Chapter 7:Solutions and Colloids

Spencer L. SeagerMichael R. Slabaugh

www.cengage.com/chemistry/seager

Jennifer P. Harris

SOLUTIONS• Solutions are homogeneous

mixtures of two or more substances in which the components are present as atoms, molecules, or ions.

• Particles in liquid solutions are:• too small to reflect light, thus

solutions are transparent (clear).

• in constant motion.• not settled by the influence of

gravity.• Some solutions are colored.

SOLVENT & SOLUTE• SOLVENT OF A SOLUTION

• The solvent of a solution is the substance present in the largest amount in the solution.

• SOLUTE OF A SOLUTION• A solute of a solution is any substance present in an

amount less than that of the solvent. A solution may contain more than one solute.

PHYSICAL STATES OF SOLUTIONS

• The physical state of a solution, solid, liquid, or gas, is usually the same as the physical state of the solvent.

SOLUBILITY• The solubility of a solute is the maximum amount of the solute

that can be dissolved in a specific amount of solvent under specific conditions of temperature and pressure.

SOLUBLE vs. INSOLUABLE• SOLUBLE SUBSTANCE

• This is a term used to describe a substance that dissolves to a significant extent in a solvent without stating how much actually will dissolve.

• INSOLUBLE SUBSTANCE• This is a term used to describe a substance that does not

dissolve to a significant extent in a solvent.

Sugar has limited solubility in water

IMMISCIBLE• This is a term used to describe liquids that are not soluble in

each other. The liquids don't mix together to form a solution.

DEGREES OF SATURATION• A saturated solution is a solution that

contains the maximum amount possible of dissolved solute in a stable situation under the prevailing conditions of temperature and pressure.

• A supersaturated solution is an unstable solution that contains an amount of solute greater than the solute solubility under the prevailing conditions of temperature and pressure.

• An unsaturated solution is a solution that contains an amount of solute less than the amount required to form a saturated solution under the prevailing conditions of temperature and pressure.

EXAMPLES OF SOLUTE SOLUBILITES AT 0°C

EXAMPLES OF SOLUTE SOLUBILITES AT 0°C (continued)

EFFECT OF TEMPERATURE ON SOLUBILITY

THE SOLUTION PROCESS• The solution process involves interactions between solvent

molecules (often water) and the particles of solute. • An example of the solution process for an ionic solute in water:

THE SOLUTION PROCESS (continued)

• An example of the solution process for a polar solute in water:

THE SOLUTION PROCESS (continued)

• A solute will not dissolve in a solvent if:• the forces between solute particles are too strong to be

overcome by interactions with solvent particles.• the solvent particles are more strongly attracted to each

other than to solute particles.• A good rule of thumb for solubility is “like dissolves like.”

• Polar solvents dissolve polar or ionic solutes.• Nonpolar solvents dissolve nonpolar or nonionic solutes.

GENERAL SOLUBILITIES OF IONIC COMPOUNDS IN WATER

INCREASING THE RATE OF DISSOLVING

• Crush or grind the solute.• Small particles provide more surface area for solvent attack

and dissolve more rapidly than larger particles.

• Heat the solvent.• Solvent molecules move faster and have more frequent

collisions with solute at higher temperatures.

• Stir or agitate the solution.• Stirring removes locally saturated solution

from the vicinity of the solute and allows unsaturated solvent to take its place.

HEAT AND SOLUTION FORMATION• Endothermic:

• Exothermic: Solute + solvent → solution + heat

Solute + solvent + heat→ solution

SOLUTION CONCENTRATIONS• Solution concentrations express a quantitative relationship

about the amount of solute contained in a specific amount of solution.

• Concentration units discussed include molarity and percentage.

MOLARITY• The molarity of a solution expresses the number of moles of

solute contained in one liter of solution.• The mathematical calculation of the molarity of a solution

involves the use of the following equation:

• In this equation, the number of moles of solute in a sample of solution is divided by the volume in liters of the same sample of solution.

solution of liters

solute of molesM

MOLARITY CALCULATION EXAMPLES

• Example 1: A 250-mL sample of solution contains 0.134 moles of solute. Calculate the molarity of the solution.

• Solution: The solution sample volume in liters is 0.250 L. The number of moles of solute in the sample is 0.134 mol. These two quantities are substituted into the equation given earlier:

liter

mole536.0

L 0.250

mole 134.0

solution of liters

solute of molesM

MOLARITY CALCULATION EXAMPLES (continued)• Example 2: 9.45 g of methyl alcohol, CH3OH, was dissolved in

enough pure water to give 500 mL of solution. What was the molarity of the solution?

• Solution: The solution volume was 500 mL or 0.500L. The amount of solute was given in grams, but needs to be expressed in moles. The molecular weight of methyl alcohol is 32.0 u, so 1 mol = 32.0 g. This fact can be used to convert the mass of methyl alcohol into moles:

• The number of moles of alcohol and the number of liters of solution can now be substituted into the equation for molarity:

alcohol mole 0.295alcohol g 32.0

alcohol mole 1 alcohol g 9.45

liter

mole590.0

L 0.500

mole 295.0

solution of liters

solute of molesM

PERCENT CONCENTRATIONS• Percent concentrations express the amount of solute

contained in 100 parts of solution. The parts of solution may be expressed in different units.

100total

part%

WEIGHT/WEIGHT PERCENT• Weight/weight percent, abbreviated %(w/w), is a

concentration that expresses the mass of solute contained in 100 mass units of solution. Any mass units may be used, but the mass of solute and solution must be in the same units.

• The equation used for mathematical calculations is:

100mass solution

mass solutew / w)%(

WEIGHT/WEIGHT PERCENT EXAMPLE

• Calculation example: Calculate the %(w/w) of a solution prepared by dissolving 15.0 grams of table sugar in 100 mL of water. The density of the water is 1.00 g/mL.

• Solution: The mass of the water used is 100 grams because according to the density each mL has a mass of 1.00 grams. The mass of solution is equal to the mass of water plus the mass of the sugar solute or 100 + 15.0 = 115 grams. The calculation is:

%0.13100g 115

g 15.0w / w)%(

WEIGHT/VOLUME PERCENT• Weight/volume percent, abbreviated %(w/v), is a concentration

that expresses the number of grams of solute contained in 100 mL of solution. The solute mass must be expressed in grams, and the amount of solution must be expressed in mL.

• This percent concentration is normally used when the solute is a solid and the solvent and resulting solutions are liquids.

• The equation used for mathematical calculations is:

100solution of mL

solute of gramsv)w / %(

WEIGHT/VOLUME PERCENT EXAMPLE

• Calculation example: Calculate the %(w/v) of a solution prepared by dissolving 8.95 grams of sodium chloride in enough water to give 50.0 mL of solution.

• Solution: The data are given in units that may be put directly into the equation. The calculation is:

%9.17100mL 50.0

g 8.95v)w / %(

VOLUME/VOLUME PERCENT• Volume/volume percent, abbreviated %(v/v), is a concentration

that expresses the volume of liquid solute contained in 100 volumes of solution. Any volume units may be used, but they must be the same for both the solute and the solution.

• The equation used for mathematical calculations is:

100volume solution

volume solutev) / v%(

VOLUME/VOLUME PERCENT EXAMPLE

• Calculation example: A solution is made by dissolving 250 mL of glycerin in enough water to give 1.50 L of solution. Calculate the %(v/v) of the resulting solution.

• Solution: The data give the solute volume and solution volume in different units. They must be the same. Either quantity could be changed, but the solute volume will be expressed in liters as 0.250 L. The calculation is:

%7.16100L 1.50

L 0.250v) / v%(

SOLUTION PREPARATION• Solutions of known concentration are usually prepared in one

of two ways. • In one method, the necessary quantity of pure solute is

measured using a balance or volumetric equipment. The solute is put into a container and solvent, usually water, is added until the desired volume of solution is obtained.

SOLUTION PREPARATION EXAMPLE• Calculation example: Describe how to prepare 500 mL of

0.250 M NaCl solution.

• Solution: The mass of NaCl needed must first be determined. The volume and concentration of the desired solution are known, so the equation for molarity is rearranged to solve for the number of moles of solute needed. The result is:

moles of solute = M x liters of solution = 0.250 M x 0.500 L = 0.125 mole

• Thus, 0.125 moles of NaCl is needed. NaCl has a formula weight of 58.4 u, so 0.125 moles has a mass of 0.125 x 58.4g or 7.30 grams. The solution is prepared by weighing a sample of NaCl with a mass of 7.30 grams. The sample is put into a 500 mL volumetric flask and pure water is added up to the mark on the flask.

SOLUTION PREPARATION (continued)

• In a second method, a quantity of solution with a concentration greater than the desired concentration is diluted with an appropriate amount of solvent to give a solution with a lower concentration. This type of problem is made simpler by using the following equation:

(Cc)(Vc) = (Cd)(Vd)

• In this equation, Cc is the concentration of the concentrated solution that is to be diluted, Vc is the volume of concentrated solution that is needed, Cd is the concentration of the dilute solution, and Vd is the volume of dilute solution.

SOLUTION PREPARATION EXAMPLE

• Calculation example: Describe how to prepare 250 mL of 0.500 M HCl solution from a 1.50 M HCl solution.

• Solution: According to the definitions given above, Cc = 1.50 M, Cd = 0.500 M, and Vd = 250 mL. The equation given above can be solved for Vc, the volume of concentrated solution needed:

• The solution is prepared by measuring 83.3 mL of 1.50 M HCl and pouring it into a 250 mL volumetric flask. Pure water is then added up to the mark on the flask to give 250 mL of 0.500 M solution.

mL 3.83

M 1.50

mL 250M 0.500

C

VCV

c

ddc

SOLUTION STOICHIOMETRY• As shown earlier, the number of moles of solute in a volume of

solution of known molarity can be obtained by multiplying together the known molarity and the solution volume in liters.

• Molarity is a ratio of moles of solute to liters of solution. This ratio can be written as two conversion factors:

• The conversion factor on the left is used to multiply by the molarity. It is selected to cancel the units of liters of solution and obtain the units of moles of solute.

• The conversion factor on the right is used to divide by the molarity. It is selected to cancel the units of moles of solute and obtain the units of liters of solution.

solute moles

solution liters or

solution liters

solute moles

SOLUTION STOICHIOMETRY EXAMPLE

• Calculation example: Consider the balanced equation HCl(aq) + NaOH(aq) NaCl(aq) + H2O(l)

How many mL of 0.100 M HCl solution would exactly react with 25.00 mL of 0.125 M NaOH solution?

• Solution: The number of moles of NaOH in 25.00 mL of 0.125 M solution is calculated by multiplying the molarity and the solution volume in liters:

moles of NaOH = 0.125 M x 0.02500 L = 0.00313 moles of NaOH

• According to the reaction, 1 mole of NaOH reacts with 1 mole of HCl, so 0.00313 moles of HCl will be needed.

SOLUTION STOICHIOMETRY EXAMPLE (continued)• The volume of 0.100 M HCl solution that contains 0.00313 can

be calculated by dividing the needed number of moles by the solution molarity:

• Thus, it is seen that 0.0313 liters or 31.3 mL of 0.100 M HCl solution will be required to react with the 25.00 mL of 0.125 M NaOH solution.

M

solute of molessolution liters

liters 0313.0M 0.100

moles 0.00313solution liters