13.3 Colligative Properties
PRE-HEALTH PREP
Colligative Properties
Formulas, Calculations & Examples
General Chemistry 13.3
Lesson Overview
Colligative properties—freezing point depression, boiling point elevation, vapor pressure lowering, and osmotic pressure—depend on the number of solute particles in solution rather than the identity of the solute. The van 't Hoff factor determines how many particles a solute produces upon dissociation, serving as the key multiplier in equations such as ΔTf = −iKfm, ΔTb = iKbm, Raoult's law, and the osmotic pressure formula π = iMRT. Worked examples demonstrate both qualitative ranking of solutions and quantitative calculations, including the connection between osmotic pressure and reverse osmosis water purification.
What Are the 4 Colligative Properties?
Colligative properties are properties of a solution that change depending on the concentration of solute particles—not the identity of the solute itself. In general chemistry, four colligative properties appear throughout solution chapters and appear regularly on exams:
- Freezing point depression — adding solute lowers the freezing point
- Boiling point elevation — adding solute raises the boiling point
- Vapor pressure lowering (described by Raoult's law) — vapor pressure decreases
- Osmotic pressure — pressure required to stop osmosis increases
The Key Idea Behind All Colligative Properties
Colligative properties are largely independent of what the solute is. What matters is how many solute particles are present in solution. The more particles disrupting the intermolecular forces of the solvent, the greater the change in each colligative property.
What Is the van 't Hoff Factor?
Because colligative properties depend on the total number of particles in solution, you need to account for the fact that electrolytes dissociate into multiple ions when they dissolve. The van 't Hoff factor (i) captures this: for strong electrolytes, it equals the number of ions produced per formula unit. Getting the van 't Hoff factor right is the first step in every colligative properties calculation.
Strong Electrolytes: Determining the van 't Hoff Factor
Strong electrolytes include soluble ionic compounds, strong acids, and strong bases. Each formula unit dissociates completely, and you simply count the total number of ions produced.
Example: NaCl (i = 2)
Sodium chloride is a soluble ionic compound. When it dissolves:
NaCl → Na+ + Cl−
One sodium ion plus one chloride ion gives a total of 2 ions, so i = 2.
Example: Ba(OH)2 (i = 3)
Barium hydroxide is a strong base. When it dissolves:
Ba(OH)2 → Ba2+ + 2 OH−
One barium ion plus two hydroxide ions gives a total of 3 ions, so i = 3.
Non-Electrolytes: van 't Hoff Factor of 1
Non-electrolytes do not dissociate at all, so their van 't Hoff factor is simply i = 1. Non-electrolytes are typically molecular compounds that are neither acids nor bases.
Example: Methanol, CH3OH (i = 1)
Methanol is a molecular compound made entirely of nonmetals (C, H, O) and is not an acid or a base. It dissolves as intact molecules with no dissociation, so i = 1.
Compounds with Polyatomic Ions
When a compound contains polyatomic ions, remember that polyatomic ions stay together as a single unit—they do not break apart into individual atoms.
Example: Al(NO3)3 (i = 4)
Aluminum nitrate dissociates into ions, but each nitrate ion remains intact:
Al(NO3)3 → Al3+ + 3 NO3−
One aluminum ion plus three nitrate ions gives a total of 4 ions, so i = 4.
Why the van 't Hoff Factor Matters for Colligative Properties
Consider one mole of methanol (i = 1) versus one mole of aluminum nitrate (i = 4) in water. The aluminum nitrate solution contains four times as many particles and will therefore produce four times the change in any colligative property. The van 't Hoff factor is the multiplier that connects solute molality to the actual particle concentration driving these effects.
| Compound | Type | Dissociation | i |
|---|---|---|---|
| NaCl | Soluble ionic compound | Na+ + Cl− | 2 |
| Ba(OH)2 | Strong base | Ba2+ + 2 OH− | 3 |
| CH3OH | Non-electrolyte | No dissociation | 1 |
| Al(NO3)3 | Soluble ionic compound | Al3+ + 3 NO3− | 4 |
2 Formulas: Freezing Point Depression & Boiling Point Elevation
The first 2 colligative properties work as a pair: as you add more solute, the freezing point drops and the boiling point rises. Think of it this way—the liquid range expands at both ends. Both effects are calculated using the same variables; only the constants and signs differ.
The Freezing Point Depression and Boiling Point Elevation Formulas
Many textbooks present these equations without the van 't Hoff factor built in:
ΔTb = Kb × m
Here, m is intended to be the total molality of all particles. A cleaner approach is to incorporate the van 't Hoff factor explicitly:
ΔTb = i × Kb × m
In these equations:
- i = van 't Hoff factor
- Kf = freezing point depression constant (for water: 1.86 °C/m)
- Kb = boiling point elevation constant (for water: 0.512 °C/m)
- m = molality of the solute
Watch the Negative Sign in the Freezing Point Depression Formula
The freezing point depression equation includes a negative sign because freezing points go down. Some textbooks omit this sign and expect you to remember the direction. Including it in the formula guarantees that your ΔTf comes out negative—meaning the freezing point decreased—so there is no ambiguity.
Interpreting Kf and Kb for Water
For water, Kf = 1.86 °C/m means that every 1 molal of particles lowers the freezing point by 1.86 °C. Similarly, Kb = 0.512 °C/m means every 1 molal of particles raises the boiling point by about half a degree. These constants differ for every solvent—for most problems, you will look them up in a table.
Comparing Solutions: Qualitative Rankings
Before diving into calculations, let's tackle the conceptual comparison questions that appear frequently on exams. These questions ask you to rank solutions by boiling point or freezing point without doing any arithmetic. The strategy is always the same: identify the effective particle concentration for each solution.
Equal Molalities, Different van 't Hoff Factors
Suppose you are given four aqueous solutions, all at the same molality, and asked: which has the highest boiling point, and which has the highest freezing point?
Since the molality is identical for all four, the deciding factor is the van 't Hoff factor. The solution with the largest i produces the most particles and therefore the biggest change in both boiling point elevation and freezing point depression.
Example: Ranking 4 Solutions at Equal Molality
| Solute | i | Relative Effect |
|---|---|---|
| CH3OH | 1 | Smallest change |
| NaCl | 2 | ↑ |
| Ba(OH)2 | 3 | ↑ |
| Al(NO3)3 | 4 | Largest change |
Highest boiling point: Al(NO3)3 — its boiling point rises the most.
Highest freezing point: CH3OH — its freezing point drops the least, so it remains the highest.
Common Exam Trap: "Highest Freezing Point"
The highest freezing point belongs to the solution whose freezing point has gone down the least—that is, the one with the fewest particles. Many students get this backwards. The solution with the biggest change in freezing point depression has the lowest freezing point, not the highest.
Different Molalities: Finding the Biggest i × m Product
When the molalities are not equal, you must compute the product i × m for each solution. The largest product corresponds to the greatest total particle concentration and therefore the greatest change in any colligative property.
Example: Ranking 4 Solutions at Different Molalities
| Solute | i | Molality (m) | i × m |
|---|---|---|---|
| NaCl | 2 | 2.0 m | 4.0 |
| CH3OH | 1 | 1.5 m | 1.5 |
| Ba(OH)2 | 3 | 1.2 m | 3.6 |
| Al(NO3)3 | 4 | 0.9 m | 3.6 |
Highest boiling point: NaCl (i × m = 4.0) — not aluminum nitrate! The higher molality of NaCl compensates for its smaller van 't Hoff factor.
Highest freezing point: CH3OH (i × m = 1.5) — the fewest total particles means the smallest freezing point depression.
How to Calculate Freezing & Boiling Point Changes
Freezing Point Depression Calculation: 1 Molal NaCl
Find the freezing point of a 1.00 m NaCl aqueous solution. This is one of the most common freezing point depression problems in general chemistry.
Step-by-Step Solution
Apply the freezing point depression formula:
ΔTf = −(2)(1.86 °C/m)(1.00 m)
ΔTf = −3.72 °C
To find the new freezing point, add the change to the normal freezing point of water:
Why the Final Step Matters
The result of −3.72 °C looks the same as the change only because water happens to freeze at 0 °C. For any other solvent—say one with a normal freezing point of 20 °C—the new freezing point would be 20 + (−3.72) = 16.28 °C. Always add ΔT to the normal temperature to get the actual freezing point.
Boiling Point Elevation Calculation: 1 Molal NaCl
Find the boiling point of the same 1.00 m NaCl aqueous solution.
Step-by-Step Solution
ΔTb = (2)(0.512 °C/m)(1.00 m)
ΔTb = +1.02 °C
Add the change to the normal boiling point of water:
Values You Need to Know
You are expected to know that water freezes at 0 °C and boils at 100 °C. For any other solvent, the normal freezing and boiling points will be provided. The Kf and Kb constants are typically given in a table—you do not need to memorize them.
Why Freezing Points Drop & Boiling Points Rise
Why Freezing Point Depression Happens
When a pure substance freezes, its molecules arrange into a regular, repeating crystal lattice. This works smoothly because every molecule has the same shape and fits neatly into the structure.
When you introduce a solute, those foreign particles have a different shape. They don't fit into the crystal lattice the way the solvent molecules do. The result? You have to cool the liquid more than usual to force crystallization despite these imperfections. That is why the freezing point drops—this is freezing point depression.
Identity Doesn't Matter—Only Quantity
It doesn't matter what the solute is—as long as it's different from the solvent, it disrupts the crystal lattice. This is why freezing point depression depends only on particle concentration, not particle identity. This is the defining characteristic of all colligative properties.
Why Boiling Point Elevation Happens
The common (but incorrect) explanation is that solute particles increase intermolecular forces, making the liquid harder to boil. That's not what's really happening.
A substance boils when the chemical potential of the liquid equals the chemical potential of the gas. When you dissolve a solute, the process of mixing lowers the energy of the liquid phase. Now the liquid sits at a lower energy than the gas at what used to be the boiling temperature—so the substance doesn't boil yet.
To reach the boiling point, you must heat the solution a little hotter than normal so the liquid's energy climbs back up to match the gas. The result is boiling point elevation.
The Short Version
Dissolving a solute stabilizes the liquid phase (lowers its energy). You need extra heat to overcome that stabilization, so the boiling point increases. This is fundamentally related to the entropy of mixing, though the full thermodynamic treatment goes beyond general chemistry.
Raoult's Law & Vapor Pressure Lowering
The Formula and Concept
Raoult's law describes vapor pressure lowering—how the vapor pressure above a solution decreases when a nonvolatile solute is added. The equation states:
Where:
- Psolution = vapor pressure above the solution
- Xsolvent = mole fraction of the solvent
- P°solvent = vapor pressure of the pure solvent
The logic is straightforward: if 90% of the moles in solution are water, you get 90% of the vapor pressure that pure water would produce. If 80% are water, you get 80%. The mole fraction is simply that percentage expressed as a decimal. Vapor pressure lowering occurs because fewer solvent molecules are at the surface to escape into the gas phase.
Vapor Pressure Lowering Calculation: Water and Methanol
Find the vapor pressure of water at 25 °C above a solution containing 162 g of water and 32 g of methanol. Pure water has a vapor pressure of 24.0 torr at 25 °C.
Step-by-Step Solution
Step 1: Convert grams to moles.
Moles of CH3OH = 32 g ÷ 32 g/mol = 1.00 mol
Step 2: Calculate the mole fraction of water.
Step 3: Apply Raoult's law to find the vapor pressure lowering.
The vapor pressure drops from 24.0 torr (pure water) to 21.6 torr because the water is no longer pure—methanol molecules occupy some of the solution, reducing the fraction of water molecules that can escape into the vapor phase. This is vapor pressure lowering in action.
Limitations of Raoult's Law
Raoult's law works best when the solute is present in small amounts (dilute solutions). As the solute concentration increases, deviations from ideal behavior become more pronounced. More advanced treatments exist for concentrated solutions, but those fall outside the scope of general chemistry.
What Is Osmotic Pressure?
The Osmotic Pressure Formula and Its Connection to the Ideal Gas Law
Osmotic pressure uses the Greek letter π (pi) to distinguish it from gas pressure. The osmotic pressure formula is:
Where:
- i = van 't Hoff factor
- M = molarity of the solute
- R = gas constant (0.08206 L·atm/mol·K)
- T = temperature in Kelvin
Looks Familiar?
Since molarity = mol/L, you can rearrange to get πV = nRT, which mirrors the ideal gas law (PV = nRT). The resemblance is not a coincidence—osmotic pressure and gas pressure share a deep mathematical relationship. Just remember to use π (not P) for osmotic pressure.
How Does Osmosis Work?
Osmosis is the diffusion of water through a semi-permeable membrane from a region of higher water concentration (lower solute concentration) to a region of lower water concentration (higher solute concentration). The membrane allows water molecules to pass through but blocks solute particles.
Which Way Does Water Flow in Osmosis?
Water flows toward the more concentrated solute solution. Equivalently, water flows from where it is more pure to where it is less pure. Both statements describe the same direction—choose whichever framing makes more sense to you.
The Osmotic Pressure Setup
Picture two compartments separated by a semi-permeable membrane, each topped with a piston. One side contains pure water; the other contains 1 M NaCl. Pure water is approximately 55.5 M, so it is far more concentrated in water than the NaCl side.
Water naturally flows from the pure side toward the NaCl side. This causes the NaCl-side piston to rise and the pure-water piston to drop. The osmotic pressure is the amount of external pressure you must apply to the NaCl side to prevent this flow—to keep both pistons perfectly stationary.
Osmotic Pressure Calculation: 1 M NaCl at 25 °C
Find the osmotic pressure at 25 °C of a 1.00 M NaCl aqueous solution.
Step-by-Step Solution
π = (2)(1.00 mol/L)(0.08206 L·atm/mol·K)(298 K)
π = 48.9 atm
Units check: Kelvins cancel, moles cancel, liters cancel, leaving atmospheres—a pressure unit, as expected.
3 Outcomes: Osmosis, Equilibrium, and Reverse Osmosis
What happens when you apply different amounts of external pressure to the NaCl side depends on whether that pressure is less than, equal to, or greater than the osmotic pressure:
| Applied Pressure | Compared to π | Result |
|---|---|---|
| 30 atm | Less than 48.9 atm | Normal osmosis proceeds—water still flows toward NaCl side |
| 48.9 atm | Equal to π | Equilibrium—no net water movement |
| 100 atm | Greater than π | Reverse osmosis—water is forced back toward the pure side |
Real-World Application: Reverse Osmosis Water Purification
Reverse osmosis (RO) water purification works by applying pressure greater than the osmotic pressure, forcing water away from the dissolved solutes and through the membrane. As water is removed from the concentrated side, the solute concentration increases, and so does the osmotic pressure. Eventually the osmotic pressure rises to match the applied pressure, and the process stops. At that point, the concentrated brine is flushed and the cycle restarts.
8 Key Takeaways: Colligative Properties
- Colligative properties depend on the number of solute particles, not their identity. The 4 colligative properties are freezing point depression, boiling point elevation, vapor pressure lowering, and osmotic pressure.
- The van 't Hoff factor (i) equals the number of ions a strong electrolyte produces upon dissociation. Non-electrolytes have i = 1. Always account for i before plugging into any colligative property equation.
- Freezing point depression formula: ΔTf = −iKfm. Adding solute disrupts the crystal lattice, requiring a lower temperature to freeze.
- Boiling point elevation formula: ΔTb = iKbm. Dissolving a solute stabilizes the liquid phase, so more heat is needed to reach the boiling point.
- For qualitative comparison questions, compute the i × m product for each solution. The largest product gives the biggest change. Remember: the highest freezing point belongs to the solution with the smallest change.
- Raoult's law and vapor pressure lowering (Psolution = Xsolvent × P°solvent) shows that vapor pressure decreases in proportion to the mole fraction of solvent. It works best for dilute solutions.
- Osmotic pressure (π = iMRT) mirrors the ideal gas law and measures the pressure needed to prevent water from flowing across a semi-permeable membrane toward higher solute concentration.
- Reverse osmosis occurs when applied pressure exceeds osmotic pressure, forcing water away from the concentrated side—this is the principle behind RO water purification systems.