Chemistry Test 4

Front 1

Chapter 13:

Gases

Back 1

The atmosphere consists manily of nitrogen (N 2) and oxygen (0 2-) and both supports life and acts like a waste receptacle for exhaust gases.

The two main sources of pollution are transportation and the production of electricity.

Gases in the atmosophere shield us from harmful radiation from the sun and reflect heat radiation back toward the earth to keep it warm.

Front 2

Pressure:

Back 2

Gases uniformly fill any container, it is easily compressed, mixes completely with other gases, and exerts pressure on it's surroundings.

Front 3

Barometer:

Back 3

A device that measures atmospheric pressure. A glass tube filled with mercury and inverted in a dish of mercury will measure atmosopheric presure. Atmospheric presure pushes on the mercury in the container and forces the mercury up the column until it equals the pressure of the atmosphere.

Front 4

Atmospheric pressure:

Back 4

Results from the mass of air being pulled toward the center of the earth by gravity.

It varies with altitude and is lower at high altitudes because of less air pushing down on the earths surface.

Front 5

Units of pressure:

Back 5

mm HG or torr and standard atmosphere
1 std atmosphere = 1.00 atm = 760.0 mm Hg = 760.0 torr

pascal (Pa) is the standard SI unit for pressure
1 standard atmosphere = 101,325 Pa

lbs per square inch -psi- used in engineering sciences
1.0 atm = 14.69 psi

The various equivilancies can be used to convert between the units like we have done.

Manometer-device used for measuring the pressure of gas in a container.

Front 6

Pressure and volume: Boyle's law

Back 6

Robert Boyle was an irish scientist in the 1600's that observed that as pressure increased, the volume of a trapped gas decreased.

Boyle's Law-Gases at constant temperature and constant amount

PV=k (pressure, volume, a constant)

Boyle's Law-can also be stated as:

P1V1=P2V2 (where p1 and v1 are the original pressure and temp and p2 and v2 are the final pressure and temp.

Example: A 1.5 liter sample of gaseous CCl2F2 at a pressure of 56 torr will occupy what volume at a new pressure of 150 torr?
p1=56 torr
v1=1.5 L
v2=?
p2=150 torr

You can set this up two ways:

V2=V1 x P1/P2=1.5 L x 56 torr/150 torr = 0.56 L

or P1xV1/P2=V2

*temperature and moles are constant*

Front 7

Volume and temperature: Charles's Law

Back 7

Jacques Charles in the late 1700's to earlyu 1800's showed that a given amount of gas (at constant pressure) increases with the temperature of the gas.

Absolute zero-the lowest possible temperature where gases extrapolate to zero volume is at -293 C.

Charles's law (constant pressure and gas)

V=bT (when temp is in kelvins, V is volume and b is proprtionally constant).

can also be expressed: V1/T1=V2/T2

Example: Calculate the volume of 2.0 liter sample of air collected at 298 k and then cooled to 278 k. The pressure is held constant (add 273 for got from C to k).

Either way:

V1/T1=V2/T2 and cross multiply or V1xT2=V2xT1 or V1xT2/T1=V2

Initial conditions:
T1=298 k
V1=2.0L
T2=278 k
V2=?

V2=T2 x V1/T1=278 k x 2.0L/298 k=1.9L

Front 8

Volumes and moles: Avagadro's Law

Back 8

The volume of a gas is directly proportional to the number of moles, if the temperature and pressure remain constant.

V=an (where v is volume and n is number of moles of gas)

For a gas at constant temperature and pressure, the volume is directly proportional to the number of moles of gas.

v=volume, n=moles, p=pressure, t=kelvin (k)

Avagadro postulated the law in 1811. The law can be written as:

V1/n1=V2/n2

Example: A 12.2 liter sample containing 0.50 mol of oxygen gas, O2, at a pressure of 1 atm and a temp of 25 C is converted to ozone, O3, at the ssame pressure and temperature. What will be the volume of ozone formed?

The balance chemical equation is 3 O2 (g) -> 2 O3 (g)

0.50 mol O2 x 2 mol 03/3 mol 02=0.33 mol 03

Initial conditions:

n1=0.50 mol
v1=12.2 L
n2=0.33 mol
v2=?

rearranging the equation:
v2=v1 x n2/n1=12.2 L x 0.33 mol/0.50 mol=8.1 L

thus the volume decreased, as would be predicted.

v1 x n2=v2 x n1

Front 9

Ideal Gas Law:

Back 9

Summary of gas laws derived from experimental observations:

Boyles Law: PV=k

Charles' Law: V=bT (p and n are constant)

Avogadros Law: V=an (constatnt T and P)

These gas laws can be combined to form the ideal gas law:

PV=nRT (where P is pressure in atm, v is volume, n is the number of moles, r is the universal gas content = 0.08206 L atm/k mole, and t is temperature in k.

A gas that obeys the above equation is said to behave ideally and is an ideal gas. For the problems in this course, you should assume that they behave like ideal gases.

Ex: A sample of hydrogen gas, H1, has a volume of 8.56 L at a temperature of 0 C and a pressure of 1.5 atm. Calculate the number of moles of H2 present in this gas sample . Assume that the gas behaves ideally.

First convert t=0 C to k by adding 273. Temp is 273 k.

n=PV/RT=1.5 atm x 8.56 L/0.08206 L atm/K mol x 273 k=0.57 mol

1.5(8.56)/0.08206(273k)=n 0.57=moles

The ideal gas law can be used to solve all types of gas law problems.

Front 10

Daltons law of partial pressures:

Back 10

Dalton's Law: For a mixture of gases in a container, the total pressure exerted is the sum of the partial pressure of the gases present.

Partial pressure of a gas: The pressure that the gas would exert if it were alone in a container.

Dalton's Law: P total=P1+P2+P3...where the numbers represent individual gasses in a mixture.

The fact that the pressure exerted by and ideal gas is affected by the number of gas particles and is independent of the nature of the gas particles tells us:

1. The volume of the individual gas particles (atom or molecule) must not be very important

2. The forces among the particles must not be very important.

Example: If the total pressure of a mixture of a water and oxygen gas is 754 torr and the partial pressure of water is 21 torr, what is the partial pressure of the oxygen gas?

754 torr=21 torr + partial oxygen
733 torr=p oxygen.

Front 11

Laws and Models: A review

Back 11

The ideal gas law is the most useful because it relates al the important gas properties.

Under certain conditions, like high pressure and/or low temperatures, gases do not obey the ideal gas law.

Laws do not tell us why nature behaves the way it does.

Theories or models are speculations about how individual atoms or molecules cause the behavior of macroscopic systems.

Any model is an approximation and is destined to be modified.

Front 12

The Kinetic Molecular Theory of Gases:

Back 12

The kinetic molecular theory is a relatively simple model to explain the behavior of an ideal gas, based on speculatoins about the behavior of the individual particles (atoms or molecules) in a gas.

Front 13

Postulates of the Kinetic Molecular Theory of Gases:

Back 13

1. Gases consist of tiny particles (atoms or molecules).

2. These particles are so small, compared with the distance between them, that the volume (size) of the individual particles can be assumed to be negligible (zero).

3. The particles are in constant random motion, colliding with the walls of the container. These collisions with the walls cause the pressure exerted by the gas.

4. The particles are assumed not to attract or to repel each other.

5. The average kinetic energy of the gas paricles is directly proportional to the kelvin temp of the gas.

KE(kinetic energy)=1/2mv 2, where m is the mass of the particle and v is the velocity.

(NOT ON TEST

Real gases at high temperatures and/or low pressures conform to these assumptions.

Front 14

The Implications of the Kinetic Molecular Theory: The meaning of temperature.

Back 14

Temperature is a measure of the motions of the gas particles and the Kelvin temperature of a gas directly proportional to the average kinetic energy of the gas particles.

At high temperatures, the gas paricles move very fast and hit the walls of the container frequently; whereas, at low temps, the particles move slowly and collided with the walls of the container much less.

The relationship between pressure and temp:
-pressure and temp are directly related; as temp is increased for a given sample of gas, the pressure increased if the volume is not changed.

The reltionship betweeen volume and temperature:
-volume and temperature are directly related; as temperature is increased for a given sample of gas at a constant pressure, the volume will increase.

Front 15

Gas Stoichiometry:

Back 15

In dealing with the stoichiometry of reactions involving gases, it is useful to define the volume occupied by 1 mole of a gas under specified conditions.

Molar Volume of a Gas: the volume of a mole of an ideal gas at 0 C (273 k) and 1 atm pressure.

Molar Volume=22.4 liters at STP

STP=standard temp and pressure are 0 C and atm pressure.

Front 16

CHAPTER 14:

Liquids and solid:

Back 16

The solid and liquid states are more similar than the liquid and gaseous states. It takes more energy to go from a liquid to a gas, than from a solid to a liquid.

Property Gas Liquid Solid

Density low high very sightly

Compressibility high slight very slightly

Form completely fill container fills container, but doesn't change volume maintains it's shape

Front 17

Water and It's Phase Changes:

Back 17

-Water behaves differently from other liquids in that the density of ice is lower than that of water and that is why ice floats. (liquid, gas and solid)

-Normal boiling point-the temp (c) at which a liquid becomes a gas (vapor) at 1 atm pressure.

-Normal Freezing Point: The temp (C) at which a liquid becomes a solid at 1 atm pressure.

-Heating/Cooling Curve for water:
*Going from left to right, heat is added at a constant rate. Going from left to right, heat is removed at a constant rate.

Front 18

Energy Requirements for the Changes of State:

Back 18

Physical changes-Changes of state from solid to liquid and from liquid to gas.

Intramolecular Forces-Bonding forcew that hold the atoms of a molecule together (forces within the molecule).

Intermolecular Forces-forces that occur among molecules that cause them to aggregate to forma solid or a liquid (forces between molecules).

Molar Heat of Fusion-The energy required to melt 1 mole of substance.

Molar Heat of Vaporization-The energy required to change 1 mole of liquid to a vapor.

Ex: Liquid to Solid: Calculate the energy required to melt 8.5 g of ice at 0 C. The molar heat of fusion for ice is 6.02 kJ/mol.

Solution: The molar heat of fusion is the enrgy required to melt 1 mole of ice. We need to convert 8.5 g of ice to moles of water. One mole of water, H 2 0, weighs 18.0g

8.5g H20 x 1 mole H20/18.0 g H20=0.47 mole H20

Then we can calculate the amount of energy required to melt 0.47 mol of ice

0.47 mol H20 x 6.02 kJ/mol H20=2.8 kJ

Example: Liquid to Gas: Calculate the energy (in kJ) required to heat 25 g of liquid wter from 25 C to 100 C and change it to steam at 100 C. The specific heat capacity of liquid water is 4.18 kJ C and the molar heat of vaporization of water is 40.6 kJ/mol.

Step 1: Heating to Boiling
-energy required: specific heat capacity x mass of water x temperature change
-energy required: 4.18 J/g degrees 25 g x 75 C=7.8 x 10 3 J=7.8 kJ

Step 2: Vaporization
-convert grams water to moles water
25g H20 x 1 mole H20/18 g H20=1.4 mol H20
-40.6 kJ/mol H20 x 1.4 mole H20=kJ
-Total energy is the sum of the two steps.
7.8 kJ + 57 kJ = kJ

Front 19

Intermolecular Forces:

Types of Intermolecular Forces

Back 19

Dipole-dipole attraction: Attraction between the positive end of one dipole to the negative end of another dipole. Only about 1% as strong as covalent or ionic bonds and they become weaker as the distance between the dipoles increases.(middle amount of strength).

Hydrogen Bonding: Particularly strong dipole-dipole forces that occur between molecules in which hydrogen is bound to a hightly electronegative atom, such as nitrogen, oxygen, or fluoring. (Reason why water had such a high boiling point, compared to other Group VI hydrides like H 2 S, H 2 Se, and H 2 T e). This is the strongest type of bond.

London Dispersion Forces: The weak and short-lived attraction between instantaneous dipoles formed between nonpolar molecules due to the temporary uneven distribution of charge as electrons move around the nucleus. This is the weakest of the forces.

Hy

Front 20

Evaporation and Vapor Pressure

Back 20

Vaporization or evaporation: the escape of molecules in a liquid from the liquid's surface and into the gaseous phase.

Condensation: The process by which vapor molecules form a liquid.

Vapor Pressure: The pressure of the vapor present at equilibrium with it's liquid (rates of condensation and evaporation are equal).

The vapor pressure of a liquid is determined by the intermolecular forces that act among the molecules.

Volatile Liquids: have high vapor pressures and evaporate rapidly. (If you can smell it, it has high vapor pressure).

Front 21

The Solid State: Types of Solids

Back 21

Crystalline Solids: Solids with a regular arrangement of their components, like sodium chloride (NaCl), quartz (SiO2), and iron pyrite (FeS2).

1. Ionic solids-ions are present and are free to move when dissolved in water and conduct a current. (sodium chloride) This is a combination of a metal and a non-metal.

2. Molecular Solids: Molecules are present and are free to move when dissolved in water, but do not conduct a current. (sugar) Covalent bonding; doesn't form an ion.

3. Atomic Solids: Atoms are present and only of one element. (graphite and diamond both contain carbon) One element.

Front 22

Bonding in Solids:

Back 22

Ionic Solids (like batteries): Stable substances with high melting points that are held together by the strong forces that exist between oppositely charged ions (e.g. NaCl). Intramolecular ->metals and non metals (+)(-).

2. Molecular Solids: Solids held together by the relatively weak (covalent bonding) intermolecular forces between molecules and have relatively low melting points (e.g. ice, dry ice=solid CO 2). Remember: Intermolecular bonds are dipole, hydrogen and london dispersion.

Atomic Solids: Solids made up of atoms interacting with each other and can have very diverse properties. (e.g. diamond, graphite).
-Network Solid: Atomic solids that contain large molecules (diamond).
-Metals: Atomic solids which have a regular array of metal atoms in a sea of valence electrons. The valence electrons are shared among the atoms in a non directional way and are quite mobile. The mobile electrons can conduct heat and electricity.

Front 23

Alloys: A substance that contains a mixture of elements and has metallic properties.

Back 23

1) Substituional alloy: An alloy in which some of the host metal atoms are replaced by other metal atoms of similar size.

2) Interstitial Alloy: An alloy in which some of the interstices (holes) among the closely packed metal atoms are occupied by the atoms much smaller than the host atoms (e.g. steel).

Front 24

CHAPTER 15:

Solutions

Back 24

Solution: a homogenous mixture, a mixture in which the components are uniformly intermingled. (A solution can be a gas (the atmosphers), a liquid (coffee), or a solid (brass=mixture of copper and zine)).

Solvent: Substance in a solution present in the largest amount.

Solute: Substance in the solution present in the lesser amounts. Usually solid, but doesn't have to be.

Ex: 95 g H 2 0 (solvent) and 5 g NaCl (solute).

Aqueous Solutions: Solutions with water as the solvent.

Front 25

Solubility: Types of solutions (homogenous mixture; dissolves in it)

Back 25

*Gas solutions resulting from gases mixing-air or natural gas.
*Liquid solutions resulting from the liquids mixing-ethanol in water or anti-freeze in water.
*Solid solutions resulting from various metals mixing-brass.
*Liquid solutions resulting from a gas dissolved in a liquid-carbonated water.
*Liquid solutions resulting from a solid dissolved in a liquid-sugar solution.

Front 26

Sodium Chloride Dissolving in Water: Water has a dipole.

Back 26

When an ionic compound dissolves in water, it breaks up into individual cations and anions, which are dispersed in the water.

When a polar molecular compound dissolves in water, the molecules are dispersed in the water by the attraction of the polar ends of the molecule with the oppositely charged polar end of the water molecule. If hydrogen bonding is possible, than this type of bonding will also form.

Non polar molecules do no dissolve in water, which is polar. Ethanol-polar solubles dissolve polar solubles. Non polare dissolves non polar.

Common Table Sugar (sucrose) and Petroleum Hydrocarbon: OH is very electronegative and it is polar. Non polar is not very electronegative (gas and oil).

Solubility Rule: Like dissolves like, or polar solvents dissolve polar molecules and non polar solvents dissolve non polar molecules.

Front 27

Solution Composition:

Back 27

Terms associated with concentration of a solution:

Saturated-a solution that contains as much solute as will dissolve it at that temperature (sugar and water heated until you can't dissolve any longer. Cool, it will stick in it).

Unsaturted-A solution that has not reached the limit of solute that will dissolve in it.

Concentrated-A solution containing a large amount of solute (a lot).

Dilute-A solution containing a relatively small amount of solute (a little)

Know that is the solubility is 50+ it's saturated and 49.9 and below, it's unsaturated.

Front 28

Solution Composition: Mass Percent

KNOW THIS FORMULA

Back 28

Mass percent (weight percent)-mass of the solute present in a given mass of a solution.

Mass percent=mass of solute/mass of solution x 100%

Ex: A solution is prepared by mixing 1.00g of ethanol (solute), C 2 H 5 OH (solvent) with 100.0 g of water.

Calculate the mass percent of ethanol in this solution.

Mass percent C 2 H 5 OH=grams of C 2 H 5 OH/grams of solution (solvent and solute) x 100%

=1.00g C 2 H 5 OH/100.0g H 2 0 + 1.00 g C 2 H 5 OH

=1.00g/101.0 g x 100%

20 g ethanol, 100 g h2o

mass % = wt solute/wt solution

.20/100 + 20 x 100

20/120 = .167 x 100%

Front 29

Solution Compositon: Molarity

KNOW THIS FORMULA

Back 29

Concentration-the amount of solute in a given volume of a solution

Molarity (M)-the number of moles of solute per volume of solution in liters.

M=Molarity=moles of solute/liters of solution=mol/L

*Moles per liter=molarity

Ex: Calculate the molarity of a solution prepared by dissolving 11.5 g of solid NaOH in enough water to make 1.50 L of solution.

-First calculate the number of moles of NaOH.
11.5g NaOH x 1 mol NaOH/40.0 g NaOH=0.288 mol NaOH

-Then divide by the volume of the solution in liters.

Molarity=moles of solute/liters of solution=0.288 mol NaOH/1.50 L solution=0.192 M NaOH.

Standard solution: a solution whose concentration is accurately known. Usually an accurately weighed sample is transferred completely to a volumetric flask (accurately marked volumes) and enough solvent is added to bring the solution up to the mark on the neck of the flask. _

Front 30

Dilution:

KNOW THIS FORMULA

Back 30

Dilution: The process of adding more solent to a solution (remember when doing problem that only water is added).

Dilution formula: M1 x V1=M2 x V2, where m1 is molarity before dilution, v1 is volume before dilution, m2 is molarity after dilution and v2 is volume after dilution.

Ex: What volume of 16 M sulfuric acid must be used to prepare 1.5 L of a 0.10 M H 2 SO 4 solution?

16 M x V1=0.10 x 1.5 L
16 x V1=0.15L
V1=9.4 x 10 -3L or 9.4 ml

or V1=M2 x V2/M1

Front 31

Steps for solving stoichiometric problems involving solutions:

Back 31

1. Write the balanced chemical equation for the reaction, and for reactions involving ions, write the net ionic equation.

2. Calculate the moles of reactants.

3. Determine which reactant is limiting.

4. Calculate the moles of other reactants or products, as required.

5. Convert to grams or other units, if necessary,.

Front 32

Neutralization reactions:

Back 32

A neutralization reaction is an acid base reaction in which just enough of a strong base (or conversely, stong acid) is added to react exactly with the strong acid (or conversely, strong base) in solution.

A net reaction where there is one hydrogen per molecule and one hydroxide per molecule.

H+ (aq) + OH-(aq) -> H 2 0(l)

Solution stoichiometry is calculating volume in neutralization reactions.

What volume of a 0.100 M HCl solution is needed to neutralize 25.0 ml of a .350 M NaOH solution?

Skip the M, use N.

Ma xVa = Mb x Vb, so:

Va=Mb x Vb/Ma

0.350 m x 0.025 L/0.10 = 0.0875 L HCl

Front 33

Solution Composition: Normality

Back 33

One equivilant of an acid-the amount of that acid tha cxan furnish 1 mole of H+ ions.

One equivilant of a base-the amount of that base that can furnish 1 mole of OH- ions.

Ex: HCl, mm 36.5g, equiv. wt=36.5 g (1 H+ ion).

H 2 S0 4, mm 98.0g, equiv. wt=49.0 (98/2, 2 H+ ions).

NaOH, mm 40.0g, equiv. wt.=40 g (1 OH- ion).

Front 34

Molar masses and equivilant weights of common strong acids:

Back 34

ACIDS:
HCl: mm 36.5g, equiv. wt. 36.5g
HNO 3: mm 63.0g, equiv. wt. 63.0g
H 2 SO 4: mm 98.0, equiv. wt. 49.0 g

BASES:
NaOH: mm 40.0g, equiv. wt. 40.0g
KOH: mm 56.1g, equiv. wt. 56.1g

Front 35

Calculate the equivilant weight of phosphoric acid:

Back 35

Phosphoric acid gives 3 H+ ions per molecule. Thus, 1 equivilant of H 3 PO 4 (the amount that can furnish 1 mole of H+) is one third of a mole. The equivilant weight equals 1/3 of it's molar mass.

Equiv. wt.=molar mass/3

Front 36

Normality:

Back 36

The number of equivilants of solute per liter of solution.

Normality (N)=number of equiv./1 L of solution=equiv/liter=equiv/L

Solve problems in the same manner as molarity problems, using equivilents instread of moles.

Ex: Calculate the normality of a sulfuric acid solution containing 86g of H 2 S0 4 per liter of solution.

Equiv. wt. H 2 SO 4=mm (98/2)

86 g H 2 SO 4 x 1 equiv. H 2 S0 4/49.0 g H 2 SO 4=1.8 equiv H 2 S0 4

N=equiv/L=1.8 equiv H 2 S0 4/49.0 g H 2 S0 4

Front 37

Neutraization Equation:

Back 37

One equivalent of acid reacts with one equivalent of base. (One proton reacts one hydroxide=normality)

N acid x V acid=N base x V base

Ex: What volume of a 0.075 N KOH solution is required to react with 0.135 L of 0.45 N H 3 P0 4?

0.45 equiv/L x 0.135 L=0.075 equiv/L x Vbase

0.81 L = V base

or, N a x V a/N b = V b

0.45N x 0.135L/.075=0.81L

Front 38

CHAPTER 16: Acids and Bases

Arrhenius Model:

Back 38

Acids taste sour and bases have a bitter tast and feel slippery.

(OH-)

Arrhenius Model: Arrhenius was the first person to recognize the nature of acids and bases.

Arrhenius Acid is a compound that produces hydrogen ions in aqueous solution.

HCl(g) -> H+(aq) + Cl-(aq) is a strong acid.

Arrhenius Base is a compoind that produces hydroxide ions.

NaOH(s) -> Na+(aq) is a strong base.

The Arrhenius concept of acids and bases is limited because it only has one type of base, OH- ion.

Front 39

Bronsted-Lowry Model:

Back 39

(H+)

Danish chemist Johannes Bronsted and English chemist Thomas Lowry suggested a more general definition of acids and bases.

Bronsted-Lowry acid is a compound that is a proton donor.

Bronsted-Lowry Base is a compound that is a proton acceptor.

When an acid dissolving in water can be represented as an acid (HA) donating a proton to a water molecule to form a new acid (the conjugate acid) and a new base (the conjugate base).

HA(aq) + H 2 0(l) -> H 3 0+(aq) + A-(aq)

Front 40

Conjugate Acid Base Pair:

Back 40

Two substancew related to each other by the donating and accepting of a single proton.

The reaction shown above has two conjugate acid-base pairs: HA (acid) and A- (base) and H 2 0 (base) and H 3 0+ (acid)

This model emphasizes the significant role of the polar water molecule in pulling the proton from the acid.

Hydronium Ion: simplified to H+, is a water molecule with an added proton, H 3 0+.

Ex. of conjugate acid base pairs:

HF is acid, F- is conj. base
NH 4 + is acid, NH 3 is conj. base
NClO 4 is acid, ClO 4 - is the conj. base
H 3 P0 4 is acid, H 2 PO 4 - is conj. base

Front 41

Acid Strength:

Back 41

When an acid dissolves in water, the acid transfers a proton to water to form a new acid, H 3 0+, which can also react with the newly formed base and the reverse reaction can also occur.

HA(aq) + H 2 0(l) <=> H 3 0+(aq) + A-(aq), reverse can usually occur.

Strong acids are completely ionized or completely dissociated.

HA(aq) + H 2 0(l) -> H 3 0+(aq) + A- (aq), works only in on direction.

Weak acids have the reverse reaction dominate and most of the HA molecules remain intact.

HA(aq) + H 2 0(l) <- H 3 0+(aq) + A-(aq)

A solution can conduct a current in proportion to the number of ions in it.
-strong acids: strong electrolytes and conduct an electric current.
-weak acids: weak electrolytes and conducts an electric current only weakly.

Front 42

Relationship of Acid Strength and Conjugate Base Strength for Dissociation Reaction:

Back 42

HA(aq) + H 2 0(l) <=> H 3 0+(aq) + A-(aq)

If the relative acid strenth is very strong, the relative conjugate base is very weak.

If the RAS is strong, it's RCBS is weak.

If the RAS is weak, it's RCBS is strong.

If the RAS is very weak, it's RCBS is very strong.

A strong acid contains a relatively weak conjugate base and a weak acid contains a relatively strong conjugate base.

Front 43

Ways to describe acid strength:

Back 43

We can list them by property, strong acid or weak acid.

In the acid ionization (dissociation reaction), the forward reaction predominates in the SA or the reverse reaction predominates in a WA.

The strength of the conjugate base compared with that of water is that the SA, A- has a much weaker base than H 2 0 and A- has a much stronger base than H 2 0 in a weak acid.

Front 44

Oxyacids:

Back 44

Oxyacids are acids in which the acidic hydrogen is attached to an oxygen atom.

Phosphoric Acid- H 3 PO 4 ((OH)3 P=O) is a strong acid

Nitrous Acid-O=N-OH is a strong acid

Acetic Acid-CH 3 CO 2 H is a weak acid.

Strong acids completely dissociate.

Front 45

Organic Acids:

Back 45

Organic acids have a carbon backbone, commonly the carboxyl group.

Front 46

Water as an Acid and a Base:

Back 46

Amphoteric Substances can behave either as an acid or as a base.

Water is the most common amphoteric substance. One water molecule is acting as an acid and the other one as a base.

H 2 0(l) + H 2 0(l) <=> H 3 0+(aq) + OH-(aq). The equilibrium lies to the left. A little hydronium and a little hydroxide (aq).

Front 47

Know this info about water, acids and bases:

Back 47

In pure water at 25 C, the concentrations of H 3 0+ and OH- are: [H 3 0+]=[OH-]=1.0 x 10 -7 M.

At 25 C the product of H 3 0+ and OH- concentrations is a constant. [H 3 0+] [OH]=1.0 x 10 -14=Kw, where Kw is te ion product constant.

To simplify the notation, we often write H 3 0+ as H+ and then: [H+] [OH-]=1.0 x 10 -14=Kw, where Kw is the ion product constant.

In any aqueous solution at 25 C, no matter what it contains, the of [H+] and [OH-] must always equal 1.0 x 10 -14.

Front 48

For Aqueous Solutions:

Back 48

In a neutral solution, [H+]=[OH-]

In an acidic solution, [H+]>[OH-]

In a basic solution, [H+]<[OH-]

In each case, however, Kw=1.0 x 10 -14

Front 49

Calculating Ion Concentrations in Water

Back 49

Calculate [H+] concentration for 1.0 x 10 -5 M OH-. Assume 25 C.

[H+] [OH-]=1.0 x 10 -14
[H+] x 1.0 x 10 -5=1.0 x 10 -14

[H+]=1.0 x 10 -14/1.0 x 10 -5=1.0 x 10 -9 M

This solution is basic, since the concentration of OH- is greater than the concentration of H+.

Front 50

The pH scale:

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To express small numbers conveniently, chemists often use the p scale, which is based on common logarithms (base 10 logs). In this system, if N represents some number, then:

pN=-log N=(-1) x log N or, the p means to take the log of the number that follow and multiply the result by -1.

The pH scale is a convenient way to represent solution acidity. pH=-log[H+]

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Steps for calculating pH on my calculator:

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#, ee, chg sign, =, log, equals, -, chg to pos.

The number of decimal places for a log must be equal to the number of significant figures in the original number.

EX: [H+]=1.0 x 10 -5 M, so pH=5.00

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Relationship of the H+ concentation of a solution to its pH.

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Since the pH scale is based on 10, the pH changes by 1 for every power of ten change in [H+]

1.0 x 10 -1 PH 1.00
1.0 x 10 -2 PH 2.00 etc.

Log scales similar ot the pH scale are used for representing the OH- concentration.

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pOH Scale:

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pOH=-log[OH-]

Ex: Calculate the pOH for a solution containing 1.0 x 10 -3 M OH.

pOH=-log[OH-]=-log (1.0 x 10 -3)=3.00

In aqueous solutions: pH + pOH = 14.00

Ex: Calculate the pOH for a pH of 6.00 solution. The answer is 8.

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Calculating [H+] from pH (anti-log)

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On my calculator:
#, chg sign, 2nd, log

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Calculating the pH of strong acid solutions:

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Calculate the pH of 0.10 M HNO 3

HNO 3 is a strong acid, so the ions in solution are H+ and NO 3-

0.10 M HNO 3 -> 0.10 M NO 3-
pH = -log[H+] = -log(0.10) = 1.00

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Buffered Solutions:

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Buffered solutions resist change in pH even when a strong acid or base is added to it.

A solution is buffered by the presence of a weak acid and it's conjugate base.

Ex: Buffer system HC 2 H 3 O 2 (acetic acid) and NaC 2 H 3 O 2 (sodium acetate) added strong acid reacts with the acetate anion to form acetic acid.

H+(aq) + C 2 H 3 O 2- ->HC 2 H 3 O 2(aq)

Added strong base reacts with acetic acid to form wate rand the acetate anion.

OH-(aq) + HC 2 H 3 O 2(aq)

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Characteristics of a buffer:

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1. The solution contains a weak acid HA and it's conjugate base A-.

2. The buffer resists changes in pH by reacting with any added H+ or OH- so that these ions do not accumulate.

3. Any added H+ reacts with the base A-. H+(aq) + A-(aq) -> HA(aq)

4. Any added OH- reacts with the weak acid HA. OH-(aq) + HA (aq) -> H 2 0 (l) + A-(aq)

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Important stuff!

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[H+][OH-]=1.0 x 10 -14
pH= -log [H+]
pOH= -log [OH-]
pH + pOH=14.0
[H+]=antilog (-pH)
[OH-]=antilog (-pOH)