# PROPERTIES OF SOLUTIONS

• Colligative Properties

• ## What are Solutions:

Solutions are homogeneous mixtures of two or more substances. For instance, if you add sugar to water the sugar dissolves and forms a mixture that seems the same throughout. One of the substances is called the solvent and the others are solutes. Which is which is really arbitrary. However by convention the major component is usually the solvent. In our example, the sugar is the solute and is dissolved in the solvent, water. At any rate, in a homogeneous mixture there are no aggregates of solute bigger than a single molecule or ion. These molecules of solute are evenly ddistributed throughout the solvent. Thus the sugar has broken up into sugar molecules which are randomly distributed through the water.

Solutions are usually thought of as a solid solute dissolved in a liquid solvent, as in sugar water. However solutions in a wider sense can be formed by combinations of all states of matter.

 Forms of Solutions Solute Solvent Examples Solid Solid alloys; stained glass Solid Liquid salt water; sugar water; tincture of iodine Solid Gas moth flake odor in air Liquid Liquid vodka; gasoline; perfume Liquid Solid mercury amalgam; crystal hydrates Liquid Gas moist air Gas Gas air; natural gas Gas Liquid carbonated water; household ammonia Gas Solid hydrogen in platinum or palladium

## Concentration of Solutions:

Solutions, being homogeneous mixtures, have a variable composition. That is, we can add various amounts of sugar to water and still get a solution. How can we specify the amount of sugar in a certain sample of sugar water?

When we describe a solution, we need to specify its composition in terms of how much solute there is in a unit of solvent or solution. This is called the concentration of the solution.

There are a number of different ways of defining the concentration of a solution. We will discuss:

### Percent

The word percent comes from the Latin "per centum" which means "for every hundred". Percent then means a composition of so many parts in every hundred parts of the whole.

Solution concentration can be expressed as percent by mass. This means that there are so many mass units of solute in each 100 mass units of solution.

Percent by mass, %m/m can be calculated as:

```
(mass solute)
%m/m = --------------- X 100
(mass solution)
```
Expressed as a conversion factor, percent by mass is:
```
x grams solute
x %m/m =  --------------
100 g solution
```
or the equivalent.

If the solution is measured in volume instead of mass we get percent mass/volume. Percent mass/volume, %m/v, is the number of mass units of solute for each 100 volume units of solution. Thus:

```
(mass solute)
%m/v = ----------------- X 100
(volume solution)
```
or:
```
x g solute
x %m/v = ----------------
100 ml solution
```
If both solute and solvent are liquids, it is common to use percent by volume, %v/v. Percent by volume is the volume of solute dissolved to give 100 units of volume of solution. Thus:
```
(volume solute)
%v/v = ----------------- X 100
(volume solution)
```
or:
```
x ml solute
x %v/v = ---------------
100 mL solution
```
Note that since volume depends on the temperature, %m/v and %v/v also depend on the temperature of the solution. However %m/m does not depend on temperature. When we refer to concentrations as just percentages we will understand that to mean by mass.

### For Instance:

Sugar (35.73 g) was dissolved in 256.00 of water. What is the % by mass of the resulting solution?

Note that masses are additive and thus the solution weighs 35.73 + 256.00 = 291.73 g.

```
35.73 g sugar
----------------- X 100 = 12.25 % sugar
291.73 g solution
```
Used as a conversion factor, percent allows us to convert between an amount of solution and the amount of solute dissolved in it.

How much salt is there in 345 g of a 10.0% aqueous salt solution?

This is percent by mass so:

```
10.0 g salt
345 g solution X -------------- = 34.5 g salt
100 g solution
```

How much of a 55.4 %v/v alcohol solution will contain 24.7 mL of alcohol?

```
100 mL solution
24.7 mL alcohol X --------------- = 44.6 mL solution
55.4 mL alcohol
```

### Molarity

Suppose one has a solution that is 10% sulfuric acid in water. If you had the task of reacting this sulfuric acid with say some NaOH, you might want to know how many moles of H2SO4 you had on hand in order to do some stoichiometric calculations. How many moles are there in a given amount of 10% solution?

One approach to this problem is to use solutions with their concentrations in molarity. Molarity, M, is the number of moles of solute in every liter of solution. Thus it can be used to convert between moles of solute and volume of solution. The advantage of this is that one can use reagents in the form of solutions and measure out a given number of moles by measuring out the equivalent volume of reagent.

The symbol M is read "molar". Thus 2.3 M NaOH is read "2.3 molar sodium hydroxide". The units of molarity are moles/L, i.e. "moles per liter".

```
x moles solute
x M = --------------
1 L solution
```

### For Instance:

How many moles of hydrogen chloride are there in 326 mL of a 0.100 M solution of hydrochloric acid?

```
0.100 moles HCl
326 mL solution X --------------- = 0.0326 moles HCl
103 mL solution
```

How many mL of a 0.361 M NaOH solution will contain 0.125 moles of NaOH?

```
103 mL solution
0.125 moles NaOH X ---------------- = 346 mL solution
0.361 moles NaOH
```

### Molality

Molality is defined similarly to molarity but do not get them confused. Molality, m, is the number of moles of solute dissolved in each kg of solvent. The units of molality are moles/kg. The symbol m is read "molal". Thus 2.3 m is "2.3 molal".
```
x moles solute
x m = --------------
1 kg solvent
```
Note that m is defined in terms of the amount of solvent rather than the amount of solution! It is useful for problems involving the mixing of a solution from given amounts of solute and solvent. Note that the total mass of the solution is the sum of the masses of solute and solvent. The total volume of solution can then be calculated from the density of the final solution.

### For Instance:

What is the molality of a solution of 34.7 g of glucose, C6H12O6, in 476 g of water? What is the final volume if the solution has a density of 1.02 g/mL?

```
1 mole glucose
34.7 g glucose X ---------------- = 0.193 moles glucose
180.18 g glucose

0.193 moles glucose
------------------- = 0.405 m
0.476 kg water
```
To calculate the volume we need the total mass of solution:
```
476 g + 34.7 g = 511 g solution
```
Then use the density to calculate the volume:
```
1 mL
511 g solution X ------ = 501 mL solution
1.02 g
```

### Mole Fraction

When you are dealing with a solution composed of more than two components, the concentration of each component is often expressed as mole fraction. The mole fraction of component A, XA, is the number of moles of A divided by the total number of moles of all other components. For a solution of compound A, B and C in solvent S:
```
(moles total) = (moles A) + (moles B) + (moles C) + (moles S)

(moles A)
XA = -------------
(moles total)

(moles B)
XB = -------------
(moles total)

(moles C)
XC = -------------
(moles total)

(moles S)
XS = -------------
(moles total)
```
Note that the solvent is considered just one more component. It should be obvious that any mole fraction will be less than 1 and the sum of the mole fractions for all components is equal to 1. Since mole fraction is moles divided by moles, it has no units. Mole fraction can be used for any mixture of components.

### For Instance:

A solution consists of 51.3 g of glucose (C6H12O6), 48.7 g of sucrose (C12H22O10), and 1.00 kg of water. What is the mole fraction of each component?

```
1 mole glucose
51.3 g glucose X ---------------- = 0.285 mole glucose
180.18 g glucose

1 mole sucrose
48.7 g sucrose X ---------------- = 0.142 mole sucrose
342.34 g sucrose

1 mole water
1.00 X 103 g water X ------------- = 55.5 mole water
18.02 g water

55.5 + 0.142 + 0.285 = 55.9 moles total

0.285 moles glucose
Xglucose = ------------------- = 0.00510
55.9 moles total

0.142 moles sucrose
Xsucrose = ------------------- = 0.00254
55.9 moles total

55.5 moles water
Xwater = ------------------ = 0.993
55.9 moles total
```

** !IMPORTANT! **

Symbols such as M or % represent not a single quantity but a ratio of quantities. Therefore one should never use them as single quantities in conversion problems but only as conversion factors.

For instance, you should never see something like this in a calculation:

```
1.0 M       10% NaCl            40.00 g
----------  or -------- or 5.6 m X ------- X .... etc.
3.55 moles      g NaCl             1 mole
```
In a calculation, M should be written always as mole/L, m as moles/kg, etc., so they will cancel correctly.

### Saturation

As you add a solute to a solvent you usually get to a point where no more solute will dissolve. We say that such a solution is saturated. The solubility of a solute is the maximum amount that can be dissolved in a given amount of solvent, i. e. its maximum concentration. The solubility of a compound is also then the concentration of a saturated solution.

A good rule of thumb is that if a solute can form better than a 0.1 M solution it is soluble in that solvent. If it saturates at significantly less than a 0.1 M solution it is considered insoluble. There is a grey area in the mid range where compounds are considered slightly soluble.

## Colligative Properties:

There are some properties of solutions that depend on what solute is dissolved in what solvent. For instance the taste of salt water is different from that of sugar water. A solution of HCl in water is acidic while a solution of NaOH is basic. There are, however, some properties that depend only on the amount of solutes dissolved and not on what those solutes are. Such properties are called colligative properties. We will look at:

### Vapor Pressure Depression

The pressure of the vapor existing above a liquid is called the vapor pressure. The vapor pressure depends on the temperature and the identity of the liquid. For instance the vapor pressure at 25oC for water is 23.8 torr but for methanol is 139.7 torr. At 65oC water has a vapor pressure of 185 torr but methanol has a pressure of 760 torr.

When a nonvolatile solute is added to a solvent the vapor pressure of the solution is less than that of the pure solvent. This is called vapor pressure depression. The amount of depression for a given solvent is the same irregardless of what solute is used. Vapor pressure depression depends only on the solvent and the total concentration of solutes. That is it is a colligative property.

The amount of vapor pressure depression is summarized by Raoult's Law, named after François-Marie Raoult. Raoult found that the vapor pressure of a solution was proportional to the mole fraction of the solvent:

Psoln = XsolvPsolv

or:

DP = XsolPsolv

where Psoln is the vapor pressure of the solution, Xsolv is the mole fraction for the solvent, Xsol is the total mole fraction for solutes, DP is the amount of depression, and Psolv is the normal vapor pressure of the pure solvent. Again note that the amount of depression depends on the solvent used but not on the identity of the solutes.

### For Instance:

What is the vapor pressure of a solution of 10.5 g of sucrose, C12H22O11, in 100 mL of water at 25oC?

We need to know the moles of water and sucrose to calculate the mole fraction of water:

```
1 mole sucrose
10.5 g sucrose X --------------- = 0.0.306 mole sucrose
342.3 g sucrose

1 g water     1 mole water
100 mL water X ---------- X ------------- = 5.55 mole water
1 mL water   18.02 g water

5.55 mole water
Xsolv = ------------------ = 0.995
5.55 + 0.0306 mole
```
Then by Raoult's Law:
```
Psoln = XsolvPsolv = (0.995)(23.8 torr) = 23.7 torr
```

What is the vapor pressure depression at 25oC when 6.44 g of NaCl is dissolved in 100 mL of water?

Note here that NaCl is an ionic salt and dissolves to form sodium ions and chloride ions. The vapor pressure depression depends on the total amount of dissolved nonvolatile materials so:

```
100 mL water = 5.55 moles water (see above)

1 mole NaCl
6.44 g NaCl X ------------ = 0.110 mole NaCl
58.44 g NaCl

0.110 mole NaCl = 0.110 mole Na+ ion = 0.110 mole Cl- ion
```
Thus the total moles of ions dissolved is 0.220 moles.
```
0.220 moles ions
Xsol = ------------------------ = 0.0381
5.55 moles + 0.220 moles

DP = XsolPsolv = (0.0381)(23.8 torr) = 0.906 torr
```
That is, the vapor pressure of water goes down by 0.906 torr when this amount of salt is added.

** !IMPORTANT! **

Colligative properties depend on the total concentration of solute species dissolved. Thus one has to consider what species are formed and how many when a given solute dissolves.

### Boiling Point Elevation

The boiling point of a liquid is the temperature at which the vapor pressure is equal to the external pressure. If you add a solute to a boiling solvent at the boiling point, its vapor pressure will decrease due to vapor pressure depression. Since the vapor pressure will then be less than the external pressure, the liquid will stop boiling. It will be necessary to heat the solution to a higher temperature to raise the vapor pressure back up to the external pressure and resume boiling. That is to say, the boiling point of the solution will be higher than that of the pure solvent. This is called boiling point elevation.

Boiling point elevation is governed by the formula:

DTb = kbmsol

 Solvent kb, oC kg/mol normal BP, oC Acetone 1.71 56.2 Benzene 2.53 80.1 Naphthalene 5.80 217.7 Water 0.51 100
where DTb is the change in boiling point, msol is the total molality of the solutes. The symbol kb is a constant for the particular solvent called the boiling point elevation constant. It has the units of K kg/mol or oC kg/mole. Values for kb for various solvents are given in the table to the right.

### For Instance:

What is the normal boiling point for a solution of 118 g of CaCl2 in 846 g of water?

Note that calcium chloride will dissolve to give one calcium ion and two chloride ions! Therefore the total molality of solutes is three times the molality of CaCl2. The molality of CaCl2 is:

```
1 mole CaCl2
118 g CaCl2 X -------------- = 1.07 mole CaCl2
110.98 g CaCl2
```
And so:
```
1.07 mole CaCl2
msol = 3 X ---------------- = 3.78 m
0.846 kg water

DTb = kbmsol = (0.51 oC kg/mol)(3.78 mol/kg) = 1.93 oC

BP = 100 + DTb = 101.93 oC
```

### Freezing Point Depression

The freezing point of a compound is the temperature at which it spontaneously goes from liquid to solid. This is the same as the melting point. When solutes are added to a solvent the freezing point of the solution is less than that of the pure solvent. This is called the freezing point depression. It is a colligative property in that the identity of the solute is not important, only the amount.

It should be noted that not only is the freezing point depressed when a solute is added but also the range of temperatures over which freezing occurs is broadened. A pure solvent will freeze, or melt, over a small range. It will begin to freeze as it is cooled when the temperature reaches the freezing point. By the time all the liquid is frozen the temperature will have dropped only one or two degrees. For a solution the temperature will have dropped by a greater amount by the time all the solution has frozen.

This is a consequence of the fact that as a solution freezes it tends to form pure crystals of solvent leaving the solute behind in the remaining liquid. This means that as it freezes the remaining liquid portion becomes more concentrated. That means that the freezing point of the remaining liquid is even lower than before. Thus the temperature will have to drop to freeze the entire solution. This behavior is shown in the cooling curve formed when the temperature is plotted against time as the material is cooled.

The freezing point depression is governed by a formula similar to the one used for boiling point elevation:

DTf = kfmsol

 Solvent kf, oC kg/mol Freezing Point, oC Acetone 2.40 -95.35 Benzene 5.12 5.5 Naphthalene 6.94 80.5 Water 1.86 0
where DTf is the depression in the freezing point (oC), msol is the total molality of the solutes (mol/kg), and kf is the freezing point depression constant (oC kg/mol). Some examples of these constants are shown in the table to the right.

### For Instance:

What is the freezing point of a mixture of 45.7 g of benzoic acid, C7H6O2, in 250 g of benzene?
```
1 mole benzoic acid
45.7 g benzoic acid X ---------------------
122.13 g benzoic acid

= 0.374 mole benzoic acid

0.374 mole benzoic acid
msol = ----------------------- = 1.50 m
0.250 kg benzene

DTf = kfmsol = (5.12 oC kg/mol)(1.50 mol/kg) = 7.68 oC

FP = 5.5 - DTf = -2.18 oC
```

### Osmotic Pressure

Suppose we take a solution of say sugar in water and place it in a bag made of a membrane such as cellulose acetate. We then place a tube in the opening of the bag and tie it shut around the tube. When the bag is placed in a beaker of pure water the level of liquid rises in the tube up to a certain height above the water level and stops.

What is happening here. First the cellulose acetate is a semipermeable membrane. That is it contains small holes that allow small molecules, such as water, to pass but prevents the passage of larger molecules, such as sugar. The rise in liquid level in the tube is due to the net flow of water into the bag. When the level reaches a certain height the weight of the liquid in the tube counteracts the force that is causing the flow inward. We can say that there must be a pressure across the membrane that can support the column of liquid in the tube. This pressure is called osmotic pressure and this phenomenon is called osmosis.

What causes this pressure? Something is causing more molecules of water to move through the small holes in the membrane going into the bag rather than out of the bag. One explanation is that there is a higher concentration of water outside than there is inside and therefore it is more likely that a water molecule will enter a membrane hole from the outside rather than from the inside. Another explanation is that higher concentrated solutions have higher potential energy. Therefore any process that will dilute a solution is spontaneous. Therefore there is more of a tendency for water to enter the bag and dilute the sugar solution rather than the opposite. This tendency can be counteracted by the higher potential energy of the column of liquid as it rises in the tube. This balancing of potential energies is similar to a balancing of pressures, thus the name osmotic pressure.

Osmosis occurs whenever a semipermeable membrane separates two solutions of different total concentration. This total concentration is called the osmolarity of the solution when expressed as a molarity, mol/L. When expressed as other forms of concentration it is refered to as the tonicity of the solution. The net flow is always from the region of lower osmolarity to the higher.

An important example of osmotic pressure is the behavior of living cells placed in various solutions. The cell membrane is semipermeable. Therefore when the cell is placed in a solution of different tonicity than the cytoplasm inside the cell there will be a pressure difference across the membrane. If the cell is placed in a solution of lower tonicity, a hypotonic solution, there is a net flow of water into the cell causing it to swell until it bursts. If the solution is higher in tonicity, hypertonic, water flows out of the cell and it shrivels up.

Both situations can of course lead to the death of the cell. Therefore keeping a balance in tonicity between the cells in living tissues and their surroundings is necessary for the health of the organism. A good example of this is the behavior of plants in a drought. The cells on the surface of the roots normally have a tonicity higher than that of the soil moisture. Therefore there is a net flow of water into the roots and from there the other tissues of the plant. If the soil drys out, the tonicity around the roots increases until water flows out of the plant into the soil. The cells in the plant tissues shrivel up and cause the plant to wilt. Without water the plant will eventually die.

Osmotic pressure is a colligative property in that the pressure depends only on the total concentration of solutes and not their identity. It is governed by the formula:

P = MRT

where P is the osmotic pressure across the membrane, R is the gas constant (0.08206 L atm/mol K), T is the Kelvin temperature, and M is the molarity of the solution or more precisely the difference in molarity between the two sides of the membrane. Since molarity is moles per unit volume, M = n/V, we can rearrange the formula as:

PV = nRT

which is suspiciously like the ideal gas law. That suggests that the behavior of solute molecules dissolved in a solvent has similarities with the behavior of gas molecules moving through an open space.

Since the formulas for the colligative properties include a concentration term, determination of a property like freezing point depression or osmotic pressure for a solution can allow one to determine its concentration or related values.

### For Instance:

When one dissolves 1.55 g of an unknown compound in water, 254 mL of solution is formed. What is the molarity of the solution if the osmotic pressure is 1.55 atm at 23.0 oC? What is the molar mass of the unknown?
```
P = MRT

M = P/RT = (1.55 atm)/(0.08206 L atm/mol K)(296 K)

M = 0.0638 mol/L
```
If we assume that the unknown does not dissociate into multiple ions or other species on solution then the total molarity is the moles of unknown per liter of solution:
```
0.0638 mole
0.254 L X ----------- = 0.0162 mole
1 L

1.55 g
----------- = 95.7 g/mol
0.0162 mole
```

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