium ions. What solution could you add to the unknown solution to determine the type of ion present? ***you need a solution that will form a solid

product in with one of the ions but not the other one

Al3+(aq)

K1+(aq)

Al(OH)3(s)3 OH(aq)+

K(OH)(aq)2 OH(aq)+

eg) NaOH(aq)

***if a solid precipitate forms then the unidentified ion is aluminum, if no precipitate forms, then the ion is potassium***

Solubility assignment Pg. 226 #1-6

Assignment

Solutions Unit Review

  p. 204 #1-26 omit 15, 16, 17