Chemistry 3

Any reaction where one or more electrons are transferred from one atom to another.

The atom that loses electrons is oxidized and the atom that gains electrons is reduced.

A reducing agent is an atom or molecule that donates electrons to another atom or molecule and is itself oxidized in the process. An oxidizing agent is an atom or molecule that accepts electrons and is itself reduced in the process.

In this reaction iron loses two electrons and two hydrogens each gain one electron. Iron is therefore a reducing agent and water is an oxidizing agent. To recognize this, you must be able to calculate the oxidation state of each atom.

This is the apparent charge that an atom takes on while in a molecule. The sum of the oxidation states for all of the atoms in a molecule must equal the charge on that molecule, or equal zero if the molecule is neutral. Oxidation states have an assignment order.

Any elemental atom= 0
Fluorine= -1
Hydrogen = +1
Hydrogen w/ a metal= -1
Oxygen (except peroxides)= -2
Alkali metals = +1
Alkaline earth metals= +2
Group V = -3
Group VI= -2
Group VII = -1

Hydrogen will always be a 1+ unless bonded to a metal, in which case it will be a 1-. Fluorine will always be 1-.

a) N = -3 ; H = +1 ; S = + 6; O = -2

b) Fe = +2 ; C = +4 ; O = -2 ;

c) H = +1 ; O = -1

d) Na = +1 ; H = -1

e) S = +6 ; F = -1

Electrical potentials tell us the degree to which a species “wants electrons,” or “wants to be reduced.” These potentials are given in volts and are always presented in what is called a “half-reaction.” Below are a few examples of half-reactions and their associated potential in volts:

Ag2+(aq) + 2e- –> Ag(s) E° = 0.80 V
Cu+(aq) + 1e- –> Cu(s) E° = 0.52 V
Ni2+(aq) + 2e- –> Ni(s) E° = -0.23 V
Zn2+(aq) + 2e- –> Zn(s) E° = -0.76 V

Notice that in all of these examples an aqueous metal ion is being reduced to form the associated solid metal. This is by far the most common half-reaction. The only half-reactions you are likely to see on the MCAT that do NOT begin with a metal cation are: O2, H2O and H+.

2H+ + 2e- –> H2 E ̊ = 0.00V

The E° values assigned to each half-reaction represent the relative reduction potential of that species compared to the potential of two hydrogen ions to gain two electrons to form hydrogen gas. This is called the “Hydrogen Half-Cell.” It is the standard against which all other half- reactions are rated and we define its electrical potential as E ̊ = 0.00V. The naught symbol (°) signifies that all potentials are measured under standardized conditions.

A species with a positive E ̊ is more likely to gain electrons (i.e., be reduced) than are hydrogen ions. A species with a negative E ̊ is less likely to gain electrons than are hydrogen ions.

a) This statement is false. A species with a negative reduction potential can be spontaneously reduced as long as it is paired with another species that has a more negative reduction potential. This is easily proven. Suppose species A has a potential of -1.5V and species B has a potential of -1.8V. Species B will be oxidized, so we reverse the sign and add it to the potential for species A: -1.5 + 1.8 = 0.3V. With a positive cell potential we know this pairing would proceed spontaneously in a Galvanic cell.

b) This statement is also false. Hydrogen ions do have an affinity for electrons and can be reduced. The potential of 0.00V was arbitrarily assigned to the hydrogen half-cell to facilitate an assignment of standardized potentials. This is a good example of the need for students to think—to actually think—about the logic of a statement and try to evaluate it based on other things they know. If one thinks about this only in terms of “reduction potentials” it may not be obvious that this is a false statement. However, if one asks: “Is it logical that hydrogen ions cannot be reduced? Is it logical that a hydrogen ion, with its positive formal charge, has no affinity for electrons? Have I ever seen an H+ ion get reduced? A reduced hydrogen ion would be a hydrogen without a charge: or nearly every hydrogen one would encounter in chemistry outside of acids. Notice that this little demonstration of the need to “think” involved asking oneself a series of questions—or in other words—the Socratic Method!

c) We included this question because it seems to create confusion. Reduction half-reactions can be reversed to give oxidation potentials. In other words, the half-reaction runs in the opposite direction. Notice, however, that the reverse of one of these half-reactions involves the LOSS of one or more electrons as the metal forms the associated metal cation. For some reason, it is common for students to think that reversing the sign of the reduction potential gives the voltage associated with reduction of the solid metal. For the MCAT, just remember that cations (Cu+, Fe2+, etc.) get reduced to form solid metals (Cu(s), Fe(s), etc.), and solid metals get oxidized to form cations, but solid metals are NOT reduced.

The cell potential, E°cell, is the sum of the electrical potentials for the two half-reactions that make up
an electrochemical cell. Remember the following:

Half-reactions always come in pairs—one reduction half-reaction plus one oxidation half-reaction. This makes logical sense because in order for one species to be reduced, another species must be oxidized to make those electrons available. Although listed individually in a table, it always requires two species (i.e., two half-reactions) to make a cell.

Usually, only the reduction half-reactions are given in tables. The oxidation half-reaction is the reverse of the reduction half-reaction. E° for any oxidation half-reaction is simply the negative of E° for the associated reduction half-reaction.

You CANNOT add two E° values directly off of a half-reaction table. These are all reduction half- reactions and you need one of each—one reduction and one oxidation. Therefore, you must reverse the half-reaction of the species with the lowest reduction potential and take the negative of its E° value. Only after changing the sign can you add these two together to get the E°cell.

DO NOT USE STOICHIOMETRY! One mole of Cu2+ has the same reduction potential as two moles of Cu2+.

It would have a high reduction potential. It is favorable for it to take on electrons and oxidize something else in the process. For example, MnO4- and Cr2O7-2 have high reduction potentials.

O2(g) + 4H+(aq) + 4e- –> 2H2O(l) E ̊ = 1.23 V;
This is the last step of the Electron Transport Chain.

Galvanic cells convert chemical energy into electrical energy. By taking advantage of the difference in reduction potentials between two metals, a current can be spontaneously generated along a wire that connects two metal electrodes submerged in solutions that contain metal ions.

See the labeled diagram of a Galvanic cell below. Note that the zinc half-reaction was taken from a table. In that table the half-reaction was listed as the reverse of what is shown here. The table listed a reduction potential of -0.76 V. Be sure students understand why these changes were necessary to properly label this cell.

In the cell above notice that over time there will be a buildup of negative charge in the copper vessel due to continual loss of copper cations, and a buildup of positive charge in the zinc vessel due to the continual production of zinc cations. This polarity resists the flow of electrons and would eventually shut down the cell if a salt bridge were not present. Within the salt bridge sodium ions can flow toward the copper vessel and nitrate ions can flow toward the zinc vessel, neutralizing the buildup of charge and allowing electron flow to continue. The metal cations themselves, as well as any other ions in the solutions, can also flow through the salt bridge. In an electrical sense, the salt bridge connects the circuit, allowing continual flow of electrons from electrode to electrode and then back through the salt bridge via ion diffusion.

Reduction always happens at the cathode and oxidation always happens at the anode. This is true for ALL electrochemical cells!!!!! Reduction is occurring at the cathode, electrons must be flowing TO the cathode, not away from it.

Galvanic cells are spontaneous while electrolytic cells are not.

o Cathode = (+) ; Anode = (-) True of galvanic cells only, NOT electrolytic cells!

o Cell potential is always positive. True of galvanic cells only, NOT electrolytic cells!

o A functioning Galvanic cell can be created using any two metals, regardless of their reduction potentials. If a galvanic cell is properly set up, it will always produce current. Electrons will automatically flow from the species with the lower reduction potential to the species with the higher reduction potential. However, some MCAT questions read something like this: “Which of the following species can form a spontaneous Galvanic cell with copper, where copper is at the cathode?” That creates an entirely different situation. If the question requires that copper be reduced, then it must be paired with a species with a lower reduction potential.

Remember a Galvanic cell has a positive voltage. Adding galvanic cell half reactions to get an overall voltage or “emf” for the cell is very much like adding half-reaction to get a Gibb’s free energy or enthalpy of reaction. If you’re doing the opposite in the equation of what is listed in the table, switch the sign; if not, keep the sign. The major DIFFERENCE is that you do NOT multiply by the molar coefficient as you would in thermodynamics. As a hint, since we are always dealing with redox reactions in galvanic cells, and ONLY reduction potentials are listed in the reduction tables, you’ll always have to switch the sign on one of them and leave it on the other. If not, you’re probably doing something wrong.

In galvanic and electrolytic cells, the net gain or loss of electrons for the system is zero. Electrons are not gained or lost overall, only transferred.

Essentially, a galvanic cell to which an external voltage is applied, forcing the electrons to flow in the opposite direction.

o Oxidation still occurs at the anode and reduction at the cathode.

o The species with the lower reduction potential will be reduced!

o The cell potential will always be negative.

o The sum of the externally applied voltage (Vbattery) and the negative cell potential (-E°cell) must be positive.

o Cathode = (-); Anode = (+) Note the difference compared to galvanic cells.

Reduction is occurring at the cathode, electrons must be flowing TO the cathode, not away from it.

In a Galvanic cell, because reduction happens at the cathode, the species with the higher reduction potential would be at the cathode. However, because this is an electrolytic cell we know that the electron flow will be forced in the opposite direction—toward the metal with the lower reduction potential. In this case, that is species Y, so we know that metal Y will be at the cathode.

A special type of galvanic cell; The same electrodes and solution are used in both beakers. In one beaker the metal is oxidized via its oxidation half-reaction, and in the other beaker it is reduced via its reduction half-reaction. Because the reduction potentials of oxidation and reduction half- reactions for the same species only differ by the sign of E°, E°cell = 0.00V. It appears that nothing would happen. However, all E° values are given for standard conditions (the reason for the naught symbol). You do NOT need to know those conditions for the MCAT, but one aspect of standard conditions happens to be 1M concentrations for both solutions. Concentration cells are therefore nonstandard conditions by definition. They have a positive reduction potential E (no naught symbol, signifying nonstandard conditions) if there is a difference in the molarities of the two solutions. The Nernst equation is used to calculate the cell potential based off of the E ̊ of the species and the two concentrations. Yes, you do need to know the Nernst equation—it has shown up on the MCAT at least twice before.

Nernst Equation: E = E° – (0.06/n)*log[lower]/[higher]; where n = moles of electrons transferred
(e.g., Fe3+(aq) –> Fe(s) = 3 and Ag+(aq) –> Ag(s) = 1)

Remember: By definition, E° will always equal zero for a concentration cell.

Concentration cells are spontaneous. Two containers MUST be used in a concentration cell because the only difference between the two solutions is the concentration. The same species are used, only at differing concentrations, meaning they would have exactly the same reduction potential. Only one species is used, we know that it must be being both reduced and oxidized.

∆G° = -nFE°; where n is the number of moles of electrons transferred in the balanced redox reaction, and F is Faraday’s constant.

o From the above EQN we learn: + E ̊ = negative ∆G = spontaneous reaction

o Faraday’s Constant = the charge on one mole of electrons.

Faraday vs. Farad: A Faraday is an obsolete unit of charge equal to the charge on one mole of electrons. In other words, Faraday’s constant = 1 Faraday. The Faraday has since been replaced by the Coulomb. A Farad is a unit of capacitance. It is a “summary” unit similar to a Newton. Just as we can say 1 Newton instead of saying 1 Kg*m/s2, we can say 1 Farad instead of saying 1 C2*s2/m2*kg. A Farad is the amount of capacitance necessary to hold 1 C of charge on a capacitor with a potential difference of 1 Volt.

Faraday’s constant is the charge on one mole of electrons. One mole of electrons is 6.022 x 10^23 electrons. Each electron has a charge of 1.6 x 10-19C. Therefore, the charge on one mole of electrons is: (6 x 1023)(1.6 x 10-19) = 9.6 x 104 C/mol.

PV = nRT

R = 0.0821 L*atm/mol*K or 8.314 J/mol*K

This is the theoretical model used by scientists to study and predict the “ideal” behavior of gases. Have you noticed a trend? To make precise calculations about projectile motion we had to ignore air resistance, assume constant gravitational acceleration, and so forth. In order to deal with fluids we had to assume non-viscous, non-turbulent, “ideal” fluid flow. Gas behavior is no exception. The MCAT authors have shown a clear affinity for test questions that address the contrasts between “ideal” behavior and real behavior. For the MCAT you need to know how to calculate a projectile’s motion ignoring air resistance, but you must also be able to predict how, and in what way, those predictions will be altered under real conditions. This is also true for the differences between an “ideal” and real fluid, and—in terms of the topic presently before us—the differences between “ideal” gas behavior and the behavior of real gases.

Depending on the textbook author and how you choose to lump them together, one can list anywhere from three to a dozen or more assumptions for ideal gases. Focusing on the following “ideal” conditions should suffice for the MCAT:

1) Gas molecules themselves are of negligible volume compared to the volume occupied by the gas.

2) All intermolecular forces between gas molecules are negligible. Focus on the first two assumptions; they are responsible for most of the differences between what PV=nRT predicts and how real gases actually behave. To simplify things even further: THINK OF IDEAL GAS MOLECULES AS HAVING: NO volume and NO intermolecular forces.

Other Assumptions:

3) All collisions between gas molecules are perfectly elastic
4) Gases are made up of a large number of molecules that are very far apart from one another
5) Pressure is due to collisions between gas molecules and the walls of the container
6) All molecular motion is random
7) All molecular motion follows Newton’s laws of motion
8) The average kinetic energy (KE) of gas molecules is proportional to temperature

One of the important take-home messages from these ideal gas assumptions is that the identity of the actual gas molecules is of no importance. Cl2 gas is quite different from H2 gas, but when we assume ideal gas behavior they are absolutely indistinguishable. They have the same molecular volume (i.e., zero), have the same intermolecular forces (i.e., none), and one mole of either gas occupies the same space.

This can be very counterintuitive. Suppose you have 6g of H2 gas in a 1L sealed vessel and 71g of Cl2 gas in an identical 1L sealed vessel. Which vessel will have the greater pressure? The answer might surprise you. Six grams of hydrogen gas is equal to three moles. Seventy-one grams of chlorine gas is about one mole. Assuming ideal gas behavior, the hydrogen gas will be at three times the pressure of the chlorine gas.

Gas molecules do not dissipate energy because of the assumption that all of their collisions— whether with each other or with the walls of the container—are perfectly elastic. As we recall from Physics 2, energy is conserved in a perfectly elastic collision.

A set of standard conditions true of any ideal gas said to be “at STP.” For the MCAT, unless you are specifically told otherwise, assume that all gases are ideal and start out at STP.

At STP the variables in the Ideal Gas Law are defined as follows:

P = 1 atm
V=22.4L
n = 1 mole
R = 0.0821 L*atm/mol*K or 8.31 J/mol*K
T = 273 K (0°C)

Don’t confuse STP with Standard Conditions: Standard Conditions is a set of agreed-upon conditions at which thermodynamic data, reduction potentials, or other standardized data are measured. STP and standard conditions are NOT interchangeable. As an example, temperature at STP is 0°C and temperature at standard conditions is 25°C.

An absolute temperature is any temperature measured relative to absolute zero. The Kelvin scale is measured relative to absolute zero, where absolute zero is defined as 0 degrees Kelvin—therefore all Kelvin temperatures are absolute. Absolute zero is a theoretical temperature limit where all molecular motions cease.

Whenever you see more than one of the above variables together in the same problem (i.e., T, P, n and/or V), you are most likely dealing with an Ideal Gas Law problem. There are two ways you can approach these problems:

1)Manipulating Equations:
Compare the variables for any system to the variables outlined above for STP. For any variation found, estimate the factor by which that variable has changed and use the manipulating equations skills we taught you (Intro, Strategy & Math lesson) to predict the effect that change will have on the unknown variable. For example, in a given problem you may be asked to find the volume of an ideal gas, given that the pressure is 2 atm, the temperature is 273K, and there are 3 total moles of gas. The pressure is double what it is at STP, so the volume will be cut in half. At the same time, the molar amount is 3 times greater, which would increase volume by a factor of 3. Multiply these two factors together to see that the volume will be 3/2 (1.5) times what it normally is at STP (22.4L x 1.5 = 33.6L).

2) P1V1/T1 = P2V2/T2 (a.k.a., The Combined Gas Law)
Because PV/T = nR and R is a constant, for the same number of moles of gas the ratio of PV/T must remain constant regardless of the changes made to the system. You can choose the first set of data as being STP, or as any other point where P, V and T are known. The second set of data will be different, but the ratio will always be the same. Plug in the data and solve for the unknown. Conceptually, you’re probably better off if you understand and can apply the first method. The first method is also much faster!

This is a simple manipulating equations question. Gas A has twice the moles of gas B. Using PV = nRT, we see that doubling n will double P. Gas A also occupies twice the volume of gas B, which according to the same equation, will cut pressure in half. These two factors must be multiplied by one another if they both act simultaneously: (2)(1/2) = 1. We see, therefore, that there is no net difference in the pressure of the two gases and the ratio of A to B would be 1:1 (the pressures are equal).

Boyle’s Law: P1V1 = P2V2 (assumes constant temperature)

Charles’ Law: V1/T1 = V2/T2 (assumes constant pressure)

Note: We find the two laws above to be of limited value for the MCAT because the relationships they demonstrate can be intuited from either the Ideal Gas Law or the Combined Gas Law.

The greatest deviation between ideal gas behavior and real gas behavior occurs when either a) the temperature is extremely low, or b) the pressure is extremely high. At very high pressure the molecules are pushed close together and their actual size (assumed to be zero) becomes comparable to the distance between them. At very low temperature, the interaction between gas molecules (which is not really zero, as assumed in the ideal gas law) starts to become more important. In either case, there is a deviation from ideal behavior, but with opposite effects: at high pressure, gas molecules will occupy a greater volume; and at low temperature, they will produce a smaller pressure than predicted by the ideal gas law. These effects are better represented in the Van der Waals equation.

[P + a'(n/V)2]*[(V/n) – b’] = RT

a’ is a constant that represents the actual strength of the intermolecular attractions

b’ is a constant that represents the actual volume of the molecules

The Van der Waals equation has never been on the MCAT and it should NOT be memorized. We present it here because it demonstrates a few important principles you should know for the MCAT:

1) Rules for Manipulating Equations: As was stated in the Intro, Strategy & Math lesson, you cannot use the manipulating equations skills we have presented on an equation that involves addition or subtraction—such as the Van der Waals equation.

2) Increased intermolecular attractions (a’) decrease pressure in real gases. The larger a’ is, the larger the second term will become and therefore the smaller P will be.

3) Increased molecular volume (b’) increases volume in real gases.

The ratio given by PV/nRT tells us which of two assumptions is the major cause of the deviation from Ideal Gas Law behavior:

If PV/nRT > 1 it is due mostly to the molecular volume assumption
If PV/nRT < 1 it is due mostly to the intermolecular forces assumption

In real gases volume is MORE than would be predicted by the ideal gas law because real molecules do occupy some volume. Notice that this is an INCREASE. By contrast, in real gases pressure is LESS than predicted by the ideal gas law. This is because real molecules do experience intermolecular attractions—forces that tend to slow down the molecules as they collide with the walls of the container and therefore pressure. Notice that this is a DECREASE. Looking at the ratio PV/nRT, we see that both pressure and volume are in the numerator. As a result, the “real” deviations just described that increase volume will cause PV/nRT to increase to greater than one, and “real” deviations that decrease pressure will cause PV/nRT to decrease to less than one.

Solution: By definition, any ideal gas behaves in an identical manner, regardless of the actual gas molecule present. In other words, even though chlorine gas is a larger, heavier gas than hydrogen gas, they both are treated as identical “ideal” gas particles. For this reason any ideal gas problem generally has to do only with the moles of gas present and nothing to do with the identity or molar mass of the individual gas molecules. In this case, therefore, answer D is correct.

Ptotal=P1 +P2 +P3 …

This seems pretty straightforward; the sum of the partial pressures equals the total pressure. However, if you gloss over this you will miss a very important point. If we add more of Gas 1 (P1) to an existing mixture of three gases, we have increased the partial pressure of Gas 1 and the total pressure, but have had zero effect on the partial pressure of the other gases. Partial pressure is NOT similar to mole fraction or mass percent. By adding more of Gas 1 we did decrease the mole fraction and the mass fraction of Gases 2 and 3, but we did NOT decrease their partial pressures.

Partial pressures and mole fractions are different. Mole fraction is a ratio of how much of one substance there is in a given system compared to the sum of all substances in that same system. Partial pressure is just a measure of how much of something there is and remains exactly the same no matter what other substances, or how much of them, are present.

o Diffusion: The process by which gas molecules spread from areas of high concentration to areas of low concentration due to the random motion imparted to them as a result of their kinetic energy and collisions with other molecules.

o Effusion: The diffusion of gas particles through a pin hole. A pin hole is defined as a hole smaller than the average distance a gas molecule travels between collisions.

o E1/E2 = √MW2/√MW1

E1 and E2 can represent either the effusion rate or the diffusion rate of gases 1 and 2,
respectively.

Notice that the rate of effusion or diffusion is inversely proportional to the molecular weight of the gas.

Definition: In common use, the term “phase” is used to distinguish between the solid, liquid, and gas forms of a substance. Solid, liquid and gas are more correctly called “states” (a.k.a., “states of matter”). Molecules of the same “phase” a) are in the same state, b) have the same chemical composition, and c) are structurally homogenous (e.g., In its solid state, carbon can exist as either diamond or graphite. These meet the first two criteria, but not the third, so they are different “phases” of carbon).

In a sealed vessel containing sodium chloride solution there are two phases: 1) the sodium chloride solution, and 2) the vapor above that liquid. If the solution was fully saturated and there was some solid NaCl not yet dissolved, this would be a third phase. Many students tend to think that ions in solution would be different phases from the solvent in which they are dissolved. Because a true solution is—by definition—homogenous throughout, the solute and solvent (two phases) become one phase once the solute is fully dissolved.

Because diamond and graphite are structurally different , one would expect that they will absorb a different amount of energy per gram, per rise in temperature unit.

o Solid -> Liquid = Melting
o Liquid -> Solid = Freezing

ΔH fusion = The amount of energy in Joules/mole required to go from solid to liquid or the energy that must be removed to go from liquid to solid. This describes the transition in both directions (i.e., melting and freezing).

o Liquid -> Gas = Evaporation
o Gas -> Liquid = Condensation

ΔH vaporization = The amount of energy in Joules/mole required to go from liquid to gas OR the energy that must be removed to go from gas to liquid. Again, it describes both evaporation and condensation.

o Solid -> Gas = Sublimation
o Gas -> Solid = Deposition (ΔH sublimation is defined similarly, but is not used on the MCAT exam).

Melting point is the temperature at which a substance changes state from solid to liquid. However, it is very important to realize that melting point and freezing point are exactly the same thing. You might think of freezing point as the temperature at which a liquid changes into a solid, but the value measured for mp or fp is simply a temperature, which indicates no direction of progress. For any substance, mp = fp.

Boiling point is the temperature at which a substance changes state from liquid to gas. Liquids boil when the vapor pressure of the liquid equals atmospheric pressure.

Volatile is a term used to describe the relative tendency of a substance to form a vapor. How readily a substance vaporizes is primarily a function of its vapor pressure. Therefore, if one substance is said to be “more volatile” than another, this indicates that the former has a higher vapor pressure than the later at the same temperature.

Some substances are said to be non-volatile. This would indicate that the substance does not form a vapor, or has an extremely low vapor pressure, at room temperature. This usually refers to solutes such as sodium chloride that do not contribute to the vapor pressure of a solution when dissolved in a solvent. By contrast, something like methanol would have its own vapor pressure that would add to the vapor pressure of the solvent into which it is dissolved.

You should have labeled the y-axis Pressure and the x-axis Temperature.

o The section to the far left, filling up all or most of the upper part of the y-axis is always the solid.
From there, going clockwise: solid -> liquid -> gas.
o Lines on a phase diagram represent points where the two phases on either side of the line are in equilibrium. At the triple point all three phases are in equilibrium. Beyond the critical point, there is no distinction between liquid and gas and the phase is called a fluid.

The triple point is the precise temperature and pressure at which all three phases (i.e., states) exist simultaneously in equilibrium with each other. The critical point is the precise temperature and pressure above which liquid and gas phases become indistinguishable. At this point liquid and gas phases cease to exist, merging into a single phase called a supercritical fluid. This supercritical fluid cannot be compressed back into the liquid phase by increasing pressure, nor can it be turned into a gas by increasing temperature. The critical temperature and the critical pressure are simply the temperature and pressure at the critical point. At the triple point all three phases are present. At the critical point none of the original three phases are present, only the new “supercritical fluid” phase.

The heat of vaporization is the enthalpy change associated with the transition between liquid and gas. Whatever that might be at or near the triple point, it does not exist at the critical point. By definition, at the critical point a liquid cannot be changed into a gas, so the heat of vaporization would be zero. Any value is greater than zero, therefore, the heat of vaporization will always be greater at the triple point.

A graph of temperature (T) in Kelvin or Celsius vs. heat (q) in Joules. Occasionally, time is graphed on the x-axis instead of heat (if heat is added at a constant rate the temperature vs. heat graph and the temperature vs. time graph look approximately the same).

o Calculating ΔH from a heating curve:

ΔH fusion = The change in q (x-axis) during the phase change from solid to liquid.

ΔH vaporization = The change in q (x-axis) during the phase change from liquid to gas.

o There is NO CHANGE in temperature DURING a phase change. The fact that heating curves are flat (i.e., horizontal, slope = 0) during the actual phase change demonstrates the following frequently-tested MCAT principle: Once a phase change starts, all of the energy goes into breaking intermolecular forces and none goes toward an increase in temperature!!!!!

See the heating curve below. Note that the x-axis is usually a measure of the heat added/absorbed, but time can also be represented on the x-axis. The horizontal sections of the graph represent phase changes. The first flat section will represent the phase change between solid and liquid and the second will represent the phase change between liquid and gas. If heat is on the x-axis then the length of the first horizontal section represents the heat of fusion and the length of the second horizontal section represents the heat of vaporization. The slope of the lines between these horizontal sections represents the inverse (∆T/Q) of heat capacity (Q/∆T) for that particular phase of the substance. One should observe, therefore, that different phases of the same substance usually have different heat capacities—as indicated by the differing slopes of those sections of the following graph:

For a heating curve of water the phase change from solid to liquid should be shorter than the phase change from liquid to gas. This is because to change phases from solid to liquid only some of the intermolecular forces must be broken. Recall that hydrogen-bonding (the strongest type of intermolecular attraction) is prevalent in liquid water. In order to change phase from liquid to gas, all of these intermolecular hydrogen bonds must be completely broken (because no intermolecular forces exist between molecules in water vapor).

Solution: The triple point is the place on a phase diagram where all three lines converge. At that point the substance exists as a mixture of gas, liquid, and solid all in equilibrium with one another. An increase in temperature will move the substance to the right on the phase diagram into the gas portion of the diagram. This makes Answer A the best choice.

Definition: Vapor Pressure (Vp) is the partial pressure of the gaseous form of a liquid that exists over that liquid when the liquid and gas phases are in equilibrium.

Increased temperature increases vapor pressure. This makes logical sense for a few reasons. First, increased temperature means the molecules of the liquid have a higher average kinetic energy. This indicates that a larger fraction of those molecules will have the energy necessary to escape the intermolecular forces between liquid molecules to enter the gas phase. One could also remember that liquids boil when their vapor pressure increases to the point that it equals atmospheric pressure. It obviously requires an increase in temperature to cause a liquid to boil, therefore increasing temperature must increase vapor pressure.

Addition of a non-volatile solute decreases vapor pressure. The non-volatile solutes in solution occupy a portion of the limited surface area available for vaporization. Liquid molecules must be at the surface of the liquid in order to escape into the gas phase. Therefore, one could think of the surface of the liquid as being made up of a limited number of “exit ports.” If some of those “exit ports” are filled by non-volatile solutes, this leaves less area available for vaporization and therefore decreases vapor pressure. The vapor pressure with the non-volatile solute present will be equal to some fraction of what it would be for the pure solvent: Vp = XVp° ; where Vp equals actual vapor pressure, X equals mole fraction, and Vp° equals the vapor pressure of the pure solvent.

When a volatile solute is added to a solvent it usually decreases vapor pressure for the same reason that a non-volatile solute decreases vapor pressure. As long as the vapor pressure of the solute is LESS THAN the vapor pressure of pure solvent, addition of the volatile solute will decrease vapor pressure. However, if a solute is added that has a vapor pressure greater than that of the pure solvent, then the vapor pressure of the solution will actually be higher than that of the pure solvent. This can be seen by simple examination of the formula for calculating the vapor pressure of a solution containing a volatile solute: Vp = (Xsolute*Vpsolute) + (Xsolvent*Vpsolvent).

A liquid boils when the vapor pressure of that liquid is equal to atmospheric pressure.

Vapor Pressure w/ a Non-Volatile Solute = (mole fraction of the pure solvent, X)*(Vp of the pure solvent, Vp°). Vp = XVp°

Total Vapor Pressure w/ a Volatile Solute = (mole fraction of solvent* Vp° of the solvent) + (mole fraction of the solute* Vp° of the solute). Vp,total = Vp,solvent + Vp,solute = Xsolvent Vp°solvent + Xsolute Vp°solute

Primarily used to describe the solubility or partial vapor pressure of gases dissolved in liquids. There can be some confusion here, because Henry’s Law is defined in multiple forms;

Vapor Partial Pressure of solute = (mole fraction of solute)*(Henry’s Law Constant)

Vapor Partial Pressure of solute = (concentration of solute)*(Henry’s Law Constant)

Vapor Partial Pressure of solute = (concentration of solute)/(Henry’s Law Constant)

If you examine all three of the equations just outlined, it appears mathematically impossible that they can all be true. All three equations work because there is a different Henry’s Law Constant for almost every perceivable situation. You must look up the constant that is specific to 1) the solute involved, and 2) the form of the equation you are using. Because of this complexity and potential for confusion, if Henry’s Law shows up on the MCAT they will define it for you and give you the appropriate constant. We recommend you treat this section as background information only and focus on the following principle illustrated by Henry’s Law:

The solubility of a gas in a liquid is directly proportional to the partial pressure of that gas over that liquid!!!!!

o The solubility of gases in liquids follows a trend that is exactly the opposite of the solubility of solids in liquids. For gases dissolved in liquids, increased temperature DECREASES solubility and decreased temperature INCREASES solubility.

o Increasing the vapor pressure of gas X over a liquid increases the solubility of gas X in that liquid (This is why they pressurize soda pop cans with excess CO2).

o Polar and non-polar gases easily form homogenous mixtures.

The boiling point of a liquid is elevated when a non-volatile solute is added
according to:

o ∆T = kbmi ; where kb is a constant, m is molality (NOT molarity) and i is the number of ions formed per molecule (a.k.a., The Van’t Hoff Factor; i.e., for NaCl i = 2; for CaCl2 i = 3).

The freezing point of a liquid is depressed when a non-volatile solute is added according to:

o ∆T = kfmi ; where kf is a new constant, different than kb above.

Impurities in a solid result in a slightly lower melting point and a broader melting point range. Impurities get in the way and actually make it easier for molecules to escape intermolecular forces. Most students also recall that impurities DEPRESS the freezing point of a solid. Remember that freezing point and melting point are two names for the same thing (you’re just going in different directions).

Because of the Van’t Hoff Factor, one can think of freezing point depression and boiling point elevation as being a function of not just moles of solute, but moles of solute particles. The 1.5 moles of NaCl described in the question will produce 3.0 moles of ions. The 1.25 moles of CaCl2 will produce 3.75 moles of ions. So, even though there were fewer moles of calcium chloride added, it will be the calcium chloride solution that has the lower freezing point.

A measure of the tendency of water to move from one solution to another across a semi-permeable membrane. Usually represented by the capital Greek letter pi, Π. It is the side that will receive the water via osmosis that has the higher osmotic pressure. In other words, more solute means more osmotic pressure.

II = iMRT ; where i = # of ions formed in solution, M is the solute molarity, R is the gas constant, and T is the absolute temperature.

Definition: A solution is a homogenous mixture of two or more compounds in the same phase. (We usually think of all solutions as being in the liquid, or “aqueous” phase; however, a homogenous mixture of gases is also called a “solution”.)

o Solvent vs. Solute: The solute is the substance dissolved into the solvent. Thus solvent is more abundant than solute.

o Colloids: Colloids are NOT solutions. Colloids are solvents containing undissolved solute particles that are too small to be separated by filtration, but are much larger than the solute particles in a true solution. Colloids scatter light, while true solutions do not. Examples of colloids include paint (a suspension of solid crystals in a solvent) and dust floating in air.

o Nomenclature of Polyatomic Ions: Review the ions in the list below, which you were asked to memorize (formulas and charges) in the Chemistry 1 lesson. These ions will appear regularly in solution chemistry and the MCAT will always assume that you are familiar with their charge and structure:

Hydroxide (OH−)
nitrate (NO3−)
nitrite (NO2−)
chlorate (ClO3−)
chlorite (ClO2−)
hypochlorite (ClO−)
perchlorate (ClO4−)
carbonate (CO3^ -2)
bicarbonate ( HCO3−)
ammonium ( NH4+)
sulfate (SO4^ 2−)
phosphate (PO4^ 3−)
manganite (MnO(OH))
permanganate (MnO4−)
cyanide (CN)

o Hydration: water molecules attached to ionic units in a solid.

Carbonates and sulfates are very insoluble.

The greater the charges on the ions, the harder they will be to pull apart in solution.

Solvation is a general term for the process wherein solvent molecules surround a dissolved ion or other solute particle creating a shell.

Hydration is a specific kind of solvation wherein water is the participating solvent. Water molecules, being polar, can surround both negatively and positively charged solutes by directing either their partially-negative oxygen, or partially positive hydrogen, moieties toward the ion.

The hydration number is the number of water molecules an ion can bind via this solvation process, effectively removing them from the solvent and causing them to behave more like an extension of the solute.

A hydrate is an inorganic compound in which water molecules are permanently bound into the crystalline structure. The nomenclature of a hydrate is altered to reflect the presence of water molecules. For example, anhydrous cobalt(II)chloride contains no water, but cobalt(II)chloride hexahydrate [CoCl2∙6H2O] contains six water molecules complexed with each cobalt. As we see in these two names, the term anhydrous is often applied to a compound that can form complexes with water to differentiate molecules that do not contain water from those that do. Aqueous refers to any solution for which water is the solvent.

See the diagrams below. The red arrows represent the dipole of each water molecule.

moles solute/Liter solution

moles solute/Kg solvent

moles solute/total moles solution (solute + solvent)

mass solute/total mass of solution * 100

mass solute/total mass solution * 10^6 (for ppb multiply by 10^9)

Parts per million (ppm) is NOT a measure of how many solute particles there are per 1 million total particles. This is how most students erroneously think of it. It is nothing more than mass percent multiplied by 10^4, or “mass fraction” multiplied by 10^6. The purpose of multiplying by 1 million is to make very, very small concentrations easier to work with.

ppm = mg/Kg = mg/L (since 1 L of water has a mass of 1 Kg)

= # of moles of equivalents/Liter solution.

For example, A 1M solution of H2SO4 can be referred to as a “2 Normal” solution because it produces two moles of hydronium ions per Liter of solution. By the same token, a 2M solution of H3PO43- would be a “6 Normal” solution with respect to hydronium ions. This concept ignores the decreasing acidity of each proton. Some chemists discourage the use of this measure, but it has been used on the MCAT—albeit rarely.

o For a solution to form, the intermolecular forces between the solute particles must first be broken; then any intermolecular forces between the solvent particles must be broken (to make room for the solute). Finally, new intermolecular forces are formed between the solute particles and the solvent particles.

o If the new intermolecular forces formed are greater (i.e., stronger, more stable) than the sum of the intermolecular forces that had to be broken, net energy is released and the solution is said to have a negative Heat of Solution ΔH solution < 0. This means that the dissolution process is exothermic and heat will be evolved. If the new intermolecular forces are not more stable than the old ones, the solution has a positive ΔH solution. A positive heat of solution means that energy must be added to the system to make the solute dissolve.

A negative heat of solution tells us that the dissolution releases heat. In order for this to occur, the solvent-solute bonds must be relatively stronger than the solvent-solvent and solute-solute bonds that had to be broken in order to dissolve the solute into the solvent. We would expect such a solution to have a lower vapor pressure than a solution with a positive heat of solution because—as stated at the outset—the negative heat of solution indicates strong solvent-solute interactions. It is these very intermolecular forces that must be overcome in order for solvent molecules to escape into the vapor phase. For a solution with a positive heat of solution, these attractions are relatively weaker, and therefore (at the same temperature) more of the molecules should have enough energy to enter the vapor phase.

The dissolution of a solute into a solution is accompanied by a very large, positive change in entropy. A solid or crystal is highly ordered and the break-up and solvation of that solid into individual molecules represents a significant increase in disorder.

Solubility is the amount of a solute that will dissolve in a given solvent at a given temperature. Temperature is usually specified because for most solids dissolved in liquids, solubility is directly related to temperature. On the MCAT, solubility is usually measured in either g/mL, g/100mL, or mol/L.

A precipitate is a solid formed inside of a solution as the result of a chemical reaction, such as the common ion effect. Precipitates only form when the ion product exceeds the solubility product constant, Ksp. For example, given the dissolution of iron(III)chloride in water [Equation: FeCl3(s) –> Fe3+(aq) + 3Cl-(aq)], if NaCl is added to the solution Le Chatelier’s principle predicts that the reaction will shift to the left, reforming the solid.

A saturated solution is a solution that contains the maximum amount of dissolved solute it can hold. For a saturated solution the ion product equals the Ksp.

An unsaturated solution is any solution that contains less than its maximum amount of dissolved solute. For unsaturated solutions the Ksp is greater than the ion product.

A super-saturated solution is a solution that contains more dissolved solute than predicted by the solubility product constant—in other words, the ion product exceeds the Ksp without a precipitate forming. Supersaturated solutions usually form only when a solution is held at a higher temperature during dissolution (at which Ksp would be larger) and then slowly cooled to a temperature at which Ksp is smaller.

This phrase refers to the fact that polar substances are soluble in polar solvents and non-polar substances are soluble in non-polar solvents. Polar and non-polar substances do NOT form solutions.

Exactly the same thing as Keq, Ka, and Kb. Like all the other Ks, remember the following:

1) Leave out pure liquids and pure solids (this will make all Ksp equations only have a numerator – if you have something in the denominator of a Ksp equation, you’ve made a mistake).

2) Temperature is the only thing that changes Ksp (the MCAT does not include activity coefficients).

3) Ksp can only be observed in a saturated solution. This is because saturation is the point at which the dissolution reaction has reached equilibrium. In other words, it’s just like all other equilibrium constants—you cannot measure them anywhere other than at equilibrium.

This seems to be a repeated point of confusion among many students. Solubility is a measure of “how much” of a solute can be dissolved in a given solute. For example, the solubility of iron(III)chloride in water is 74.4g/100mL. The solubility product constant, or Ksp, is defined as the product of the dissolved ions in a saturated solution (i.e., at equilibrium) raised to their coefficients in the balanced equation. Ksp and solubility are directly related to one another (i.e., a large Ksp indicates a large solubility), but are not the same thing. An analogous comparison would be to ask “how much” of a strong acid will dissociate in 100mL of water and then compare that to the Ka of that acid. Because Ksp and solubility are not identical, a ranking of Ksp values for various substances may or may not match the order of a ranking of solubilities for those same substances. As an example, the solubility of NaCl is approximated by the square root of the Ksp (i.e., Ksp = [x][x]), but the solubility of CaCl2 is approximated by the cube root of one- fourth of the Ksp (i.e., Ksp = [x][2x]2 = 4×3). Finally, the two quantities have different units.

Also referred to as the “Solubility Product.” The ion product has the same relationship to Ksp as Q does to Keq. Plug in the values for the actual concentrations of each species at some point other than equilibrium (i.e., for an unsaturated or supersaturated solution).

If the product is greater than Ksp, you know a precipitate will form. If it is less than or equal to Ksp, then you know that no precipitate will form. If the ion product happens to be exactly equal to Ksp, then the solution must be exactly saturated (i.e., at equilibrium).

Write out the Ksp expression.

Substitute into the expression the value given for Ksp.

Substitute a factor of x into the equation for the concentration of each ion, using 2x, 3x, etc., if more than one mole of each ion is produced (Hint: Ask yourself, “If x moles of the reactant are dissolved, how many moles of each ion will be produced?”).

Solve for x. Your answer, “x” is the “solubility” of that particular specie.

The Common Ion Effect is a specific application of Le Chatelier’s principle to solution chemistry. Consider the dissolution of Iron(III)Chloride in water: FeCl3(s) –> Fe3+(aq) + 3Cl-(aq). Suppose that enough solute is added to saturate the solution. If sodium nitrate is then added to this solution it would have no effect. However, if NaCl were added, the presence of extra chlorine ions from NaCl would—according to LeChatelier’s Principle—drive the reaction to the left resulting in precipitation. In this example, chloride is considered a “common ion” and the precipitation as a result of its addition is what is referred to as the “Common Ion Effect.” Other ions, such as sodium and nitrate—that do not shift the equilibrium—are considered “spectator ions.”

Addition of a common ion will cause precipitation. If a spectator ion is added no precipitation will result.

Remembering the following general rules will give you a quick shortcut to the right answer on several problems.

1) All compounds containing the following are SOLUBLE: nitrate, ammonium, and all alkali metals (Group IA).

2) All compounds containing the following are INSOLUBLE: (unless paired with something from the “always soluble” list above) carbonate, phosphate, silver (Ag), mercury (Hg), and lead (Pb).

When we ask, “What is the ‘solubility’ of BaCl2 in water?” we are asking how many moles of BaCl2 will dissolve per liter of water. The variable x represents the number of BaCl2 molecules that break apart into Ba2+ and 2Cl-. This is the solubility by definition. The fact that 2 moles of chlorine ions are formed per mole of BaCl2 is irrelevant. Similarly, if we were examining the solubility of Fe2(SO4)3, “x” would still give the solubility. Even though the right side of our Ksp expression would have a 2x and 3x term, “x” still represents the number of moles of the parent molecule that broke apart.

Solution: Adding a solute to a solution would decrease the freezing point, making answer A plausible; but the solution is saturated, so no more sodium chloride can go into solution and its concentration (and thus the freezing point) will stay the same. Adding solute to a solution decreases the vapor pressure and increases the boiling point, making B also plausible, but wrong for the same reason as above. Answer C is equivalent to answer A because fp = mp. Answer D is therefore the correct answer; although it would be more correct to say that the sodium chloride did not dissolve in the first place, rather than that crystals appeared (implying that a precipitate formed).