This section will focus on what you need to know about solutions,solution concentrations, and colligative properties in order to besuccessful on the
SAT II Chemistry test. This material is closely tiedin with the material from the first half of this chapter and “TheStructure of Matter.”
Properties of Solutions
A
solution is a homogenous mixture of two or more substances that exist in a single phase. There are two main parts to any solution. The
soluteis the component of a solution that is dissolved in the solvent; it isusually present in a smaller amount than the solvent. The
solventis the component into which the solute is dissolved, and it is usuallypresent in
greater concentration. For example, in a solution of saltwater, salt is the solute and water is the solvent. In solutions wherewater is the solvent, the solution is referred to as an
aqueous solution.
A solution does not have to involveliquids. For instance, air is a solution that consists of nitrogen,oxygen, carbon dioxide, and other trace gases, and solder is a solutionof lead and tin. The general rule of thumb for solutions is the ideathat
like dissolves like. Polar, ionic substances are solublein polar solvents, while nonpolar solutes are soluble in nonpolarsolvents. For example, alcohol and water, which are both polar, canform a solution and iodine and carbon tetrachloride, which are bothnonpolar, make a solution. However, iodine will not readily dissolve inpolar water.
In a solution, the particles are reallysmall—anywhere from 0 to 100 nm. They never settle on standing, theycannot be separated by filtering, and light will pass through asolution unchanged. One type of mixture that is not a solution is knownas the colloid. In a
colloid, particles are between 100 and1000 nm in size—still too small for our eyes to distinguish, butparticles this small will not settle. As is the case in solutions, theparticles cannot be filtered, but they do scatter light. Some examplesof colloids include gelatin, fog, smoke, and shaving cream. Anothertype of mixture that is not considered a solution is known as asuspension.
Suspensions have much larger particles: usuallyover 1000 nm. Particles in a suspension will settle on standing, canoften be separated by a filter, and may scatter light, but they areusually not transparent. Some examples of suspensions are muddy water,paint, and some medicines, like Pepto-Bismol.
The Solution Process
In order for a solute to be dissolved in asolvent, the attractive forces between the solute and solvent particlesmust be great enough to overcome the attractive forces within the puresolvent and pure solute. The solute and the solvent molecules in asolution are expanded compared to their position within the puresubstances.
Theprocess of expansion, for both the solute and solvent, involves achange in the energy of the system: this process can be eitherexothermic or endothermic. After dissolving, the solute is said to befully solvated (usually by dipole-dipole or ion-dipole forces), and when the solvent is water, the solute is said to be hydrated.The separation of the solute particles from one another prior todissolving is an endothermic process for both solvent and solute (steps1 and 2), but when the solute and solvent combine with each other, thisis an exothermic process (step 3). If the energy released in step 3 isgreater than the energy absorbed in steps 1 and 2, the solution formsand is stable.
The term solubilityrefers to the maximum amount of material that will dissolve in a givenamount of solvent at a given temperature to produce a stable solution.By looking at the plot of solubilities below, you can see that mostsolids increase in solubility with an increase in temperature.
K k
Gases, however,
decrease in solubility with an increase in temperature.
Degrees of Saturation
When referring to solutions, there are threedegrees of saturation—unsaturated, saturated, and supersaturated. If asolution is
unsaturated, the solvent is capable of dissolving more solute. When the solution is
saturated,the solvent has dissolved the maximum amount of solute that it can atthe given temperature. At this point we say that the solution is in astate of
dynamic equilibrium—the processes of dissolving and precipitation are happening at the same rate. A
supersaturatedsolution is one in which the solvent contains more solute than it cantheoretically
hold at a given temperature. Supersaturated solutions areoften formed by heating a solution and dissolving more solute, thencooling the solution down slowly. These solutions are unstable andcrystallize readily.
Concentration Terms
Solutions are often referred to as being concentrated or dilute. These two terms are very general. While
concentrated indicates that there is a lot of solute dissolved in the solvent (perhaps the solution is near to being saturated) and
diluteindicates that a small amount of solute is dissolved in the solvent, weoften need to be exact with quantities in chemistry. The units ofconcentration that you should be familiar with for the SAT II exam arereviewed below.
Molarity (
M )
The
molarity of a solution is ameasure of the number of moles of solute per liter of solution. This isthe most common concentration unit used in chemistry. For instance, youmight see an expression that looks like this:
[NaCl] = 0.75
which means that 0.75 mole of NaCl isdissolved per 1.00 L of solution. The brackets around the numberindicate that the concentration is expressed in terms of molarity.Let’s now run through how you calculate the molarity of a solution.
Example
Calculate the molarity of a solution prepared by dissolving 20.0 g of solid NaOH in enough water to make 100 mL of solution.
Explanation
Convert grams to moles:
Then convert mL to liters:
Then divide:
Dilution
Dilution is the process of taking a moreconcentrated solution and adding water to make it less concentrated.The more concentrated solution before the dilution is performed isknown as the
stock solution. You can relate the concentrationof the stock solution to the concentration of the diluted solutionusing the equation below:
M1V1 = M2V2
where
M is molarity and
V is the volume, in liters, of the solution. Try the following example using this equation.
Example
What volume of 6.0 M sulfuric acid (H2SO4) must be used to prepare 2.0 L of a 0.10
M H2SO4 solution?
Explanation
Just plug the numbers into the formula! Be careful to read closely.
M1V1 = M2V2
(6.0 M) (V1) = (0.10 M) (2.0 L)
V1 = 0.033 L
or 33 mL should be measured out and then diluted by adding enough water to make 2.00 L total volume.
Mass Percent (Weight Percent)
The mass percent of a solution is anotherway of expressing its concentration. Mass percent is found by dividingthe mass of the solute by the mass of the solution and multiplying by100; so a solution of NaOH that is 28% NaOH by mass contains 28 g ofNaOH for each 100 g of solution. Here’s the equation:
;\*g0i(P%Mp.d K Now try a problem involving the equation:
Example
A solution is prepared by mixing 5.00 g ethanol (C2H5OH) with 100.0 g water. Calculate the mass percent of ethanol in this solution.
Explanation
Plugging the values we were given into the mass percent equation, we get:
Molality (
m)
The
molality of a solution is a measure of the number of moles of solute per kilogram of solvent. Whereas the
molarity of a solution is dependent on the volume of the solution, the
molalityis dependent on the mass of the solvent in the solution. Do not getthese confused, and when you see either term on the SAT II Chemistrytest, double-check to make sure which one they’re talking about—thewords look similar, too! Try an example:
Example
A solution is prepared by mixing 80.0 g ofsodium hydroxide (NaOH) with 500.0 g of water. Calculate the molalityof this solution.
Explanation
Convert grams of solute to moles:
Convert grams of solvent to kg:
Divide:
Electrolytes
Certain solutions are capable of conducting an electric current and these solutions are referred to as
electrolytes.Generally speaking, we say that there are three classes of electrolytes(solutions that conduct a current): acids, bases, and salts.
- Strong electrolytes consist of solutes that dissociatecompletely in solution. Strong acids, strong bases, and soluble saltsare in this category. (We will discuss acids and bases in chapter 6.)
- Nonelectrolytes are substances that are predominantlycovalently bonded, generally will not produce ions in solution, andtherefore are considered nonconductors.
- Weak electrolytes consist of solutes that dissociateonly a little in solution. Weak acids, weak bases, and slightly solublesalts are in this category.
The greater the degree of dissociation ofthe solute, the greater the conductivity of the solution. Consider twoacid solutions that have the same concentration—hydrochloric acid andacetic acid. Hydrochloric acid ionizes completely, while only about 2%of the acetic acid molecules ionize. If a conductivity apparatus wereused to test the two solutions, HCl would conduct an electric currentto a much greater degree because there is more available charge insolution. Below is a figure showing the ionization of barium chloride;as you can see, the Ba+ and Cl- ions are floating free in solution, and this makes barium chloride an electrolyte.
Colligative Properties
Properties of solutions that depend on the number of solute particles present per solvent molecule are called
colligative properties.The concentration of solute in a solution can affect various physicalproperties of the solvent including its freezing point, boiling point,and vapor pressure. For the SAT II you will only need to be familiarwith the first two.
Freezing Point Depression
The freezing point of a substance is definedas the temperature at which the vapor pressure of the solid and theliquid states of that substance are equal. If the vapor pressure of theliquid is lowered, the freezing point decreases.
Why is a solution’s freezing point depressedbelow that of a pure solvent? The answer lies in the fact thatmolecules cluster in order to freeze. They must be attracted to oneanother and have a spot in which to cluster; if they act as a solvent,solute molecules get in the way and prevent them from clusteringtightly together. The more ions in solution, the greater the effect onthe freezing point. We can calculate the effect of these soluteparticles by using the following formula:
DTf = Kf 
msolute 
i
where
D
Tf = the change in freezing point
Kf = molal freezing point depression constant for the substance (for water = 1.86ºC/
m)
m = molality of the solution
i = number of ions in solution (this isequal to 1 for covalent compounds and is equal to the number of ions insolution for ionic compounds)
Boiling Point Elevation
As you learned earlier in this chapter, theboiling point of a substance is the temperature at which the vaporpressure equals atmospheric pressure. Because vapor pressure is loweredby the addition of a nonvolatile solute, the boiling point isincreased. Why? Since the solute particles get in the way of thesolvent particles trying to escape the substance as they move aroundfaster, it will take more energy for the vapor pressure to reachatmospheric pressure, and thus the boiling point increases. We cancalculate the change in boiling point in a way that’s similar to how wecalculate the change in freezing point:
DTb = Kb
msolute
i
where
Kb = molal boiling point elevation constant (for water = 0.51˚C/
m)
Now try a problem that deals with freezing point depression and boiling point elevation.
Example
Calculate the freezing point and boiling point of a solution of 100 g of ethylene glycol (C2H6O2) in 900 g of water.
Explanation
Calculate molality:
Freezing point depression = (m)(Kf)(i)
Tf = (1.79)(1.86)(1) = 3.33ºC
Freezing point = 0ºC - 3.33ºC = -3.33ºC
Boiling point elevation = (m)(Km)(i)
Tb = (1.79)(0.51)(1) = 0.91ºC
Boiling point = 100ºC + 0.91ºC = 100.91ºC
