Chapter 13 – Properties of Solutions

Homework:33, 34, 35, 38, 39, 41, 44, 46, 47, 50, 55, 59, 63, 65, 66, 67,

69, 72, 74

13.4 – Ways of Expressing Concentration Concentration of a solution can be

expressed qualitatively or quantitatively Terms dilute and concentrated are

qualitative statements Dilute meaning a relatively small concentration of

solute Concentrated meaning a relatively large

concentration of solute Our focus will be on the quantitative

descriptions of concentration

Mass Percentage One of the simplest quantitative

methods of expressing concentration Given by

So if I had a solution of 36% HCl by mass It would contain 36 g of HCl for each 100 g

of solution Note: NOT per 100g of solvent

100solution of mass total

solutionin component of mass component of % Mass

ppm Oftentimes, very dilute solutions are given in

terms of parts per million (ppm) ppm is given by

What this means If I have a solution whose concentration is 1 ppm It has 1 g of solute for every million (106) g of

solution Or 1 mg of solute per kg of solution

610solution of mass total

solutionin component of masscomponent of ppm

Because density of water is 1 g/mL 1 kg of a very dilute aqueous solution will

have a volume very close to 1 L So, 1 ppm also means 1 mg/L of solution

ppm often used to express maximum concentration of toxic or carcinogenic substances in the environment In US, maximum arsenic in water is 0.010

ppm

ppb A solution that is VERY dilute is

often measured in terms of parts per billion (ppb) 1 ppb means 1 g of solute per billion

(109) g of solution Or 1 microgram (μg) per liter of

solution So the allowable concentration of

arsenic in water could also be 10 ppb.

Mole Fraction Concentration can also be expressed by

the number of moles of components of the solution.

Mole fraction we’ve talked about

No units, just a ratio The sum of all of the mole fractions of

the components adds up to 1

components all of moles total

component of moles component offraction mole X

Molarity The big one Molarity (M) relates moles of solute to

volume of solution

Very useful when doing stoichiometry, because of the easy mole relationship

solution liters

solute moles Molarity

Molality Molality (m) is the concentration unit that

looks at the number of moles of solute per kilogram of solvent

Don’t get it confused with molarity Molarity is moles/L Molality is moles/kg

Molality doesn’t change with temperature, while molarity does

Molality often used when the solution is used over a range of temperatures

solvent of kilograms

solute of moles Molality

Conversion of Concentration Units Sometimes the concentration

needs to be known in several different units

It is possible to convert concentration units

Example An aqueous solution of

hydrochloric acid contains 36% HCl by mass. Find the mole fraction of HCl in the solution. Find the molality of HCl in the solution

Working Mole fraction When dealing with percentages, assume

100g of the substance So 36% = 36g per 100 g of the substance So if HCl is 36g, that means that water

would be 64g To do mole fraction, we now convert the

g of each of the components to moles

HCl mol 99.0HCl g 36.5

HCl mol 1HCl g 36HCl Moles

OH mol 6.3OH g 18

OH mol 1OH g 64OH Moles 2

2

222

22.099.06.3

99.0

HCl moles OH moles

HCl molesX

2HCl

Converting to molality Need moles of HCl and mass of water Luckily, we got those from doing the mole

fraction 0.99 mol HCl and 0.064 kg H2O

m 15H kg 0.064

HCl mol 0.99 HCl ofMolality

2

O

Molality ↔ Molarity To convert between molality and

molarity we need to know the density of the solution

Example A solution contains 5.0g of toluene

(C7H8) and 225 g of benzene and has a density of 0.876 g/mL. Calculate the molarity of the solution.

To find molarity, we need mol of toluene and volume of solution

We find volume by using the density of the solution But first, we need total mass of solution

mol 054.0HC g 92

HC mol 1HC g 0.5HC Moles

87

878787

To find volume, we use density Note: We actually have 230 g of substance

225 g benzene and 5 g of toluene

Finally, to find molarity, we just divide moles of toluene by liters of solution

mL263g 0.876

mL 1g 230 Volume

Final Step – Almost There!

Note: To solve this problem we needed the following Density (allows us to move from mass of

solution to volume of solution) Mass of solute (to find moles of solute)

M21.0solution L 1

mL 1000

solution mL 263

HC mol 0.054

solutionliter

HC moles Molarity 8787

13.5 – Colligative Properties The physical properties of many

solutions differ from those of the pure solvent We add ethylene glycol to water in

radiators to lower the freezing point (antifreeze) and to raise the boiling point (so that the water won’t boil in the radiator)

This lowering of the freezing point and raising of the boiling point are physical properties These properties (in a solution) depend on

the quantity (concentration) but not the identity of the solute

These types of properties are called colligative properties.

Colligative means “depending on the collection” So these properties depend on the collective

effect of the number of solute particles

Common Colligative Properties The common (the one’s we’re going to

look at) colligative properties include: Lowering freezing point Raising boiling point Vapor-pressure reduction Osmotic pressure

For each of these, notice how the concentration of the solute affects the property we’re looking at (relative to the pure substance)

Lowering the Vapor Pressure We know that a liquid in a closed

container will create a dynamic equilibrium with its vapor At equilibrium, the pressure exerted by

the vapor is called the vapor pressure Remember, a substance that has no

measurable vapor pressure is called nonvolatile

And a substance that has a vapor pressure is called volatile

We find that the vapor pressure of solutions (if we have added a nonvolatile solute) will be lower than the vapor pressure of the solvent The amount the vapor pressure is

lowered is directly proportional to the concentration of the solute

This relationship is given by Raoult’s law

Raoult’s Law PA = partial pressure of the vapor from

the solution In other words: vapor pressure of the

solution XA= mole fraction of the solvent PºA= vapor pressure of the pure solvent

AAA PXP

Example The vapor pressure of water is 17.5

torr at 20ºC. If we add glucose to the water, so that the mole fraction of water is 0.800, then the new vapor pressure of water will be PH2O= (0.800)(17.5 torr) = 14.0 torr Doesn’t matter which pressure units

we use! Just be consistent

Raoult’s law assumes that the solutes are non-volatile, molecular compounds. Not always the case Will deal with ionic compounds (and

their effects) when we discussion freezing and boiling points

Deviation from Prediction An ideal solution obeys Raoult’s

law. Real solutions best approximate ideal

solutions when solute concentration is low and when solute and solvent have similar molecular sizes and types of intermolecular attractions Understand that this occurs, but assume

ideal solution unless told otherwise

Boiling Point Elevation Remember, adding solute lowers vapor

pressure Also remember, boiling occurs when

vapor pressure = atmospheric pressure So we need a higher temperature to

give us more vapor pressure when we add a solute

So, the boiling point of a solution is higher than that of a pure liquid

The increase in boiling point (relative to the pure solvent) is given by the term ΔTb

Is directly proportional to the number of solute particles per mole of solvent molecules

Molality is the number of moles of solute per 1000 g of solvent, which shows a fixed number of moles of solvent.

Thus, ΔTb is directly proportional to molality

ΔTb = Kbm

The magnitude of Kb (called the molal boiling-point-elevation constant) depends on the solvent in question Typical solvents listed on pg. 551 For water, Kb = 0.51ºC/m So a 1 m aqueous solution of sucrose

(sugar) would boil at 0.51ºC higher than pure water

The boiling-point elevation is proportional to the concentration of the solute particles Doesn’t matter if they are ions or molecules Note: When an ionic compound dissolves, total

molality increases When 1 mol of NaCl dissolves, 2 moles of solute form

(1 mol of Na+ and 1 mol of Cl-) So if we have a 1 m aqueous solution of NaCl, we

actually have a total of 2 m in solute particles So double the effect of boiling point increase

Freezing-Point Depression The freezing point of a solution is

the temperature at which the first crystals of pure solvent begin to form In equilibrium with the solution Lower vapor pressure effects freezing

point too Lower vapor pressure (we’ve added

solute) means the freezing point is lower than that of a pure solvent

ΔTf = Kfm Like the boiling-point elevation, the

decrease in freezing point, ΔTf is directly proportional to the molality of the solute

The values of Kf, the molal freezing-point-depression constant for several common solvents are also found on pg. 551

For water, Kf = 1.86ºC/m So 1m aqueous solution of sucrose will freeze

1.86ºC lower than pure water While the same rule for ionic compounds

applies here A 1m aqueous solution of NaCl will freeze 3.72ºC

below the freezing point of the pure solvent

Example Calculate the freezing point of a

solution containing 0.600 kg of CHCl3 and 42.0g of eucalyptol (C10H18O), a fragrant substance found in the leaves of eucalyptus trees.

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Kf = 4.68ºC/m normal freezing point = -63.5ºC Find molality!

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Example 2 Which of the following solutes will

produce the largest increase in boiling point upon addition to 1 kg of water? 1 mol of Co(NO3)2

2 mol of KCl 3 mol of ethylene glycol (C2H6O2)

Osmosis Certain materials are

semipermeable Often in biological systems and

synthetic substances When in contact with a solution,

allows some molecules to pass through them, but not others

Generally allow small solvent molecules (like water) to pas through

But blocks larger solute molecules or ions

Consequences of Osmosis Consider a membrane that only allows

solvent particles to pass If placed between two solutions of different

concentration, solvent molecules move in both directions through the membrane

However, this process is called osmosis, and the net movement of solvent is always toward the solution with the higher SOLUTE concentration

So the less concentrated solution loses solvent, and more concentrated solution gains solvent

What does this mean?

If we wanted to stop this from happening, we apply pressure on the side of the more concentrated solution

The pressure required to prevent osmosis by pure solvent is the osmotic pressure, π

The osmotic pressure obeys a law similar in form to the ideal gas law

πV=nRT V is the volume of the solution n is the number of moles of solute R is the ideal-gas constant T is temperature in Kelvin scale

From this equation, we can write

Where M is the molarity of the solution R value determines which units your

pressure will be in

MRTRTV

n

If two solutions of identical osmotic pressure are separated by a semi-permeable membrane, no osmosis will occur The two solutions are called isotonic

If one solution is of lower osmotic pressure, that solution is hypotonic with respect to the more concentrated solution The more concentrated solution is hypertonic

with respect to the dilute solution

Biological Systems See! I know biology! The membranes of blood cells are

semipermeable Placing a red blood cell in solution that is

hypertonic relative to the solution inside the cell causes water to come out of the cell

Causes cell to shrivel (crenation) Placing a red blood cell in a solution that is

hyptotonic relative to the solution inside the cell causes water to move into the cell

Causes cell to rupture (hemolysis) When giving IV infusions to people, IV solution

must be isotonic with the solution inside the cell

Example What is the osmotic pressure at

20ºC of a 0.0020 M sucrose (C12H22O11) solution.

Determination of Molar Mass Any of the colligative properties

can be used to determine the molar mass of a solute.

Molar Mass from Boiling Point Elevation A solution of unknown nonvolatile

non-ionic solute was prepared by dissolving 0.250 g of the substance in 40.0 g of CCl4. The boiling point of the solution was 0.357ºC higher than that of the pure solvent. What is the molar mass of the solute?

To find molar mass, we need grams of solute and moles of solute

Step 1: Solve for molality Step 2: Use molality to solve for mol

solute Step 3: Use given data and mol

solute to find molar mass

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Molar Mass from Osmotic Pressure A solution contains 3.50 mg of

protein dissolved in enough water to form 5.00 mL of solution. The osmotic pressure of the solution at 25ºC was found to be 1.54 torr. Calculate the molar mass of the protein.

Step 1: Solve for molarity Step 2: Using molarity, solve for

number of moles Step 3: Divide number of grams

given by moles

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