Solutions and Gases for the MCAT: Everything You Need to Know
Learn essential MCAT topics covering solutions and gases, complete targeted practice problems, and review step-by-step answers to boost exam performance.
(Note: This guide is part of our MCAT General Chemistry series.)
Table of Contents
Part 1: Introduction to solutions and gases
Part 2: Solubility of ions
a) Solubility rules
b) Concentration units
c) Colligative properties
Part 3: Solubility of Reactions
a) Common ion effect
b) Solubility equilibria
Part 4: Gas Law Equations
a) Ideal gas law
b) Ideal gas law derivatives
c) Partial pressure
Part 5: High-Yield Terms
Part 6: Passage-Based Questions and Answers
Part 7: Passage-Based Questions and Answers
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Part 1: Introduction to solutions and gases
What happens as you stir sugar into a cup of coffee or salt in water? Perhaps predictably, the solid sugar or salt dissolves into its components: charged ions that are invisible to the naked eye. As ions become dissolved, they can change the chemistry of the aqueous solution they are in. Of course, this has special implications for compounds in biological systems, including acids and bases. In this guide, we will cover everything you need to know about solutions and gases for the MCAT. You'll run into some familiar variables, such as the equilibrium constant. However, this time it'll be in the context of dissociation reactions. As you read through this article, challenge yourself to connect the topics discussed here with topics you've encountered during your MCAT prep.
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Part 2: Solubility of ions
a) Solubility rules
Solubility refers to the ability of a solute to dissolve in a particular solvent. For example, vitamins A, D, E, and K are considered fat-soluble vitamins because they dissolve in fat and are stored in adipose tissue. When the solute is in an equilibrium between its dissolved and undissolved state, the solution is considered to be saturated. Any additional solute added will be insoluble and form a precipitate, the solute’s undissolved solid form. These are referred to as precipitation reactions.
The solvent used and the temperature of a solution can impact the solubility of a substance. For instance, polar solvents (such as water) tend to dissolve other polar or ionic compounds. Nonpolar solvents (such as oil or fat) tend to dissolve other nonpolar compounds (such as vitamin A).
In the context of the MCAT, water is the most commonly seen solvent. The table below summarizes several solubility rules you should be familiar with for test day. Note that there are several exceptions to each of these solubility rules.
Table 1 Water solubiluty rules and exceptions
| Water Solubility Rules |
|---|
| Salts containing Group I elements (alkali metals) are soluble |
| Salts with the ammonium ion (NH4+) are soluble |
| Salts with the acetate ion (CH3COO-) are soluble |
| Salts with the nitrate ion (NO3-) are soluble |
| Salts containing the sulfate ion (SO42-) are soluble Exceptions: SrSO4, PbSO4, BaSO4, Ag2SO4, CaSO4 |
| Sulfites (SO32-) are insoluble Exceptions: Sulfites with Group I elements and ammonium |
| Sulfides (S2-) are insoluble Exceptions: Sulfides with Group I elements and ammonium |
| Halides are soluble (e.g., Iodine, Chlorine, Bromine) Exceptions: Fluoride and halides containing Pb2+, Ag+, and (Hg2)2+ |
| Hydroxide salts with ammonium, Group I elements (alkali metals), and certain Group II elements (Ca2+, Sr2+, and Ba2+) are soluble All other hydroxide salts are insoluble |
| Phosphates (PO43-) are insoluble Exceptions: Phosphates containing Group I elements and ammonium |
| Carbonates (CO32-) are insoluble (e.g., CaCO3, FeCO3, and SrCO3) Exceptions: Carbonates containing Group I elements and ammonium |
This list of solubility rules may be fairly intimidating to memorize! However, as a general rule of thumb, ammonium, acetate, nitrate, and sulfate compounds are soluble. Sulfites, sulfides, calcium, and transition metal compounds tend to be insoluble.
When compounds are dissolved in solution, new compounds can be formed through displacement or synthesis reactions. If such a reaction produces an insoluble compound, the resulting product will enter its solid form and precipitate from the reaction.
Figure 1 Common types of chemical reactions.
b) Concentration units
When a solution is made, a solute is said to be dissolved in a solvent. Thus, a solute tends to be a solid compound that is placed into a liquid-phase solvent. The quantity of solute that is placed within a volume of solvent is referred to as concentration. Solutions that have a smaller ratio of solute to solvent are described as dilute. Those that have a larger ratio are considered concentrated.
The concentration of a solution can be expressed in multiple ways. The most common unit used is molarity (M), which gives the number of moles of solute per liter of solution. Similarly, molality (m) gives the number of moles of solute per kilogram of solvent. Note that these two units have similar symbols but are very different!
In contrast, the units of osmoles refer to the number of moles of distinct solute particles. For example, for every 1 mole of potassium chloride (KCl) in a solution, there are 2 osmoles of solute particles. Osmolarity refers to the number of osmoles of solute per liter of solution, while osmolality refers to the osmoles of solute per kilogram of solvent.
Lastly, mole fraction gives the ratio of moles of one substance to moles of all substances within a solution or mixture.
Table 2 Summary table of ways to determine concentration
| Unit | Symbol | Formula |
|---|---|---|
For more information on this topic, be sure to refer to our guide on moles and stoichiometry.
c) Colligative properties
The colligative properties of a solvent are the physical properties of a solution that may change based on the changes caused by mixing the solute and the solvent.
These colligative properties include the boiling and freezing points of a solution. Boiling point elevation and freezing point depression are two equilibrium points that are dependent on the molality of a solution. A solution containing a solute and solvent will have a higher boiling point than the pure solvent. How much the boiling point is raised can be calculated using the formula:
ΔTb: Change in boiling point Kb: constant for solvent (given)
i: van’t Hoff factor
m: molality
Similarly, solutions have a lower freezing point than their pure solvent. This can be calculated using the formula:
ΔTf: Change in freezing point Kf: constant for solvent (given)
i: van’t Hoff factor
m: molality
The van’t Hoff factor (i) is a whole number that describes the number of particles the solute dissociates into when mixed with the solvent. For example, potassium chloride (KCl) has a van’t Hoff factor of 2 because KCl dissociates into a potassium ion and chloride ion.
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Part 3: Solubility of reactions
a) Common ion effect
In addition to the factors discussed earlier in this article, solubility is also impacted by the common ion effect. This describes the reduction in solubility of a salt when it is mixed in a solution that contains one of its ions.
where A and B are the constituent ions.
If some quantity of the constituent ions is already present in the solution, the le Chatelier’s principle will be applied as additional reactant is added. The presence of the dissociated ion, or product, will shift the equilibrium to the left. As a result, any additional solute will tend to remain in its undissociated form.
For example, if potassium chloride (KCl) is dissolved in a solution that contains potassium ions, less of it will dissolve than if it was added to pure water.
The common ion effect can be leveraged in laboratory separations. Through forming complex ions—which are often metallic or otherwise insoluble—a solid precipitate is formed. This precipitate can easily be separated from the aqueous solution and dried.
For more information on this topic, be sure to refer to our guide on reactions and separations.
b) Solution equilibria
Note that the Ksp is simply the equilibrium constant (Keq) of the dissociation reaction! Ksp does not have a denominator because the dissociating reactant is a pure solid. Recall that pure liquids and solids are not expressed in an equilibrium constant. (For more information on this, be sure to refer to our guide on chemical equilibrium and kinetics.)
A related quantity is the ion product, or IP. If we consider Ksp as an analog to Keq, then IP can be seen as an analog to Q, the reaction quotient. Its expression is virtually identical to Ksp.
While Ksp tells us the equilibrium position of a dissociation reaction, IP tells us the reaction’s current position. Therefore, if IP is equal to Ksp, then the reaction is at equilibrium, and the solution is considered saturated. However, if IP is less than Ksp, then the solution is unsaturated, and additional solute may be dissolved. If IP is greater than Ksp, then the solution is supersaturated, and a solid precipitate will form at the bottom of the container.
The molar solubility of a substance is the molarity of it as a solute in a saturated solution. In other words, it indicates the maximum quantity of a solute that can be dissolved in a solution. It can be represented by a plethora of variables, such as x, s, or c.
A research technician adds NaCl (Ksp = 1.3 x 10-21) to a 4 x 10-4 M solution of KCl until it is saturated. What is the final concentration of sodium?
As a first step, be sure to write down the relevant dissociation reactions:
NaCl ⇋ Na+ + Cl-
In the already formed solution of KCl, there are 4 x 10-4 M of chloride ions. Due to the pre-existing presence of chloride ions in the solution, only some of the sodium chloride will dissociate.
Let’s represent this molar solubility for sodium chloride as “x”. Hence, the final concentration of chloride ions will be 4 x 10-4 + x, and the final concentration of sodium will be x. These values can be used in the Ksp expression for the dissolution of sodium chloride:
1.3 x 10-21 = [x][4 x 10-4 + x]
Since our Ksp value is extremely small, the additional concentration of chloride ions, x, is negligible. Thus, an approximation can be made.
Finally, this expression can be simplified and solved for x.
x = 3.3 x 10-18
Thus, the final concentration of sodium ions is 3.3 x 10-18 M.
These solubility calculations can be exceedingly useful in calculating the concentrations of hydrogen ions, and thus the resulting pH, of weak acids and bases. For more information on this, be sure to refer to our guide on acids and bases.
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Part 4: Gas law equations
a) Ideal gas law
What is a gas? You may have learned that gas is a state of matter that occupies the entire volume and adapts the shape of the container it is in.
The simplest way to study gases is to use the ideal gas approximation. Under this approximation, ideal gases are hypothetical gases that occupy no volume and lack intermolecular forces.
These hypothetical assumptions can be nearly met under certain conditions, such as high temperature and low pressure (high volume). Under moderately high pressure or low volume, intermolecular attractions become significant. Under moderately low temperatures, the kinetic energy decreases as the speed of gas molecules decreases. Thus, in both scenarios, the gas will occupy a volume that is less than what the ideal gas law predicts.
At pressures greater than or equal to 300 atmospheres, low volume, or low temperatures, the volume of individual gas particles is no longer negligible. Therefore, the gas will occupy a volume that is more than what the ideal gas law predicts.
The ideal gas law allows us to describe the properties of a gas sample. It is expressed as:
PV = nRT
P: Pressure
V: Volume
n: number of moles
T: Temperature
R: ideal gas constant
In the context of gas law problems, R will likely be expressed as 0.0821 L*atm / mol*K. However, on test day, you may see R expressed as 8.314 J/K*mol. Always be mindful of the units!
Deviations from the ideal gas law can be described using van der Waals’ equation:
P: Pressure
V: Volume
R: Ideal gas constant
T: Temperature
The quantities a and b are constants specific to different gases. They are included in the equation to account for the volume occupied by individual gas molecules.
When performing calculations with gases, the problems will often be in the context of Standard Temperature and Pressure, or STP. Under such conditions, the temperature is 273 K (0˚ Celsius), and the pressure is 1 atm. At STP, one mole of an ideal gas occupies 22.4L.
b) Ideal gas law derivatives
The ideal gas law has many derivative formulas that focus on specific variables. These are all laws that can be derived by assuming the same sample of an ideal gas experiences two sets of slightly different conditions.
Charles’s law states that the volume and temperature of a gas are proportional. To put it simply, if volume increases, so does the temperature. This law is expressed as:
k = V/T
k: Proportionality constant
V: Volume
T: Temperature
Boyle’s law states that the volume and pressure of a gas are inversely proportional. If the pressure increases, the volume of the sample must decrease, and vice versa. This law is expressed as:
k = PV
k: Proportionality constant
P: Pressure
V: Volume
Gay-Lussac’s law states that pressure and temperature are proportional. This law is expressed as:
k = P/T
k: Proportionality constant
P: Pressure
T: Temperature
Finally, Avogadro’s law states that the volume of and the number of moles of a substance within a sample are proportional. This law is expressed as:
k = V/n
k = Proportionality constant
V: Volume
n: Number of moles
c) Partial pressure
When multiple gases are present in a sample, the pressure that each one exerts individually is its partial pressure. It can be expressed by Dalton’s law of partial pressures:
PT: Total pressure
PA: Partial pressure exerted by species A
PB: Partial pressure exerted by species B
From its partial pressure, the mole fraction of a gas in a mixture using the following equation:
PA: Partial pressure exerted by gas A
PT: Total pressure
XA: number of moles of A / total moles of gas
Let’s work through an example, using an application of Dalton’s law.
A sample of gas (PT = 4.0 atm) contains hydrogen, nitrogen, and oxygen. Given that hydrogen exerts a partial pressure of 1 atm and nitrogen exerts a partial pressure of 2 atm, calculate the partial pressure and mole fraction of oxygen.
First, solve for the partial pressure of oxygen using Dalton’s law of partial pressures.
POxygen = PTotal - PHydrogen - PNitrogen
POxygen = 4 atm - 1 atm - 2 atm
POxygen = 1 atm
Then, use the partial pressure of oxygen to calculate the mole fraction.
XOxygen = POxygen /PTotal
XOxygen = 1 atm / 4 atm
XOxygen = 0.25
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Part 5: High-yield terms
Avogadro’s law: a gas law that describes the directly proportional relationship between the volume and number of moles of a gas
Boyle’s law: a gas law that describes the inverse relationship between the volume and pressure
Charles’s law: a gas law that describes the directly proportional relationship between the volume and temperature of a gas
Colligative properties: physical properties of a solution based on the changes caused by mixing the solute and the solvent
Common ion effect: reduction in solubility of a salt when it’s mixed in a solution that contains one of its ions
Concentrated solutions: solutions that have a large solute to solvent ratio
Dalton’s law of partial pressures: states that the total pressure of a gaseous sample is the sum of the partial pressures exerted by each gas in the sample
Dilute solutions: solutions that have a small solute to solvent ratio
Gay-Lussac’s law: a gas law that describes the directly proportional relationship between the pressure and temperature of a gas
Mole fraction: number of moles of one substance per moles of all substances
Molality (m): moles of solute per kilograms of solvent
Molar solubility: molarity of a solute in a fully saturated solution
Molarity (M): moles of solute per liters of solution
Normality: number of equivalents of interest per liter of solution
Osmolality: osmoles of solute per kilograms of solvent
Osmolarity: osmoles of solute per liters of solution
Precipitation reactions: Occur when additional solute is added to a solution in saturation, resulting in the formation of a precipitant
Saturation: state at which a solution’s solute is in an equilibrium between its dissolved and undissolved state
Solubility: the ability of a substance to dissolve in a certain solvent
STP: Standard Temperature and Pressure; refers to a temperature of 273 K (0˚ Celsius), and pressure of 1 atm
Van’t Hoff factor: The number of particles a solute dissociates into when mixed with a solvent.
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Part 6: Passage-based questions and answers
Oxygen accounts for roughly 21% of atmospheric air. Hyperbaric oxygenation therapy (HBOT) is a therapeutic intervention that places patients in a chamber with elevated concentrations of atmospheric oxygen. The use of HBOT may increase endothelial oxygenation and stimulate vascular endothelial growth factor (VEGF), a specific and potent growth factor implicated in angiogenesis. Thus, HBOT is hypothesized to encourage the wound healing process.
Researchers wish to determine the impact of five days of HBOT on levels of VEGF in diabetic ulcer patients. In this study, twelve diabetic ulcer patients received 30 minutes of HBOT 3 times a day for 5 days (HBOT group). A second group (non-HBOT) was composed of 10 diabetic ulcer patients who did not receive HBOT. The VEGF level in both groups was measured on days 1 and 5. In the HBOT group, the mean level of VEGF was 1241.3 pg/ml on day 1 and 1242.3 on day 5. For the non-HBOT group, the mean level of VEGF was 1239.4 on day 1 and 1240.6 on day 5. The results of this experiment are shown in Figure 1.
Figure 1 VEGF Levels after 5 days in HBOT and non-HBOT groups
Question 1: Assuming that air is composed of only nitrogen and oxygen, what percentage of atmospheric air is composed of nitrogen?
A) 21%
B) 50%
C) 80%
D) 79%
Question 2: What is the partial pressure of oxygen in normal air?
A) 21 mmHg
B) 760 mmHg
C) 160 mmHg
D) 210 mmHg
Question 3: What can the researchers most reasonably conclude from the results of this experiment?
A) Hyperbaric oxygen therapy for 5 days does not increase VEGF levels in diabetic ulcer patients
B) Hyperbaric oxygen therapy for 5 days increases VEGF levels in diabetic ulcers patients
C) Hyperbaric oxygen therapy for 5 days decreases VEGF levels in diabetic ulcer patients
D) Hyperbaric oxygen therapy for 5 days is hazardous to human health
Question 4: What was the purpose of the non-HBOT group?
A) The non-HBOT group served as a control group
B) The non-HBOT group served as a second experimental group to test another variable
C) The non-HBOT group served as a reference for ideal VEGF levels in diabetic ulcers patients
D) The non-HBOT group served as a backup group in case the HBOT group produced inconclusive results
Answer key for passage-based questions
Answer choice D is correct. The passage states that oxygen makes up approximately 21% of air. Dalton’s law of partial pressures states that the partial pressures of constituent gases in a mixture sum in a linear fashion. Since the question stem states that atmospheric air is only composed of nitrogen and oxygen, simply subtract 21% from 100%. This results in nitrogen accounting for 79% of atmospheric air (choices A, B, and C are incorrect).
Answer choice C is correct. The passage states that oxygen makes up approximately 21% of air. Therefore, it exerts 21% of the pressure. One atmosphere of pressure is equal to 760 mmHg; thus, 21% of this value is equal to 160mmHg (choices A, B, and D are incorrect).
Answer choice A is correct. In the experimental group, HBOT, oxygen levels did not increase significantly on day five (choice B is incorrect). In fact, there is no statistically significant difference between the oxygen levels in any of the groups (choice C is incorrect). There is no evidence to conclude that HBOT for 5 days is hazardous to human health (choice D is incorrect).
Answer choice A is correct. The non-HBOT group was also composed of diabetic ulcer patients but did not receive the therapy. Thus, they are a control group (choice A is correct). This patient group was not used to research any additional variables (choice B is incorrect). The passage does not state what the ideal VEGF levels are for diabetic ulcer patients, nor if the non-HBOT group had those levels (choice C is incorrect). The non-HBOT group could not provide any supplemental information for results from the HBOT group as they did not receive the therapy (choice D is incorrect).
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Part 7: Standalone questions and answers
Question 1: What is the molarity of a solution composed of 1L of water and 6 grams of NaCl?
A) 0.1 M
B) 1 M
C) 10 M
D) 100 M
Question 2: A sample of gas initially has a volume of 2 L and a temperature of 300 K. If the pressure is kept constant and the temperature is increased to 340K, what is the final volume of the gas sample?
A) 1.6 L
B) 2.3 L
C) 3.0 L
D) 3.4 L
A) Ksp = [Ca][Cl]
B) Ksp = [Ca][Cl]/[CaCl2]
C) Ksp = [Ca][Cl]2
D) Ksp = [Ca][Cl]2/[CaCl2]
Question 4: A 5 L container is filled with 3,200 g of oxygen gas. Assuming a temperature of 293K, what is the pressure exerted by the gas within the container? (Note: R = 0.0821 L*atm/mol*K)
A) 154 atm
B) 481 atm
C) 921 atm
D) 2,405 atm
A) 2.5 x 10-13 M
B) 3.5 x 10-13 M
C) 4.0 x 10-13 M
D) 5.0 x 10-13 M
Answer key for standalone questions
Answer choice A is correct. First, convert the mass of sodium chloride into moles. The molar mass of a compound can be found by summing the molar masses of its constituent elements in proportions equal to the stoichiometric coefficients.
6 grams NaCl * (1 mol / 58.44 grams) = 0.1 mol NaCl
To find the molarity, simply divide the moles by the volume of the solution.
0.1 mol / 1L = 0.1M
2. Answer choice B is correct. This question requires the application of Charles’s Law, which states that volume and temperature are proportional.
2L / 300K = V2 / 340K
V2 = 340K (2L / 300K)
V2 = 2.3 L
3. Answer choice C is correct. The reaction for the dissolution of calcium chloride can be written as:
This can be used to write an expression for the equilibrium constant, which is equivalent to Ksp. The Ksp expression does not include solids, and so does not include any quantities in the denominator (choices B and D are incorrect). Additionally, each species concentration must have an exponent equal to its stoichiometric coefficient in the chemical formula (choice A is incorrect).
4. Answer choice B is correct. First, convert the mass of oxygen gas to moles.
The given values can then be substituted into the ideal gas law, then used to solve for P.
P = (nRT)/V
P = (100mol O2 * 0.0821 L·atm/mol·K * 293K) / 5 L
P = 481 atm
5. Answer choice D is correct. First, set up an expression for the solubility constant. Be sure to exclude the concentrations of any solids or solvents.
Ksp = [K+][Cl-]
Next, use x to represent the molar solubility of the dissolved species. Note that according to the dissociation reaction, for every x moles of dissolved K+ ion, x moles of Cl- must also be dissolved.
2.5 x 10-25 = x2
√2.5 x 10-25 = x
5.0 x 10-13 = x
Thus, the final concentration of dissolved chlorine will be 5.0 x 10-13 M.
