ap chemistry study guide (part 2 of 3) - edl€¦ · exceptions to octet rule octet rule - the...

AP Chemistry Study Guide (Part 2 of 3)

Chapter 8 - Bonding Ions An element that reacts to form noble gas configuration When 2 nonmetals react, they share electrons in covalent bonds Ionic compounds are solid crystals They align themselves to achieve maximum attraction between opposite charges and to minimum repulsion between like ions Calculating Lattice Energy Lattice Energy = k(Q1Q2)/t k is a constant Q is charge on ions r is shortest inter-nuclear distance between cation and anion Lattice Energy > with more highly charged ions Partial Ionic Character No such thing as a completely ionic bond Calculate the percent ionic character through percent dipole

Percent dipole =

x 100

Ionic compounds is defined as any substance that conducts electricity when melted Ionic Bonds Can be broken down into a multistep process involving 1 or all of the following: Hf° of ionic substance -Sublimation energy (solid to gas) -Bond dissociation energy -Ionization energy -Electron affinity -Lattice energy Also referred to as the Born-Haber Cycle Energy of Ionic Attraction or Repulsion Calculated by Coulomb's Law:

E = 2.31 x 10-19Jnm(Q1Q2)/r

-Q is charge on ions -r is distance between ions centers in nanometers -If ions have same charge, energy will show repulsion with + sign (endothermic) -If ions have opposite charge, energy will show attraction with - sign (exothermic) Covalent Bonding Electrons attracted to nucleus of the other Electrons repel each other Electrons are shared, but not evenly Dipole Moments A molecule with a center of + and center of - charge is dipolar and has a dipole moment Will line up in the presence of an electric field


The Be has a partial + charge and the Ar has a partial - charge. The arrow is pointing toward the Ar which means the compound favors pulling in the negative direction

How It's Drawn

Covalent Bond Energies When calculating the energy of a covalent bond, we only look at the average energy. What really happens is: CH4 ---> CH3 + H H = 435 kJ/mole CH3 ---> CH2 + H H = 453 kJ/mole CH2 ---> CH + H H = 425 kJ/mole CH ----> C + H H = 339 kJ/mole The average is 413 kJ/mole, which is what we use for CH4 (Methane) The more bonds there are, the shorter the bond length Exceptions to Octet Rule Octet Rule - The center atom cannot hold more than 8 electrons Column IIA, IIIA molecules -Less than octet, electron deficient model Third row and larger exceed octet -Extra electrons go on center atom Resonance/Formal Charge Sometimes more than 1 structure for a molecule or ion (All polyatomic ions are resonant) Formal charge can be calculated by: Valence Electrons - Nonbonding electrons - 1/2 bonding electrons Example of resonant structure with formal charges: XeO3

What Molecules Have Them? Any molecule with a polar bond As long as geometry does not cancel it out Geometry and Polarity 3 shapes will cancel them out Linear (Center atom with 2 of the same thing attached) Planar Triangles (Column IIIA with 3 same things) Tetrahedral (Center atom with 4 same things)

Formal Charge (Upper Left):

Xe: 8 - 2 -

= 3

O: 6 - 6 -

= -1

Formal Charge (Upper Right):

Xe: 8 -2 -

= 2

O: 6 - 6 -

= 3 (single bonded)

O: 6 - 4 -

= 0 (double bonded)

Formal Charge (Lower):

Xe: 8 - 2 -

= 0

O: 6 - 4 -

= 0

Ideal formal charge is 0 or very close to 0. Whatever resonance structure contains the atoms with formal charges closest to 0 is ideal structure.

AP Chemistry Study Guide (Part 2 of 3) VSEPR Tells us the shape of the molecule Shape greatly affects its properties Molecules take shape that puts electron pairs as fair away from each other as possible Lone pairs of electrons take more space Multiple bonds counts as 1 pair Tells us the angle, underlying structure, and the actual shape

Chapter 9 - Hybridization and Molecular Orbitals Molecular orbital - An overlap of 2 atomic orbitals, has room for 2 electrons with opposite spins -Number of molecular orbitals is equal to the number of atomic orbitals combined Two type of molecular orbitals: Sigma σ bonds - Occur between atoms Pi π bonds - Occur above and below atoms Sigma and Pi Bonds All single bonds are sigma bonds All double bonds are 1 sigma and 1 pi bond All triple bonds are 1 sigma and 2 pi bonds When a sigma bond occurs, there is a bonding molecular orbital and an antibonding molecular orbital -Antibonding represented with a " * " -Bonding orbitals have lower potential energy than antibonding -Reason: The antibonding has a low electron density, leaving only repulsion between nuclei

Bond Order Determines the stability of the proposed molecule. -0 or a negative value means that the molecule has no stability To determine bond order, use equation:

So, for the H2 example: We have 2 electrons bonding and 0 electrons anti bonding

AP Chemistry Study Guide (Part 2 of 3) Pi bonds

sigma bonds: σPx and σ*Px pi bonds: πPy π*Py and πPz π*Pz

Hybridization Blends s, p, and/or d orbitals at valence electrons Based off experience Explains the geometry of atomic orbitals Applied on VSPER table General Rules: Linear - sp Trigonal Planar - sp2

Tetrahedral - sp3 Trigonal bipyramidal - sp3d Octahedral - sp3d2

Magnetism Ferromagnetic - Strongly attracted by a magnet Paramagnetic - Attracted by a magnet because of unpaired electrons Diamagnetic - Weakly repelled by magnet because of paired electrons

When p and s orbitals interact the energies of π p and σ p are reversed σ s and σ s* will no longer be equally spaced


Diamagnetic vs. Paramagnetic: The reason B2 is paramagnetic is because the energy level for π p and σ p flip. Why they flip: Because the σ p bond concentrates electrons in the same area as the σ s, there is an increase in repulsion which leads to a higher potential energy. However, the π py and π pz bonds are lower in energy because they are formed outside the increased concentration area. Proof: Remember that all triple bonds form 1 sigma bond and 2 pi bonds. Looking at the expected configuration, and knowing we can't skip over a segment, we see that it forms 2 sigma bonds and 0 pi bonds which would disagree with the bond rule.

The Molecular Orbital Model Assumes: A new entity is formed with positively charged nuclei and electrons. Electrons are inside the orbital Correctly predicts bond strength, magnetism, and polarity Involves the whole molecule, not the individual atoms Only involve the outer orbitals Homonuclear Diatomic Molecules Be2:

B2:

Diamagnetic because there are no

unpaired sets of electrons. The bond

and antibond both have a pair of

electrons

AP Chemistry Study Guide (Part 2 of 3) Let's Screw With Our Brains...

Notice that N2 has the π p bonds at a lower energy level. Once it hits O2, the π p and σ p flip

Assume: They all double bond, except Li2

Li2 and Be2 are both diamagnetic because all of the electrons bond B2 is paramagnetic because each Boron only has 1 electron in the p orbital Remember: The overlap of the σ p and σ s cause the π p orbitals to have lower potential energy C2 is diamagnetic because each Carbon has 1 pair of electrons N2 is diamagnetic because each Nitrogen has 1 pair of electrons AND lone electron. Because the π p bonds are already filled, the 2 lone electrons will bond together in the σ p Why does configuration order flip? Remember: π bonds have more energy than σ bonds. Remember: There is still that concentrated area of electrons containing σ p and σ s Remember: The π bonds existed outside that concentrated area Well... Since the orbital is half filled, it is big enough for those π bonds to enter the concentrated area. Remembering π bonds > σ bonds In Energy terms , we can go back to the expected configuration order as seen on the previous page with B2 This can also be explained through repulsion energy. As the orbital gets bigger, the repulsion between the σ p and σ s is decreasing to the point that the π p will be stronger than the σ p O2 is paramagnetic because each Oxygen technically has 1 electron leftover after filling out the half filled orbital with the new expected configuration. These electrons will now be placed in the π*2p bonds F2 is diamagnetic because each Fluorine has 1 pair of electrons which will complete the π*2p orbitals in the new expected configuration Ne2 is diamagnetic because each Neon has 1 pair of electrons AND 1 lone electron. With every orbital filled, except the σ*2p, the 2 lone electrons will bond together to complete the configuration


Chapter 10 - Liquids and Solids The 2 Forces Intramolecular forces - Occurs inside molecules, the atoms bonding Intermolecular forces - Occurs between molecules - Holds molecules together in condensed states Types of Intermolecular Forces Covalent bonding: Strong Ionic bonding: Weak Dipole - Dipole: 1% as strong as covalent, dependent upon distance between dipoles and how close electric field is. Hydrogen bonding: Especially strong dipole - dipole force when Hydrogen bonds with highly electronegative element (F, O, or N) London Dispersion: Nonpolar force. Weaker than other forces. Molecular forces < atomic bonds Dipole - Dipole Forces Allow for gases to liquefy easier Relatively unimportant Hydrogen Bonding Forces Especially strong dipole - dipole force Makes the boiling point higher because it keeps atoms bonded more strongly Requires high amount of energy to make a gas

Each water molecule can make up to 4 Hydrogen bonds

The drop in boiling point occurs because the hydrogen is bonding at a point where the hybridization of the orbitals is at its least repulsive moment, so it will be easier to melt.

The 2 H-O bonds account for 2 of the

hydrogen bonds, but when around

other molecules containing a central

atom of O, N, or F. The δ+ of the 2

Hydrogens will create another

hydrogen bond

Notice that each water molecule will

have 2 intermolecular forces and 2

intramolecular forces.

AP Chemistry Study Guide (Part 2 of 3) London Dispersion Forces Takes into account that electrons are not evenly distributed at every moment in time There is an instantaneous dipole When the instantaneous dipole occurs, it induces a dipole on the atom next to it, starting a chain reaction Weak and short-lived, but last longer at low temperatures

Van der Waal's Forces Include London Dispersion forces and anything involving dipole interactions Order of strength: London Dispersion < Dipole < Hydrogen bonds < Covalent/Ionic bonds Liquids Gets many of its properties due to the internal attraction of atoms -Surface tension, beading, capillary action, and viscosity -Viscosity: The resistance to flow Stronger intermolecular forces cause these properties to increase Surface Tension Molecules at top are pulled inside Molecules in middle are pulled in all directions

Beading Those droplets of water that form when a car is waxed is an example of beading The reason: A polar substance (water) is on a nonpolar surface (wax), which results in cohesive forces, but no adhesive forces. The water will not be attracted to the wax at all, so the water molecules forming will just sit on top of the wax -Cohesion: Attraction between molecules of the liquid -Adhesion: Attraction between liquid molecules and a surface

The picture to the left represents an instantaneous dipole moment surrounding a non-dipole atom. At that instantaneous dipole moment, the non-dipole atom will become dipole as shown in the picture on the right.

The ultimate goal of surface tension is to minimize the surface area. The picture to the left explains why there is a meniscus when we deal with liquids... It helps create the minimum surface area

AP Chemistry Study Guide (Part 2 of 3) Capillary Action Glass is a polar substance and will attract water molecules The meniscus in a capillary depends upon which force is stronger: -If cohesive forces are stronger, a convex meniscus will form -If adhesive forces are stronger, a concave meniscus will form

Viscosity The larger the force, the more viscous the substance will be Structure helps determine viscosity Example: Cyclohexane vs. Hexane

Solids Two types of solids: amorphous and crystalline Amorphous - Lots of disorder in the structure Crystalline - Consistent arrangement (order) within the structure Crystals and Types of Unit Cells Lattice - 3D grid that describes location of pieces in a crystalline solid Lattice point - A point in which one unit cell is connected to another unit cell Unit cell - Smallest repeating unit of the lattice Simple cubic Body-centered cubic Face-centered cubic

Cyclohexane will flow better than hexane because cyclohexane's structure is more compact than hexanes.


Special Atomic Solid - Carbon 3 arrangements: Coal, diamond, and graphite Coal: Amorphous solid Diamond: Insulates heat and electricity Graphite: Slippery, conducts electricity Structure dependent upon atomic arrangement of Carbon Diamond Carbon is sp3 hybridized which gives the molecule a tetrahedral shape with each Carbon being bonded to 4 other Carbons Strong σ bonds give the molecule it's hardness Graphite Carbon is sp2 hybridized which gives the molecule a flat structure with 120° angles between each carbon in a 6 member ring Buckyball C60 - Combination of 5-point and 6-point Carbon rings


' Molecular Solids Form when molecules occupy the corners of lattices Each molecule will have different forces between them Forces depend upon size of the molecule and strength/nature of dipole moments Molecules Without Dipole Most gases at 25°C London Dispersion Forces Larger molecules (such as I2) can be solids without dipoles

Ionic Solids Have huge melting and boiling points -Reason: The atoms are locked in the lattice (they will be hard and brittle) Poor conductors and good insulators -ONLY good conductors when they are melted or in a solution (i.e. salt vs. salt water) Vapor Pressure Vaporization - Changes from a liquid to solid at or above boiling point Evaporation - Changes from a liquid to a solid below boiling point -How is this possible? Pressure. (Will be explained later under "Dynamic Equilibrium" section) Hvap - Energy required to vaporize 1 mole at 1 atm Remember back to Thermochem: Vaporization is an endothermic process -Reason: It requires energy to break the intermolecular forces Condensation Condensation - Changes from a gas to a liquid -Attempts to achieve dynamic equilibrium with vaporization rate in a closed system -Remember back to Thermochem: A closed system is when molecules (matter) can't escape -"Dynamic"; Molecules constantly changing phase - "Equilibrium"; Rates of vaporization and condensation are the same Dynamic Equilibrium - Melting Point Two VERY important things to remember when considering dynamic equilibrium 1.) Vapor pressure will affect how the system acts, such as the phase change from a solid to a gas and how quickly the system will achieve equilibrium (Remember Thermochem?) 2.) Vapor pressure and temperature have a direct relationship (Remember gas laws?) First off, what is vapor pressure?: Vapor pressure is the pressure exerted by the vapor above the liquid when the system has achieved equilibrium Second, how is it going to affect phase changes?: With a higher vapor pressure, the molecules will be less attracted to whatever it is they are usually attracted to and go straight to the top of the liquid or solid.

Left to Right: Carbon as tetrahedral molecule (Diamond) Carbon as a flat, 120° angle structure Carbon as a buckyball Diagram showing the pi bonding of Cgraphite

Vapor - Gas molecules of a substance, when the substance is below it's critical temperature Critical temperature - The temperature in which a phase of a substance ceases to exist (i.e. water molecules existing as a gaseous state in a room temperature flask)

AP Chemistry Study Guide (Part 2 of 3) Third, how will temperature affect vapor pressure?:

Best explained through experimentation: 1.) You put solid water (ice) in one flask, and liquid water in another flask. The 2 flasks are covered and they are connected with a stop cock tube that allow the system to be interchangeably open/closed. 2.) Some of the molecules will gradually escape the surface of the solid and turn into water vapor Solid - gas phase is called sublimation This will only occur when the temperature is above freezing point since it is a solid - gas phase change and requires some sort of energy to break the intermolecular forces 3.) If the system is not already at equilibrium (equal rates of vaporization/condensation) then some of the liquid molecules will turn into water vapor (vaporize) and turn back into a solid state until the system reaches equilibrium. Capable of turning directly into a solid from gas state (deposition) through pressure. Won't pressure affect the temperature? No. The reason why temperature won't be affected is because the pressure will be increased in one quick motion, a "surge", which will ultimately lead to those intermolecular forces reforming but in a different state. The stop cock connecting the flasks will allow for this quick pressure change, by letting an external force such as a piston to apply pressure to the system. The piston will quickly enter the system when the stop cock is open and create enough pressure for the intermolecular forces to attract again, shaping the vapor molecules into a solid.

What is this even? The label "number of molecules" refers to the number of liquid molecules.

All this is saying is that when the temperature is higher, there will be more Kinetic Energy, and a less amount of liquid molecules. In further detail: Well, if there is a higher temperature then there is more kinetic energy. With this, the intermolecular and intramolecular forces will be broken, causing a higher vapor pressure. If there is a high vapor pressure, then that means the molecules will turn into a gas faster making there be less liquid molecules.

Number of Molecules

Kinetic Energies

AP Chemistry Study Guide (Part 2 of 3) Wait...what? Okay, so just a brief summary of the experiment: Remember: The ultimate goal is for the system to reach dynamic equilibrium. Step 1.) Molecules of same substance but different states are placed in two closed flasks, but the stop cock allows it to be both a closed or open system on demand. Step 2.) Depending on the initial temperature, either the solid will sublimate and then condense, or the liquid will vaporize then deposit. Temperature: If temperature is ABOVE freezing point then the SOLID will begin the dynamic equilibrium reaction. If temperature is BELOW freezing point then the LIQUID will begin the dynamic equilibrium reaction. Step 3.) If ABOVE freezing point, then liquid molecules will vaporize and after pressure is applied, will deposit into solid. If BELOW freezing point, then solid molecules will sublimate and after pressure is applied, will condense into liquid This will continue until the system has achieved dynamic equilibrium. What We Can Conclude The temperature in which no external pressure is needed to achieve dynamic equilibrium is considered the melting point. The point in which the rate of vaporization is equal to the rate of condensation will be the point of dynamic equilibrium, which will also be the melting point. The higher the initial vapor pressure of the substance, the quicker the system will achieve equilibrium. Phase Diagrams

Left Picture: Phase diagram of water Right picture: Phase diagram of water with dashed lines Tm - Melting point temperature Expt 1: Pressure is 1 atm T3 - Triple point Expt 2: Pressure is 2.0 torr Tb - Boiling point temperature Expt 3: Pressure is 4.58 torr Tc

- Critical point temperature Expt 4: Pressure is 225 atm Pc - Critical point pressure P3 - Triple point pressure

All these diagrams are telling us is that at a certain pressure and temperature, there will be a certain phase. BUT....Like always, it's never that easy.

AP Chemistry Study Guide (Part 2 of 3) Phase Diagrams Continued Okay, so here are basic definitions to help read these graphs and an explanation of each experiment. Remember that this is only for water, the values will be different for every molecule Melting point temperature: 0°C Boiling point temperature: 100°C Critical point: The point in which a state of matter for a given substance will cease to exist Critical point pressure: Pressure required to liquefy vapor at the critical temperature 218 atm Critical point temperature: Vapor cannot be liquefied no matter how much pressure is applied 374°C Triple point: The point where all 3 states of matter exist at the same time Triple point pressure: .0060 atm Triple point temperature: .0098°C For the experiments, water is set up in a container looking something like this:

Background: Ice is placed in a container at -20°C. The container is sealed with a moveable piston that will apply pressure.* * Does not apply to Experiment 4 Experiment 1: Pressure of the piston is constant at 1 atm with no vapor pressure present. The container is heated until the temperature reaches 0°C where the ice turns into liquid and energy is added. As we can see on the graph, if we continue to heat it, it will go into the gas phase Experiment 2: Pressure is constant at 2.0 torr with the container being heated. Once it hits -10°C, the ice will sublimate into vapor. The reason for this is because the vapor pressure of water is always above 2.0 torr, so if vapor pressure cannot exist in that state, then that state will also not exist. With the graph, we see that the line does not pass through the liquid state. Experiment 3: Pressure is at 4.58 torr with the container being heated. Once the temperature hits .0098°C, all 3 states of matter will be present. This is called the triple point. The reason this can occur is because all 3 phases have the same vapor pressure of 4.58 torr. Looking at the graph, we can see that the line goes through the point where all 3 phases intersect. Experiment 4: This experiment we start with liquid water in the container at 300°C and constant pressure of 225 atm. The high pressure allows for the liquid to be present at such a high temperature. As the container is heated further, the liquid will gradually turn into vapor and eventually hit a point where it is neither liquid or vapor. This happens because the temperature is beyond the critical point of water Looking at the graph, we can see that the line is above the critical point and is neither a liquid or gas.

AP Chemistry Study Guide (Part 2 of 3) Phase Diagrams Continued The graph will look different for every molecule Every molecule has a different triple point Every molecule has a different critical point because each phase for that molecule is dependent upon the pressure in which a phase change will occur.

Chapter 11 – Solutions and Solubility Review

Percent mass:

Mole fraction:

- Also works with pressures - Can be used with more than just 2 compounds

Molality:

-Abbreviated by “ ” Normality: Molarity x # of active pieces -For redox active pieces will be # of electrons, H+ ions, and OH- ions Energy of Making Solutions = Energy change in making a solution (3 basic steps)

1.) Break apart the solvent Have to overcome attractive intermolecular forces > 2.) Break apart the solute Have to overcome attractive intermolecular forces > 3.) Mixing solvent and solute Depends on what we are mixing

-If molecules are attracted to each other, the will be large and negative -If molecules are weakly attracted to each other, the will be small and negative -This explains the “like dissolves like” rule -General rule: Double digits is considered small, Triple digits is considered large

Type of Solvent and Solutes If is small and positive, a solution will still form because of entropy There are many more ways for solutions to be mixed than to be separated

Structure and Solubility Water soluble molecules must be polar, which means they will have dipole moments To be soluble in nonpolar solvent, the molecule must be nonpolar Hydrophobic – Water hating ends dissolve in grease Hyrophilic – Water loving ends dissolve in water Water molecules can surround and dissolve in grease Pressure Effects (Gases dissolving in liquids) Change in pressure will not affect the amount of liquid or solid that dissolves Reason for this is because they are incompressible, will only affect gases The dissolved gas in at equilibrium with the gas above the liquid Increasing pressure will cause the gas molecules to dissolve faster (Equilibrium is disturbed) Everyday Example: Soda Carbonation: CO2(g) molecules with small amounts of liquid attached. It is indeed not a special bubbly material When flat, there is more gas and less liquid If we tap the top, carbon dioxide molecules rise to the top because there was a temporary change in pressure that disturbed the equilibrium and needed to balance back out

AP Chemistry Study Guide (Part 2 of 3) Henry’s Law C = kP C is the concentration of dissolved gas k is a constant for a particular solution P is the pressure of a dissolved gas above solution Stronger attraction between dissolved gas and gas above solution higher constant Temperature Effects Increasing temperature will affect the rate at which a solid will dissolve, but does not help us determine if it will increase the amount of solid dissolved. In other words, just because the rate is high doesn’t mean that a lot of solid dissolved Can see temperature effects by a graph of experimental data As temperature increases, solubility increases (Gases only) Vapor Pressure of Solutions Nonvolatile solvent lowers vapor pressure of solution -Nonvolatile, meaning it does not easily turn into a liquid Molecules of solvent must overcome the forces of both the solvent and solute molecules Pure water has a high vapor pressure compared to solutions containing water and other ions Because of this we can conclude that pure water will also evaporate faster However, pure water will not condense faster than when in a solution Raoult’s Law Psolution is vapor pressure of solution Applies only to an ideal solution where solute does not contribute to vapor pressure Deviation When a solvent has a strong affinity to a solute, there is Hydrogen bonding The lower the vapor pressure, the lower the solvents ability to escape Negative deviation – strong attraction between solute and solvent large and negative (Exothermic) Positive deviation - weak attraction between solute and solvent positive (Endothermic) Colligative Properties Dissolved particles affect vapor pressure, which affect phase changes These properties are ONLY dependent upon the number of particles, not what kind of particles Useful for determining the molar mass Boiling Point Elevation/Freezing Point Depression The more non-volatile the solute, the lower the vapor pressure will be Higher boiling point, lower freezing point… Equation: Change in temperature = Boiling point elevation constant (given and determined by solvent) = Molality of solute Equation:

= Freezing point depression constant (given and determined by solvent)

Electrolytes in Solution/Van Hoff Factor Since colligative properties only depend on the number of particles -Ionic compounds should have bigger effect when they dissolve because of disassociation -Example: NaCl -Individual Na+ and Cl- ions fall apart

AP Chemistry Study Guide (Part 2 of 3) -1 mole of NaCl makes 2 moles of ions Electrolytes will have a great effect on the freezing point and boiling point (Makes more pieces) This is known as the Van’t Hoff factor, represented as “i”:

Actual value will be less at any given instance because ions will be paired Ion pairing will increase with concentration which means i will decrease with increasing concentration Osmotic Pressure RT

It s just an uppercase pi = Osmotic pressure (in atm). This is the pressure difference between solute and solvent

M = Molarity of solution (Do not confuse with molality) R = Gas constant (.0821 Latm/K mole) T = Temperature in Kelvin Helps to know what the osmosis even is. Osmosis – The flow of a solvent into a solution through a semi-permeable membrane

-If there is a different liquid level, there will be a different hydrostatic pressure (pressure involving ions mixing with water) on the solution, not the solvent -The minimum pressure that stops osmosis is the osmotic pressure of the solution

Equations Involving the Van Hoff Factor

Equation Normal w/ Van Hoff Freezing point depression

Boiling point elevation Osmotic pressure RT i RT

It’s quite simple: To include the Van Hoff factor, just multiply the normal equation by “i”. The only hard part is remembering how to find it. Other Definitions Crenation – Cells are placed in hypertonic solution, lose water, and shrink Hemolysis – Cells are placed in a hypotonic solution, gain water, and swell (potentially bursting) Colloids – Two substances, different phases, slowly separate won’t separate through filtration - Usually colloids will have high molar masses/relatively large sizes (1000nm) Suspensions – Two substances, same phases, easily separate (separate through filtration) - Suspensions have sizes greater than 1000nm (visible to naked eye) and are murky/opaque Tyndall Effect – Process where light is reflected off very small particles in a colloid or suspension in a transparent medium (Best example, headlights on a foggy night)

Chapter 12 – Chemical Kinetics The collision theory states that in order for 2 particles to chemically react, they must have enough energy and collide with proper orientation. It is important to know this to help understand kinetics. Factors that Affect the Rate of a Chemical Reaction Surface Area (Opportunity for collision) Concentration (Frequency) Temperature (Frequency) Catalyst (Effective collisions) Nature of Reactants (Effective collisions) Surface Area -Sugar cube vs Packet of sugar

Packet of sugar will dissolve faster because the molecules are smaller and will require less time for the bonds to break.

AP Chemistry Study Guide (Part 2 of 3) Concentration -Magnesium in various concentrations of acid Mg placed in 12M HCl, 6M HCl, or 2M HCl --- Dissolves fastest in 12M HCl Temperature -Fruit spoiling, spoils faster at room temperature compared to in a fridge - Alka Seltzer dissolving in hot water vs. cold water

- Reactions will occur faster at hotter temperatures because the heat is producing more energy that will break the bonds faster, allowing for the reaction to occur quicker - A general rule is that the rates of a reaction will double every 10°C

Catalysts -A substance that speeds up a reaction without being consumed -Enzyme: Large molecule used as a catalyst in biological reactions -Homogeneous catalyst: Same phase as reactants in reaction -Heterogeneous catalyst: Different phase as reactants in reaction Heterogeneous Catalyst Steps:

1.) Adsorption and activation of reactants (Reaction starts) 2.) Migration of absorbed reactants on surface (Reactants spread to maximize reaction) 3.) Reaction of the absorbed substances “Final touches” of reaction 4.) Escape, or desorption, of products (Reaction completes into products)

Homogeneous Catalyst Steps: Best shown through example of reaction mechanism

1.) NO(g) +

O2(g) NO2(g)

2.) NO2(g) NO(g) + O(g) 3.) O2(g) + O(g) O3(g)

Final)

O2(g) O3(g)

-Highlighted things are called intermediates -Intermediates are molecules that are produced and then immediately used up in the following reaction

Reaction Mechanisms Basically just the steps of a chemical reaction Important to understand that chemical reactions take several steps to complete Chemical reactions don’t show how reactants become products, reaction mechanisms do Helps us find the way for the reaction to occur in the easiest way possible Can change through a catalyst

Important Terms Intermediate: Formed in first step, used up in the next Molecularity: Number of species that must collide to produce the reaction indicated by that step

-Expressed as unimolecular, bimolecular, trimolecular, etc. Elementary Step: A reaction whose rate law can be written from its molecularity

Requirements Sum of elementary steps must match the overall balanced equation Mechanism must agree with the experimentally determined rate law Slowest step determines the rate law Integrated Rate Laws Zero Order - Rate is constant First Order - Concentration and rate both double Second Order - Concentration doubles, rate quadruples Half-Life - How long it takes for the concentration to reach half its initial concentration

Determining the Order 1.) [X] in later trial/[X] in earlier trial = L 2.) Rate of later trial/Rate of earlier trial = M 3.) L^n = M By later/earlier trial, I mean of the 2 chosen. It doesn't necessarily mean that it HAS to be the latEST/earliEST trials. Anyways, enough ranting, let's finish this.

AP Chemistry Study Guide (Part 2 of 3) Rate Law Graphs Zero Order: First Order:

Second Order:

Arrhenius Equation Derivation of Arrhenius Equation k = Ae-Ea/RT ln ln (Use when not given frequency factor)

k = Rate constant ln

(Use when not given frequency factor)

A = Frequency factor Ea = Activation energy Ea = -R x slope (Used for graphical data) T = Temperature in Kelvin R = .0821 Latm/Kmole

Order Rate Law Integrated Rate Law

Axis for straight line Half-Life

Zero Rate = k [A] = -kt + [A]0 [A] vs time t

A k

First Rate = k[A] ln[A] = -kt + ln[A]0 ln[A] vs time t

.

k

Second Rate = k[A]2

A kt

A

A vs time t

AP Chemistry Study Guide (Part 2 of 3) Final Notes on Kinetics As temperature increases, so will k Ea and E are dependent of temperature Catalysts will lower Ea and increases rate of both kproducts and kreactants

Chapter 13 Chemical Equilibrium Beginning Notes Always write the right arrow on top and left arrow on bottom The equilibrium constant is identified as Kx (Where x is basically any letter in the English alphabet except "x") Know when to use Kx

K will be constant at any temperature Equilibrium involves ONLY, ONLY, ONLY aqueous solutions and gases CAPITALIZE THE "K"!!! Big "K" won't have units, unlike "k" for rate laws, because it is all grown up Calculated by concentration of products over concentration of reactants ALWAYS write the skeletal equation before plugging in actual concentration numbers

Special K Table Alright, that may or may not have made sense, so I will do a quick example. 3NaOH(aq) + K3PO4(aq) 3KOH(aq) + Na3PO4(aq)

Keq =

Note that the coefficients are raised as exponents to that specific compound

Relation Between Keq and Kp Just 1 simple equation:

Kp and Keq are self explanatory R - Universal gas law constant (.0821 Latm/Kmole) T - Temperature in Kelvin n - Change in coefficients (Products - Reactants) Reaction Quotient (Q) It's calculated the same exact way as K, I won't waste my time writing it out again The purpose? To compare Q and K K will be given so, there are 3 things to remember depending upon the Q and K value: 1.) Q = K; System is at equilibrium, no shift 2.) Q > K; System favors products, shifts left to balance out 3.) Q < K; System favors reactants, shifts right to balance out Trouble forgetting? Look at the "<" or ">" sign and whatever side has 2 points is the side it will shift towards. Le Châtlier's Principle Principle: When an external stress is applied to a system at equilibrium, it will correct itself to undo the stress Factors that will affect equilibrium: Concentration Pressure/Volume Temperature Catalysts

Keq or Kc - Basically a "universal" constant (applies to any equilibria) Kp - Equilibrium constant involving pressure Ka - Equilibrium constant involving acids Kb - Equilibrium constant involving bases Kw - Actual constant at room temperature (1.0 x 10-14) KaKb = Kw Ksp - Equilibrium constant involving slightly soluble salts Kf - Equilibrium constant involving complex ions (ligands)

aA(aq) + bB(aq) cC(aq) + dD(aq)

[Capital letter] = Concentration of product/reactant Lowercase exponents = Corresponding coefficient of reaction


Factor Scenario Shift

Concentration Increase reactants; Decrease

products Right

Decrease reactants; Increase products

Left

Pressure/Volume

Increase pressure; Decrease volume

Side with least gaseous moles

Decrease pressure; Increase volume

Side with most gaseous moles

Temperature

Increase temperature Endothermic - Right

Exothermic - Left

Decrease temperature Endothermic - Left Exothermic - Right

Concentration Simple and logical. You increase concentration of reactant or product, it will react more, so shift to the opposite side. Pressure/Volume Remember the gas laws? They are back yet again. Must know that pressure and volume are inversely related (Increase pressure. decrease volume...vice versa). As long as you know just 1 of the 4 possibilities relating to pressure or volume, you can figure the other 3 out. Generally, the easiest one to remember is, increase pressure, shifts to side with the least amount of gaseous moles. It's as simple as counting the coefficients of the left and right. Temperature A bit more tricky. Still easy. You just need to know whether the reaction is endothermic or exothermic. Yes, thermochem is back yet again. So, logically, if the reaction is endothermic, meaning the heat is on the reactants side, and temperature is increased... the forward reaction will be quicker, so there will be a shift to the right. Remember that temperature can be written as a H. Positive is endo, negative is exo. Catalysts This is as easy as it gets. There is no shift. A catalyst will lower the activation energy for both the forward AND reverse reaction which means they will still both be equal still. It will only increase the rate.

*REMEBER: Solids and liquids DO NOT count in equilibrium*

Reverse Reactions Let's compare forward and reverse: Forward: aA(aq) + bB(aq) cC(aq) + dD(aq) Reverse: cC(aq) + dD(aq) aA(aq) + bB(aq) The reverse reaction is indeed, the opposite of the forward reaction. The only similarity is that the right arrow is still on top. However, with the reverse reaction comes a reverse Kc, known as K'c It's exactly what you would think it would be:

What if it isn't at Equilibrium?!?!?!? If it isn't at equilibrium, then we will get to explore the wonderful land of ICE charts. You will need to learn to love these, because they won't leave you alone...ever Usually, we are given a Kx value and initial concentrations. The purpose is to find the concentrations of the solutions at equilibrium I - Initial, C - Change, E - Equilibrium Best shown through example.

AP Chemistry Study Guide (Part 2 of 3) In depth look at ICE Charts .50 M H2(g) reacts with .50M I2(g) to create HI(g) at 430°C. Kc = 54.3

H2(g) + I2(g) 2HI(g)

Initial .50M .50M 0.0M Change -x -x +2x

Equilibrium .50-x .50-x 2x Okay, so...the first step should be to write the balanced chemical equation. Sometimes it will be given, and sometimes it won't. Then, in the margin, or somewhere with relative organization, write I, C and E in a column downwards on the left side of the equation Then write down the initial concentrations under the specific chemical in the reaction and in the Initial (I) row. If you aren't given a concentration for a specific chemical, assume it is 0 Next, simply write the change. It is typically x on the reactants or products and x on the other side. Note that in this particular example, the 2x came from the coefficient (2HI), not because we have 2 -x's on the other side. Basically, match the change with the coefficient. If it's 1, it's a x. If it's 2, it's a 2x, and so on. Finally, write equilibrium by adding initial with change. Now that the ICE chart is set up, we need to plug it into the equilibrium expression. Remember, ALWAYS write the general equation first:

c HI

H I

Now plug in, using the equilibrium values

. x

. x

Then use basic algebra to solve for x

. x

. x .

x

. x . . x x . . x x .

Square root everything Cross multiply Simplify Divide Now we can find any concentration at equilibrium. [H2] = [I2] = .50 - x = .50 - .393 = .107M [HI] = 2x = 2(.393) = .786M Sadly, this is just a very basic ICE chart, it can get a lot more complicated and will be discussed at the end of this guide, with the rest of the conceptual questions that should be answerable through this guide.

Chapter 14/15 - Acids, Bases, Titration, and Equilibrium Before doing anything, we must know what an acid is and what a base is. Acids Produce H+ or H3O+ ions in water (H3O+ is known as the hydronium ion) Taste sour, corrodes metal, contains electrolytes Turns blue litmus paper red, pH less than 7 Reacts with bases to form salt and water (with some exceptions) Strong acid, weak conjugate base, vice versa Bases Produce OH- ions in water Tastes bitter and chalky, slippery and soapy Turns red litmus paper blue, pH greater than 7 Reacts with acids to form salt and water (with some exceptions) Strong base, weak conjugate acid, vice versa

AP Chemistry Study Guide (Part 2 of 3) It's quite useful to know what a strong acid and a strong base is Strong Acids HCl, HBr, HI, HNO3, HClO4, and H2SO4 Strong acids will completely dissociate and separate completely into its ions (explained later) Strong Bases Any column IA alkali metal attached to a hydroxide ion

Strong bases will also completely dissociate into its ions Definitions There are also 3 different definitions as to what an acid/base is:

Definition Acid Base Arrhenius Substance that produces H+ or

H3O+ ion in water Substance that produces OH- ion

in water

Brønsted-Lowry Proton donor Proton acceptor

Lewis Accepts and electron pair Donates an electron pair

Brønsted-Lowry will actually tell us the conjugate acid and conjugate base. So basically, the proton in this theory is an H+ atom that lost its electron. So the conjugate acid will be an acid with one less H+ and the conjugate base will be an acid with an extra H+ This is best shown through example: So, let's say we have the reaction: H2SO4(aq) + H2O(l) ---> H3O+

(aq) + HSO4-(aq) (The dissociation of sulfuric acid)

We know the acid is H2SO4(aq), which means H2O is a base in this reaction. Starting with the acid, we know the conjugate acid will lose a H+ ion, so HSO4

- is the conjugate acid That means the base, gains a H+ ion, so H3O+ is the conjugate base We can also organize them as pairs: H2SO4 ----- HSO4

- (Acid) H2O ----- H3O+ (Base) It's important to remember that water is still special and can do something called autoionization. This fancy word just means that it can give or lose a H+ ion to become H3O+ or OH-....This means water can be an acid or a base. Molecularity of Acids/Bases Like almost every other topic in AP Chemistry, acids and bases can be related to bonding. The bonds helps us understand what makes a strong acid or base, and how to order acids by increasing strength and more. I will start with the simpler one. Electronegativity and Acids Let's look at HF, HCl, HBr, and HI. All of them being from group VIIA With the knowledge that electronegativity decreases down a group, and also to be a strong acid, it must completely dissociate into its ions, we can see that the order from weakest to strongest is: HF < HCl < HBr < HI. The reason for this is because the more electronegative an element is, the stronger the bond will be. Since F- is the most electronegative element, it will not want to let go of H+ that easily. So that means it won't want to dissociate completely into its ions, making it a weak acid. So, the less electronegative the ion is attached to H+ the stronger the acid will be. We can then look at the electronegativity going across a period: CH4, NH3, H2O, and HF Now ordering from weakest to strongest: CH4 < NH3 < H2O < HF This is because of the polarity of the bond. CH4 is nonpolar and will definitely not want to give up that H+ ion Oxidation Numbers If we look at 2 sets of acids: HNO2 and HNO3, we can see which acid is stronger (despite the fact that we know HNO3 is one of the strong acids)

AP Chemistry Study Guide (Part 2 of 3) The bond looks something like this:

Now we can take a quick look at bases, for they are much easier because they form ionic bonds. Any group IA or IIA metal (with the exception of Be) attached to OH- will be a strong base. The reason for this is because they have a low oxidation number and will have a weak attraction to the OH- Acid-Base Reactions and Equilibrium Now we get into the fun stuff, this is different from redox reactions, because we aren't looking at the electrons and the reactions are at equilibrium, not one-way. Neutral Solutions This is the easiest: Strong acid + Strong base will create a neutral salt and water NaOH(aq) + HCl(aq) NaCl(s) + H2O(l)

If we feel it is necessary to expand upon that, we can make another reaction from the products: NaCl(s) + H2O(l) Na+

(aq)+ Cl-(aq)

This is a dissociation reaction showing that NaCl will break into its ions when mixed in water. This occurs because it is a Group IA metal (Part of a strong base) attached to a conjugate base of a strong acid (HCl, conjugate base is Cl-). If you critically think, you shouldn't have a problem understanding that 2 reactions can occur at the same time. Acidic or Basic Favored This is completely logical and shouldn't have to be explained, but I will anyways. Weak acid + strong base will create a basic solution Strong acids + weak base will create an acidic solution We can dig in deeper though to see just why this will happen: Let's take the reaction: NH4Cl(s) + H2O(l) NH4OH(aq) + HCl(aq) From this we can see that it is an acidic solution, because we have a strong acid on the products If we look at the net ionic equation: NH4Cl(s) NH4

+(aq) + Cl-

(aq) This will show us the important ions and conveniently happens to be the dissociation reaction as well (They're the same thing). Both the ions are soluble. The chloride can't be broken up anymore; however, the ammonium ion can. So we go to the "sub-reaction": NH4

+(aq) NH3(aq) + H+

(aq) Notice that an H+

(aq) ion is produced; henceforth, acidic. Now let's do the same thing for a basic-favored solution

Generalizations for Acids: Across a period - Strength will increase Down a group - Strength will increase The stronger the acid, the weaker the conjugate base In non-monoprotic acids (acid with more than 2 elements including H), the higher the oxidation number on the central atom, the stronger the acid will be They have covalent bonds

So first, I will just clarify that just because there is an O-H bond, that doesn't mean it is immediately a base. Now, this can be looked at in 3 ways, and they all mean the same thing: 1.) The more bonds there are on the central atom (N), the stronger the acid will be 2.) The higher the oxidation state on the central atom, the stronger the acid will be 3.) The higher the ability for the central atom to attract electrons to itself, the stronger the acid will be.

Generalizations for Bases: Group IA and IIA attached to OH- ion, strong base Exception is Be(OH)2 Anything else will be a weak base Strong bases will have ionic bonds, weak bases will have covalent bonds such as NH4OH or H2O.

AP Chemistry Study Guide (Part 2 of 3) Reaction: NaCH3COO(s) + H2O(l) NaOH(aq) + CH3COOH(aq) Note that there is a strong base and weak acid, so we know it is going to be basic-favored, yet let's dig deeper, just for lols. Net Ionic Equation: NaCH3COO(s) Na+

(aq) + CH3COO-(aq)

The sodium ion can't break apart anymore, but the acetate ion should because it is polyatomic, so let's put it in water, which conveniently already happens to be in the system "Sub-reaction": CH3COO-

(aq) + H2O(l) CH3COOH(aq) + OH-(aq)

The acetate ion pulled the H+ from the water, and made some more of the weak acid, acetic acid. Which left the OH-

(aq) ion all by itself, which just so happens to be part of a strong base. Therefore, this reaction is basic.

So these equations can be related to something called the Common Ion Effect. The real definition of this is that when a common ion is added to an equilibrium reaction, there will be a shift in equilibrium. Sound familiar? It should. It's an application of Le Châtlier's Principle. It's best understood through example. Let's look at the same reaction we used before: NH4Cl(s) NH4

+(aq) + Cl-

(aq)

We will also have another reaction: NH3(aq) + H2O(l) NH4+

(aq) + Cl-(aq)

These reactions would be given in a word problem form...something like this: What direction would equilibrium shift if you add NH4Cl(s) to an ammonium solution? So, first do the equilibrium reaction because you need to know what reaction you are adding to the NH4Cl to. So it is ammonia solution which means water is involved. I won't bother rewriting the reaction, so just know that any compound in a solution will be the equilibrium. And for the compound you are adding, the reaction will be a dissociation reaction. Now we can see that by adding solid NH4Cl, to the equilibrium reaction, we are technically adding NH4

+ which

will shift equilibrium to the left. Common Ion: NH4

+ Acid-Base Reactions and Calculations With the exception of the first equation in the upper left, the equations on the left and the equations on the right are in some way related. That long equation in the upper right will be discussed later. Before doing any actual examples and calculations, we must clear up some minor details. First, you need to know what a major species is: Secondly, you need to know...sig figs?

It is pretty safe to say that with a strong acid and weak base reaction, the ion that splits up further into the "sub-reaction" from the net ionic equation will not require water. Yet, in a strong base, weak acid reaction it will.

pH + pOH = 14 pH = pKa + log

p log

[H+][OH-] = 1.0 x 10-14 KaKb = 1.0 x 10-14 (Kw) at room temperature -log[H+] = pH 10-pH = [H+] -log[OH-] = pOH 10-pOH = [OH-]

Major Species: Know whether or not the acid or base is strong or weak If it's a strong acid/base, the major species will be the ions of that acid or base because they will completely dissociate If it's a weak acid/base, the major species will simply be that weak acid or base H2O is ALWAYS a major species

Acid/Bases: Return of the Sig Figs...Sadly, you never learned all of the sig fig rules. Don't worry though, they aren't that hard. Basically, whenever converting from pH to [H+] or vice versa, the number of sig figs will be 1. If we convert from concentration to pH, the number of sig figs after the decimal in the pH is the number of sig figs of the concentration (+1 sig fig) If we convert from pH to concentration, the number of sig figs in the concentration is JUST the number of sig figs after the decimal in the pH (-1 sig fig) Say we have a concentration with 2 sig figs. The pH will have 2 sig figs after the decimal, along with however many there are before the decimal. These rules apply to [OH-] and pOH as well

AP Chemistry Study Guide (Part 2 of 3) Thirdly, titrations of weak acids/bases will involve 2 types of problems: A stoichiometry problem and an equilibrium problem. Fourth, you need to know what equivalence point is: Finally, acid/base equilibrium is very systematic: NOTE: There is a difference between acid/base titration and acid/base equilibrium, they both have equilibrium in them, but the steps are very different So, there's a very high chance that made little to no sense at all, so I will do an in-depth example: Let's do a weak base-strong acid titration, that way we can look at those Stoichiometry, equilibrium, and equivalence point...while applying the sig fig rules and calculations.

Problem: Calculate the pH in the titration of 25.0mL of 0.100M HCl with 0.100M NH3 after adding... a.) 10.00mL of 0.100M NH3 b.)25.0mL of 0.100M NH3 c.) 35.0mL of 0.100M NH3

Step 1.) List the major species Step 2.) See if the reaction goes to completion. If it forms H2O, it will go to completion Step 3.) If the reaction goes to completion: -Determine the concentration of the products -Write the major species after the reaction Step 4.) See if you are working with a basic or acidic solution by looking at a major component Step 5.) Pick the equilibrium that will control the pH -Write the equation for the reaction and K expression -Build an ICE chart Step 6.) Solve for "x" at equilibrium Step 7.) Calculate the pH and other concentrations at equilibrium

Stoichiometry for weak acid, strong base titration: The hydroxide ion + weak acid will go to completion and the concentrations of the acids and conjugate base will be determined Stoichiometry for weak base, strong acid titration: the hydrogen ion + weak base will go to completion and the concentration of the base and conjugate acid will be determined Equilibrium for weak acid, strong base titration: Position of weak acid is determined and the pH is calculated Equilibrium for weak base, strong acid titration: Position of weak base is determined and the pH is calculated from the pOH *You will be doing both of these in 1 problem but it is absolutely critical to keep them separated.*

Equivalence Point: This is defined as the exact number of moles required for the reaction. These are seen during acid/base titrations, and graphed on a titration curve. Titration curves are graphed as pH vs Volume of strong acid/base(mL). pH on the y- axis, volume on the x The location of the equivalence point on the titration curve will be dependent upon the strength of the acid and the strength of the base Strong Acid + Strong Base = Equivalence point with pH of 7 (volume is dependent upon how much of each acid/base you have) Weak Acid + Strong Base = Equivalence point with pH > 7 (volume dependent upon how much acid you have) Weak Base + Strong Acid = Equivalence point with pH < 7 (volume dependent upon how much base you have)

AP Chemistry Study Guide (Part 2 of 3) So let's start with a. Step 1.) Write the reaction: NH3(aq) + HCl(aq) NH4

+(aq) + Cl-

(aq) Or if you want to be pro and skip straight to the net ionic equation: H+

(aq) + NH3(aq) NH4+

(aq) Step 2.) Find how many moles of NH3 and HCl there are initially. This is a stoichiometry problem:

0.0250L HCl .

= .00250 mole HCl or H+

.01000L NH3 .

= .00100 mole NH3

Sub-Step 2.) Identify the limiting reactant and realize that the limiting reactant will have a final value of 0 because it will be all used up. Step 3.) Make an ICE chart if necessary, or you can be logical about it, I will do the ICE chart anyways... H+

(aq) + NH3(aq) NH4+

(aq) I .002500 .001000 0 C -x -x +x E .002500 - x 0 x .002500 - .001000 = .001500 mole of H+ Step 4.) Convert back to concentration and then find pH . mole H

. L .

.0350L came from 25.0mL + 10.0mL, the total volume So now we plug the concentration into the equation and get pH -log[H+] = pH -log(.0429M) = 1.368 pH = 1.368 (Concentration has 3 sig figs, pH has 3 sig figs after decimal) Now we do the same thing, for b. Step 1.) NH3(aq) + HCl(aq) NH4

+(aq) + Cl-

(aq) H+(aq) + NH3(aq) NH4

+(aq)

Step 2.) 0.0250L HCl .


.02500L NH3 .

= .00250 mole NH3

This is special because it is at the equivalence point, equal volumes were used. Step 3.) Okay, thanks to this being at equivalence point, there is a lot to do. In order, you would first need to find how many moles of NH4

+ there are, by building an ICE chart, or logically knowing that since it is at equivalence point, it will be the same as the initial moles of the reactants. Then you need to convert the moles of NH4

+ to concentration. After that reverse the reaction and build an ICE Chart, then solve for x.

Conversion to molarity : .

. . NH4

+

Reverse the reaction: NH4+

(aq) NH3(aq) + H+(aq)

ICE Chart with reverse reaction: NH4

+(aq) NH3(aq) + H+

(aq) I .0500M 0 0 C -x +x +x E .0500-x x x

Final or equilibrium value for NH3 is 0 because it is a limiting reactant and got completely used up. So logically, we can easily find x to be the same as the initial concentration of the limiting reactant, in this case 0.001000.

Wait, why do we reverse it? Technically, we don't "HAVE" to reverse it, it will just make the process quicker. By definition though, at equivalence point, the NH3(aq) will have completely converted into NH4

+

making the major species at this point: Cl-, NH4+ and H2O. The only major species here that can control pH

would be NH4+

as a dissociation reaction, which is exactly what is written. However if you choose not to reverse the reaction, you could use the forward reaction with the major species controlling the pOH which would be the following reaction: NH3(aq) + H2O(l) NH4

+(aq) + OH-

(aq) Either way, we will get the same answer when calculating the pH

AP Chemistry Study Guide (Part 2 of 3) Look up the Ka for NH4

+ in the appendix: 5.6 x 10-10 Solve for concentration of H+:

.

.

2.8 x 10-11 = x2 x = 5.3 x 10-6 [H+] = x = 5.3 x 10-6 Step 4.) Find the pH pH = -log[H+] pH = -log(5.3 x 10-6) pH = 5.28 (2 sig figs in concentration, 2 sig figs after the decimal in pH) Also note that because this is the equivalence point and we are adding strong acid, the equivalence point is at an acidic pH. Okay, now we do this process one more time for part c. The equivalence point is the longest one to do. Step 1.) NH3(aq) + HCl(aq) NH4

+(aq) + Cl-

(aq) H+(aq) + NH3(aq) NH4

+(aq)

Step 2.) 0.0250L HCl .


.03500L NH3 .

= .00350 mole NH3

Now, the HCl is the limiting reactant, which makes it a bit different from the first one, because that means, there is still some NH3 left after reacting with all the HCl to form NH4

+

Step 3.) .

. . NH3

.

. . NH4

+

NH4+

(aq) NH3(aq) + H+(aq)

I .0416M .0166M 0 C -x +x +x E .0416-x .0166+x x

. .

.

x = 1.4 x 10-9 [H+] = x = 1.4 x 10-9 Step 4.) Find the pH pH = -log[H+] pH = -log(1.4 x 10-9) pH = 8.85 (2 sig figs in concentration, 2 sig figs after the decimal in pH)

Below Equivalence Point:

1.) Write the (net ionic) reaction

2. ) Use stoich to find moles of strong acid/base and weak acid/base

3.) Make an ICE chart for moles or logically figure out how many moles of the strong acid/base you have left.

Logically: (Initial moles of strong acid/base - Initial moles of weak acid/base)

4.) Convert moles of strong acid/base (H+ or OH-) into molarity by dividing by total volume

5.) Calculate pH

Above Equivalence Point

1.) Write the (net ionic) reaction

2.) Use stoich to find moles of Strong acid/base and weak acid/base

3.) Determine the limiting reactant (Should be the strong acid/base) and find how many moles there are for

everything but the limiting reactant

Logically for reactants: (Initial moles of weak acid/base - Initial moles of strong acid/base)

Logically for products: Initial moles of strong acid/base

4.) Convert those to molarity by dividing by the total volume

5.) Make an ICE chart for the reverse reaction and solve for concentration of H+ or OH-

6.) Calculate the pH


Titration Curves This will be very quick, it's just to show what they look like, and we finally get some pictures

Strong Acid + Strong Base:

Weak Acid + Strong Base:

Weak Base + Strong Acid:

At Equivalence Point 1.) Write the (net ionic) reaction 2.) Use stoich to figure out moles of strong acid/base and weak acid/base It will conveniently be the same 3.) Make an ICE chart for moles or logically figure out how much of the products there is Logically: moles of products = initial moles of either reactant 4.) Convert the moles to molarity by dividing by the total volume 5.) Make an ICE chart for the reverse reaction 6.) Look up the Ka or Kb value and solve for [H+] or [OH-] 7.) Solve for pH

Halfway to Equivalence Point This is a very special point when titrating. Here’s why: If we do a weak acid/strong base titration: The concentration of the weak acid will be the same concentration as the conjugate base. If we do a weak base/strong acid titration: The concentration of the weak base will be the same concentration as the conjugate acid. The steps will be the same as “Before Equivalence Point”, but it is crucial to know this information.

Just a few notes: Whenever the x-axis is the volume of a strong acid, the curve will start high, and end low Whenever the x-axis is the volume of a strong base, the curve will start low and end high The equivalence point of a strong acid and strong base has a pH of 7 The equivalence point of a strong base and weak acid has a basic equivalence point (pH>7) The equivalence point of a strong acid and weak base has an acidic equivalence point (pH<7)

AP Chemistry Study Guide (Part 2 of 3) Buffer Solutions and Hendersen-Hasselbach Equation In order to have a buffer solution you will need a weak acid or weak base, and a salt of a weak acid or weak base. You need BOTH in order to be a buffer solution, not one or the other. Buffers are special because they have the ability to resist pH when an acid or base is added in small amounts So here is the last example for Acid/Base Equilibrium Calculate the pH of .30M NH3/.36M NH4Cl buffer system. What is the pH after the addition of 20.0mL of 0.050M NaOH to 80.0mL of the buffer solution? In a sense, this is titration-like so we will start by writing 2 equations. NH4Cl(s) NH4

+(aq) + Cl-

(aq) NH3(aq) + H2O(l) NH4

+(aq) + OH-

(aq) I .30M .36M 0 C -x +x +x E .30-x .36+x x So far so good, I just simply made an ICE chart, and now I will solve for [OH-], find pOH, then pH

. .

.

. .

.

[OH-] = x = 1.5 x 10-5 pOH = -log[OH-] pOH = -log(1.5 x 10-5) = 4.82 pH = 14 - pOH pH = 14 - 4.82 = 9.18 Well that answered the first question, now we need to answer the second one. This can be rather confusing but, we just need to split it apart and look for similarities. As I said before, this is kind of like a titration problem, but this time, we will skip the first step and go straight to the stoichiometry to moles.

. mole NH

L . L

. mole NH

. L . NH

. mole H

L . L

. mole H

. L . H

. mole NH

L . L

. mole NH

. L . NH

The .050M of OH- came from the .050M NaOH Now we can make an ICE chart for the moles NH3(aq) + H2O(l) NH4

+(aq) + OH-

(aq) I .24M .29M .01M C +.01M -.01M -.01M E .25M .28M 0 We subtracted the concentration of OH- because we need to find the pH once all of the NaOH has reacted. Now that we found the equilibrium concentrations, we can plug the equilibrium values in using the "sub-reaction" of NH4

+ because it will control the pH. We could follow through and do one for pOH, but that would take longer.

AP Chemistry Study Guide (Part 2 of 3) NH4

+(aq)

H+(aq) + NH3(aq)

. .

.

[H+] = 6.27 x 10-10 pH = -log[H+] pH = -log(6.27 x 10-10) pH = 9.203 Hendersen-Hasselbach Equation Now we will do the same problem again, but in much less time. First off, the equation is this:

pH = pKa + log

Secondly, this can only be used if the reaction follows 2 requirements: 1.) It must have the common ion effect 2.) You must be able to assume the shortcut when calculating the [H+] So for the first part we would first look up the Kb then find the Ka: KaKb = 1.0 x 10-14 Kb = 1.8 x 10-5

Ka = .

.

Ka = 5.6 x 10-10 Now to find pKa we take the negative log of Ka: -log(5.6 x 10-10) = 9.26 Finally, if you remember the conjugate base will have an additional H+ from the acid, we see that NH4

+ is the conjugate base and NH3 is the acid And with concentrations given (.30M for the acid and .36M for the conjugate base) we can put it all into the equation

pH = 9.26 + log .

. = 9.18

Then again, for the second part:

pH = -log(5.6 x 10-10) + log .

. = 9.203

This is quite useful, you just need to make sure you have the right concentrations. Percent Ionization This is very simple. It tells how much of the acid was ionized. The equation is:

[H+]E stands for the concentration of H+ at equilibrium [HA]I stands for the concentration of the acid initially Solubility Product Constant (Ksp) It is just like every other K, but looks at those ionic solids that are partially soluble Ksp will always equal concentration of cation times the concentration of the anion Ksp = [Cation][Anion] When dealing with Ksp there is 2 ways to measure the concentration: Molar solubility and solubility Molar solubility = (mole/Liter) Solubility = (g/L)

AP Chemistry Study Guide (Part 2 of 3) All Ksp values are very small, but the smaller it is, the least soluble that solid is Q and Ksp Remember Q, the reaction quotient? It's calculated the same way as all the K values are, and you will be given the Ksp, so now we will relate Q and Ksp Q < Ksp Precipitate will not form Q = Ksp Saturated solution Q > Ksp Precipitate will form Relation to pH If the solid precipitate contains an OH-, it will be less soluble in a basic solution and more soluble in acidic solution Say we are given a precipitate Mg(OH)2 and we need to find the Ksp, if we are given the pH, we can find the concentration of OH- by using the equation Complex Ions (Ligands) There is yet another K value, and this involves complex ions An equilibrium constant involving complex ions is represented as Kf

Ligand - The central molecule in a complex ion Coordination Number - The number of ligands attached to the metal ion The point of the complex ions is to attempt to increase solubility The saddest part is that ligands have a nomenclature as well, that you never learned:

|------------------------------------KNOW THESE REACTIONS--------------------------------------|

Ligand Nomenclature Ligand Name Metal Name as anion

Br - Bromo Iron Ferrate F - Fluoro Copper Cuprate O-2

Oxo Lead Plumbate OH- Hydroxo Silver Argenate CN- Cyano Gold Aurate

C2O4-2 Oxalato Tin Stannate

CO3-2 Carbonato

We will also need to remember Greek prefixes, and Roman/Binary nomenclature as well

CH3COO- Acetato NH3 Ammine H2O Aqua

Stem changes for anions: -ide --> -o -ate ---> ato -ide ---> -ito As you can see, they just fused pretty much all of the basic nomenclature into one and then added some more. It's best seen through example, but there is a general pattern to it: 1.) Start with the prefix 2.) Name the ligand(s) If there is more than 1 ligand, you place it in alphabetical order 3.) Name the metal (must know if complex ion is cation or anion) Cation: use the metals name Anion: *Refer to table* 4.) Place the Roman numeral 5.) If there is anything else, finish it with binary Let's just look at some examples: Ag(NH3)2

+ Zn(OH)4-2 FeSCN+2 Al(OH)4

-

Diamminesilver (I) Tetrahydroxozincate (II) Thiocyanatoferrate (III) Tetrahydorxoaluminate (III) AgCl + NH3 ---> Zn(OH)2 + OH- - ---> Fe+3 + SCN- ---> Al(OH)3 + OH- ---->

ap chemistry study guide (part 2 of 3) - edl€¦ · exceptions to octet rule octet rule - the...

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