11.6 Colligative Properties

Colligative Properties

Did you ever wonder why we put salt on the roads to prevent icing in winters and not sugar?  To answer this question we need to look at colligative properties of solutions.

Colligative properties are properties of solutions that depend on the number of solute particles in the solution. Since Ionic compounds dissociate in solution, the number of particles they produce in solution is greater than the number of particles produced by covalent compounds of similar concentration.

If we take a 1.0 M solution of table salt (NaCl) an ionic compound and a 1.0 M solution of table sugar (C₁₂H₂₂O₁₁) a covalent compound, they both will have a different number of particles in solution.

NaCl                                  Na⁺ + Cl^–    Eq 11.7

C₁₂H₂₂O₁₁                             C₁₂H₂₂O₁₁  Eq 11.8

As you can see in the reactions above, the number of particles present in NaCl solution will be twice the number of particles present in C₁₂H₂₂O₁₁ solution. As a result, colligative properties will be different for these two solutions.

 There are 4 different colligative properties.

1. Lowering of Vapor Pressure
2. Boiling Point Elevation
3. Depression in Freezing Point
4. Osmotic Pressure

1. Lowering of vapor pressure:

Vapor pressure is the pressure exerted by gas-phase molecules of liquid above the surface of the liquid.  As you can see in the picture on the right, we have some water molecules in the gas phase right above the surface of the liquid. These water molecules keep on bumping on the surface of liquid water. The bumping of these molecules is responsible for the vapor pressure of water.

Source: Commons.wikimedia.org     

                                                                                                               Fig. 11.18

 Fig. 11.19

Source: Commons.wikimedia.org/

Vapor pressure depend on the escaping tendency of solvent molecules from the liquid phase. If you compare the picture of these two flasks, first flask just has liquid water while the second flask has a solution.  The first flask with just water has lot of water molecules present in gas phase, which are exerting vapor pressure on the surface of liquid water. However, there is a force of attraction between solute and solvent particles in the second flask. This attraction between solute and solvent particles prevents solvent particles from leaving the liquid phase and going

to the gas phase. Since fewer solvent particles can escape from the solution, the vapor pressure of the solution goes down. Thus, we can conclude the higher the number of solute particles present in the solution, the higher will be the force of attraction between solute and solvent, and the lower will be its vapor pressure.

Raoult’s law explains this relationship of the vapor pressure of pure solvent to the vapor pressure of the solution.

P_solution = X _solvent P⁰_solvent

Here P_solution is the observed vapor pressure of solution, X _solvent is the mole fraction of solvent and P⁰_solvent is the vapor pressure of pure solvent.

We can also understand the dependency of vapor pressure on concentration of solute with respect to entropy of solution. Evaporation of solvent molecules from the liquid always leads to a large increase in entropy. Greater the volume occupied by the molecules in the gaseous state higher will be the entropy. But if the liquid solvent is initially “diluted“ with solute, its entropy is already larger to start with, so the amount by which it can increase on entering the gas phase will be less. There will accordingly be less tendency for the solvent molecules to enter the gas phase, and so the vapor pressure of the solution diminishes as the concentration of solute increases and that of solvent decreases in solution.

Fig 11.20

Source: https://www.chem1.com/acad/webtext/virtualtextbook.html

The number 55.5 mol L–1 (= 1000 g L–1 ÷ 18 g mol–1) is a useful one to remember if you are dealing a lot with aqueous solutions; this represents the concentration of water in pure water. (Strictly speaking, this is the molal concentration of H2O; it is only the molar concentration at temperatures around 4° C, where the density of water is closest to 1.000 g cm–1.)

Figure 11.20 represents pure water whose concentration in the liquid is 55.5 M. A tiny fraction of the H₂O molecules will escape into the vapor space, and if the top of the container is closed, the pressure of water vapor builds up until equilibrium is achieved. Once this happens, water molecules continue to pass between the liquid and vapor in both directions, but at equal rates, so the partial pressure of H₂O in the vapor remains constant at a value known as the vapor pressure of water at the particular temperature.

In the system on the right, we have replaced a fraction of the water molecules with a substance that has zero or negligible vapor pressure — a nonvolatile solute such as salt or sugar. This has the effect of diluting the water, reducing its escaping tendency and thus its vapor pressure.

What’s important to remember is that the reduction in the vapor pressure of a solution of this kind is directly proportional to the fraction of the [volatile] solute molecules in the liquid — that is, to the mole fraction of the solvent. The reduced vapor pressure is given by Raoult’s law (1886):

 Fig. 11.21

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From the definition of mole fraction (denoted by X), you should understand that in a two-component solution (i.e., a solvent and a single solute),

X_solvent = 1–X_solute

Problem Example 1

Estimate the vapor pressure of a 40 percent (W/W) solution of ordinary cane sugar (C₂₂O₁₁H₂₂, 342 g mol^–1) in water. The vapor pressure of pure water at this particular temperature is 26.0 torr.

Solution: Let’s begin with finding the mole fraction of solvent in this solution.

In 100 g of solution, moles of sugar =  = 0.12 mol of sugar

Moles of water =  =  3.3 mol of water

The mole fraction of water in the solution is

Since P_solution = X _solvent P⁰_solvent

P_solution =  0.96 × 26.0 torr = 25.1 torr.

Since the sum of all mole fractions in a mixture must be unity, it follows that the more moles of solute, the smaller will be the mole fraction of the solvent. Also, if the solute is a salt that dissociates into ions, then the proportion of solvent molecules will be even smaller.

Problem Example 2

The vapor pressure of water at 10° C is 9.2 torr. Estimate the vapor pressure at this temperature of a solution prepared by dissolving 1 mole of CaCl₂ in 1 L of water.

Solution: Each mole of CaCl₂ dissociates into one mole of Ca²⁺ and two moles of Cl^1–, giving a total of three moles of solute particles.

CaCl2 Ca ^{+ 2}  + 2Cl^–

Remember the number of moles of water in pure water is 55.5, so the mole fraction of water in the solution will be

The vapor pressure will be                 P_solution 0.95 × 9.2 torr = 8.7 torr.

Example of a solution with Nonelectrolyte: Find the vapor pressure of the solution at 25 ⁰C prepared by dissolving 25.7 g of table sugar C₁₂H₂₂O₁₁  (molar mass 342.34 g/mol) a nonelectrolyte in 155 g of water. Given that the vapor pressure of pure water at 25 ⁰C is 23.8 torr.

As per Raoult’s Law, we have P_solution = X _solvent P⁰_solvent

Given P⁰_solvent = 23.8 torr

Number of mols of C₁₂H₂₂O_11, n(C₁₂H₂₂O₁₁)

= 25.7 g C₁₂H₂₂O₁₁ x  =0.0751 mols

Since C₁₂H₂₂O₁₁ is a nonelectrolyte, it does not dissociate in solution.

 So moles of solute n(solute) = n(C₁₂H₂₂O₁₁) = 0.0751 mols

X _(water) =  = 0.991

P_solution = 0.991 x 23.8 = 23.6 torr

Example of a solution with an Electrolyte: Find the vapor pressure of the solution at 25 ⁰C prepared by dissolving 25.7 g of table salt NaCl (molar mass 58.44 g/mol) an electrolyte in 155 g of water. Given that the vapor pressure of pure water at 25 ⁰C is 23.8 torr.

As per Raoult’s Law we have P_solution = X _solvent P⁰_solvent

Given P⁰_solvent = 23.8 torr

Number of mols of NaCl_, n ( NaCl) = 25.7 g NaCl x  =0.440 mols

Since NaCl dissociate in solution like this:    NaCl     Na⁺ + Cl^–

It produces twice as many particles as the number of NaCl formula units,

So n(solute) = 2 x 0.440 = 0.880 mol

X _(water) =  = 0.907

P_solution = 0.907 x 23.8 = 21.6 torr

Watch this YouTube video to see an illustration of Raoult’s law.

In these examples, we took the solutes which were nonvolatile and were not contributing anything towards the total vapor pressure of the solution. However, if a solute is also volatile, it will have its vapor pressure also. In that case, we can use a modified version of Raoult’s law to find the overall vapor pressure of the solution.

P_solution = X _solvent P⁰_solvent + X _solute P⁰_solute

Here P_solution =Total vapor pressure of the solution

 X _solvent  = Mole fraction of solvent and P⁰_solvent = Vapor pressure of the pure solvent

 X _solute = Mole fraction of solute and P⁰_solute = Vapor pressure of the pure solute

A solution that follows Raoult’s law completely is called an ideal solution. However, if the force of attraction between solute and solvent particles is stronger than the force of attraction present between pure solvent particles, we see a negative deviation from Raoult’s law. Consequently, the solution’s vapor pressure will be less than the expected vapor pressure based on Raoult’s law. Similarly, if the force of attraction between solute and solvent particles is weaker than the force of attraction present between pure solvent particles, we see a positive deviation from Raoult’s law.  In that case, the solution will have a higher vapor pressure than expected vapor pressure based on Raoult’s law. Solutions that show deviation from Raoult’s law are called a non-ideal solution.

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Learning Check:

• Find the vapor pressure of a solution prepared by mixing 45 g of carbon tetrachloride CCl₄ in 75 g of benzene C₆H₆.  Vapor pressure of pure CCl₄ is  143 torr, and vapor pressure of pure C₆H₆ is 95.1 torr.

Answer: 106.3 torr

2  Boiling point elevation

The boiling point of a liquid is the temperature at which the vapor pressure of the liquid is equal to the atmospheric pressure. We saw in the first colligative property that adding a solute lowers the vapor pressure of the liquid. Therefore, in the presence of a solute, you must heat the solution to a higher temperature to reach the same vapor pressure as atmospheric pressure. In other words, you can say the boiling point of the solution will be higher than the normal boiling point. Since this property is related to the concentration of solute in solution, we can use this mathematical equation to find the elevation of boiling point.

∆T_b = i. K_b. m

Here ∆T_b is the elevation in boiling point, i is called van’t Hoff factor, which is the number of particles produced by 1 formula unit or molecule of the compound in solution. K_b is molal boiling point elevation constant of solvent and m is the molality of solute in the solution.

The molal boiling point constant K_b for some solvents is given in the table below.

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Problem Example 3

Sucrose (C₂₂O₁₁H₂₂, 342 g mol^–1), like many sugars, is highly soluble in water; almost 2000 g will dissolve in 1 L of water, giving rise to what amounts to pancake syrup. Estimate the boiling point of such a sugar solution. (Density of water = 1.00 g/ml)

Solution:        ∆T_b = i . K_b. m

 Kb _{(water )} = 0.514 K mol^–1 kg

Since sucrose is a covalent compound, it will not dissociate into ions,  So i = 1

moles of sucrose =   = 5.8 mol

Since density of water = 1.00 g / ml,

Mass of 1 L water =  1.00 g/ml x 1000 ml =  1000 g  = 1.00 Kg

The molality of the solution (m) =  5.8 m.

Using the value of K_b from the table, the boiling point will be raised by
∆T_b  = (0.514 K mol^–1 kg) × (5.8 mol kg^–1) = 3.0 K = 3.0 0C,
so the boiling point = 100 + 3.0 = 103° C.

 Example 2: Find the boiling point of the solution prepared by dissolving 25.5 g of NaCl (molar mass 58.44 g/mol) in 255 grams of water H₂O.

We know ∆T_b = i. K_b. m   Eq. 11.8

Since NaCl is an electrolyte, it will dissociate in the solution

. NaCl               Na⁺ + Cl^–

Since the number of particles produced by one formula unit of NaCl is 2, i = 2 for NaCl.

Molality (m)  =   ;

Moles of NaCl = 25.5 g NaCl x  =0.436 mols

Molality (m)  =    = 1.71 m ;

Kb for water is 0.51 Kg °C/mol

So ∆T_b = 2 x 0.51 x 1.71 = 1.74 °C,

∆T_b is elevation in boiling point, so it tells us that boiling point of solution will increase by 1.74 °C.

Therefore, the final boiling point of solution = Normal boiling point of pure solvent + ∆T_b

Boiling point of solution = 100 °C + 1.74 °C = 101.74 °C

Watch this YouTube video to see more Elevation of Boiling Point related calculations:

We can also use elevation in boiling point to find the molar mass of the unknown substance. Watch this YouTube video to understand how a change in boiling point can help us find the molar mass of the compound.

• Learning Check:Find the boiling point of water prepared by dissolving 25.5 g of CaCl₂ in 175 g of water.

Answer: 102.01 ⁰C

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• The freezing point is defined as the temperature at which the vapor pressure of the solid phase of matter is equal to the vapor pressure of the liquid phase of matter. If we take an example of the aqueous solution, the normal freezing point of pure water is 0 °C. That means at 0°C vapor pressure of ice is equal to the vapor pressure of pure water. Since, the vapor pressure of the solution is lower than vapor pressure of pure solvent, we can say vapor pressure of solution will be lower than the vapor pressure of ice at normal freezing point, hence no ice will form. As we lower the temperature, the vapor pressure of ice will go down more rapidly than that of liquid water and then a temperature will come at which vapor pressure of ice and vapor pressure of the solution will be same. That will be the new freezing point of the solution. We can use a mathematical equation to find the depression in the freezing point for the solution.

∆T_f = i. K_f. m

Here ∆T_f is the depression in freezing point, i is called van’t Hoff factor, which is the number of particles produced by 1 formula unit or molecule of the compound in solution. K_f is the molal freezing point depression constant of solvent and m is the molality of solute in the solution.

Use of anti freeze in the cars during winters lowers the freezing point of solution and prevents liquids from freezing during cold winter days.

The molal freezing point constant K_f for few solvents is given in a table below:

Example: Find the freezing point of the solution prepared by dissolving 25.5 g of BaCl₂ (molar mass 208.23 g/mol) in 150.0 grams of water H₂O.

We know ∆T_f = i. K_f. m

Since BaCl₂ is an electrolyte, and it will dissociate in the solution.

BaCl₂               Ba⁺² + 2Cl^–

Since the number of particles produced by one formula unit of BaCl₂ is 3, i will be 3 for BaCl₂.

Molality (m)  =   ;

Moles of BaCl₂ = 25.5 g BaCl₂ x  = 0.1225 mols

Molality (m)  =   = 0.816 m

K_f for water is 1.86 Kg °C/mol

So ∆T_f = 3 x 1.86 x 0.816 = 4.56 °C,

∆T_f is depression in freezing point, so it tells us that freezing point of solution will decrease by 4.56 °C.

Therefore, final freezing point of solution = Normal freezing point of pure solvent – ∆T_f

Freezing point of solution = 0 °C – 4.56 °C = – 4.56 °C

Watch this YouTube video to see another example of calculations related to freezing point depression.

Just like elevation in boiling point, depression in freezing point can also be used to calculate the molar mass of unknown compound.

Watch this you tube video to see the illustration.

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Learning Check:

Answer: -7.33 ⁰C

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• Osmosis is evident in many biological systems because cells are surrounded by semipermeable membranes. Carrots and celery become limp when they lose water, and can be made crisp again by placing them in water. The water moves into the carrot or celery cells by osmosis. Osmosis can also affect animal cells. Solute concentrations are particularly important when solutions are injected into the body. When two solutions of different concentrations are separated by a semi permeable membrane, solvent particles move from the area of low solute concentration to the area of high solute concentration. This process is called osmosis.   

We can explain this behavior with the help of Figure 11. Here we have a beaker with two solution separated by a semi permeable membrane. The compartment on the left side has a low solute concentration, while the compartment on the right side has a higher solute concentration. Since solvent particles are smaller in size, only they

will be able to cross the membrane. Solute particles being bigger in size, cannot pass through the membrane. As a result, solvent will move from left to right compartment and the volume of solution on the right side will increase. This process of movement of solvent from area of low solute concentration to area of high solute concentration is called osmosis. The pressure that governs this movement of the solvent is called Osmotic pressure. We can calculate osmotic pressure with this mathematical equation.

∏ =  i M R T

Fig. 11.22

Source: Commons.wikimedia.org/

Here ∏, (Pi) is Osmotic pressure and M is molarity of the solution, R is gas constant, and T is the temperature of the solution in kelvin. i is Van’t Hoff factor, which is basically the number of particles produced by 1 formula unit or molecule of the compound in solution.

Example: A 75.00 ml solution was prepared by dissolving 2.52 g of unknown compound (nonelectrolyte) in water. Osmotic pressure of the solution was found to be 1250 torr at 30 ⁰C. Find the molar mass of compound.

∏ =  i M R T

Since compound is nonelectrolyte, i = 1,

R = 0.08206 L.atm / mol K,

T = 30 ⁰C = 30 + 273 = 303 K

Watch another example of Osmotic pressure related problem on this YouTube video.

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Learning Check

Answer: 732 torr

Solute concentrations are particularly important when solutions are injected into the body. Solutes in body cell fluids and blood serum give these solutions an osmotic pressure of approximately 7.7 atm. Solutions injected into the body must have the same osmotic pressure as blood serum; that is, they should be isotonic with blood serum. If a less concentrated solution, a hypotonic solution, is injected in sufficient quantity to dilute the blood serum, water from the diluted serum passes into the blood cells by osmosis, causing the cells to expand and rupture. This process is called hemolysis. When a more concentrated solution, a hypertonic solution, is injected, the cells lose water to the more concentrated solution, shrivel, and possibly die in a process called crenation. These effects are illustrated in Figure below.

Fig. 11.23

Source: www.openstax.org/

Colligative Properties of Electrolytes

The colligative properties of a solution depend only on the number, not on the identity, of solute species dissolved. The concentration terms in the equations for colligative properties pertain to all solute species present in the solution.

However, the dissolution of an electroyte, is not this simple., Let’s  examine these two common examples:

dissociation NaCl(s) ⟶ Na⁺ (aq) + Cl⁻ (aq) 

ionization HCl(aq) + H₂ O(l) ⟶ Cl⁻ (aq) + H₃O⁺ (aq)

Considering the first of these examples, and assuming complete dissociation, a 1.0 m aqueous solution of NaCl contains 2.0 mole of ions (1.0 mol Na+ and 1.0 mol Cl−) per each kilogram of water, and its freezing point depression is expected to be:

ΔTf = 2.0 mol ions/kg water × 1.86 °C kg water/mol ion = 3.7 °C.

When this solution is actually prepared and its freezing point depression measured, however, a value of 3.4 °C is obtained. Similar discrepancies are observed for other ionic compounds, and the differences between the measured and expected colligative property values typically become more significant as solute concentrations increase. What causes this discrepancy?

These observations suggest that the ions of sodium chloride (and other strong electrolytes) are not completely dissociated in solution.

To account for this and avoid the errors accompanying the assumption of total dissociation, an experimentally measured parameter named in honor of Nobel Prize-winning German chemist Jacobus Henricus van’t Hoff is used.

The van’t Hoff factor (i) is defined as the ratio of solute particles in solution to the number of formula units dissolved:

Values for measured van’t Hoff factors for several solutes, along with predicted values assuming complete dissociation, are shown in Table 11.3.

Predicated and Measured van’t Hoff Factors for Several 0.050 m Aqueous Solutions

Formula unit Classification Dissolution products i (predicted) i (measured)

Table 11.3

In 1923, the chemists Peter Debye and Erich Huckel proposed a theory to explain the apparent incomplete ionization of strong electrolytes. They suggested that ions still have some attraction to each other when dissolved.  This residual attraction prevents ions from behaving as totally independent particles (Figure 11.28). In some cases, a positive and negative ion may actually touch, giving a solvated unit called an ion pair. Thus, the activity, or the effective concentration, of any particular kind of ion is less than that indicated by the actual concentration. Ions become more and more widely separated the more dilute the solution, and the residual interionic attractions become less and less. Thus, in extremely dilute solutions, the effective concentrations of the ions (their activities) are essentially equal to the actual concentrations. Note that the van’t Hoff factors for the electrolytes in Table 11.3 are for 0.05 m solutions,

at which concentration the value of i for NaCl is 1.9, as opposed to an ideal value of 2. Dissociation of ionic compounds in water is not always complete due to the formation of ion pairs.

Example 11.13

The Freezing Point of Electrolyte Solution

We know that seawater freezes at a lower temperature than fresh water, but what exactly is the freezing point of seawater?

The concentration of ions in seawater is approximately the same as that in a solution containing 4.2 g of NaCl dissolved in 125 g of water. Use this information and a predicted value for the van’t Hoff factor (Table 11.3) to determine the freezing temperature the solution (assume ideal solution behavior).

Solution

Solve this problem using the following series of steps.

Step 1. Convert from grams to moles of NaCl using the molar mass of NaCl in the unit conversion

factor.

Result: 0.072 mol NaCl

Step 2. Determine the number of moles of ions present in the solution using the number of moles

of ions in 1 mole of NaCl as the conversion factor (2 mol ions/1 mol NaCl).

Result: 0.14 mol ions

Step 3. Determine the molality of the ions in the solution from the number of moles of ions and

the mass of solvent, in kilograms.

Result: 1.2 m

Step 4. Use the direct proportionality between the change in freezing point and molal

concentration to determine how much the freezing point changes.

Result: 2.1 °C

Step 5. Determine the new freezing point from the freezing point of the pure solvent and the

change.

Result: −2.1 °C

Check each result as a self-assessment, taking care to avoid rounding errors by retaining guard digits

in each step’s result for computing the next step’s result.

Check Your Learning

Assuming complete dissociation and ideal solution behavior, calculate the freezing point of a solution of

0.724 g of CaCl2 in 175 g of water.

Answer: −0.208 °C

Colloids

In colloids, particles of one substance called dispersed phase stay suspended in another substance called dispersing medium.  The dispersed phase substance is not dissolved in the dispersing medium, only surrounded by it. You may have experienced colloids in the form of fog, where water droplets are dispersed in air. Other examples include Milk, Mayonnaise, Shaving Cream, and aerosols etc.

Colloids have particles that are much bigger in size than particles in solution. The reason these particles do not coagulate and settle out is primarily due to electrostatic repulsion.

As you can see here in this picture, particles in colloid have charged surface. Since all the colloid particles have a similar charge on the surface, these particles keep on repelling each other. This repulsion between the particles is responsible for keeping them in constant random motion called brownian motion.

                                                Fig. 11.24

Source: https://www.chem1.com/acad/webtext/virtualtextbook.html

Since these are big particles, if we shine plane-polarized light through these solutions, light gets scattered in all directions after striking these particles. This scattering of light by colloidal particles is known as Tyndal Effect. Figure 11.25 is a good example tyndall effect clearly visible in fog.  In fog, air is the dispersing medium in which water droplets are suspended (dispersed phase). When light passes through the fog, it gets scattered after striking the water droplets suspended in the air.

Fig. 11. 25

Source: Commons.wikimedia.org/

Destruction of colloidal solution can be done by heating, adding an electrolyte or simply by changing the pH of solution. In the destruction of colloid, we try to break the charged layer at the surface of the colloidal particle, and once it is broken, particles start aggregating on top of each other. Eventually, the particles get so big that they start settling out. This process of breaking the colloidal system is known as coagulation.

Fig. 11.26

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As you can see in the picture, particles in a stable colloid are tiny and farther apart from each other; while in an unstable colloid, particles are aggregating.  In coagulation, these small particles eventually get so big by aggregating that they separate from dispersing medium.

Milk is a good example of colloidal system. Putting some vinegar in hot milk can break the colloidal system and results in coagulation and separation of large aggrgates of protein particles.