Chapter 6

Solutions

Denniston Topping Caret

7th Edition

Copyright The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

6.1 Properties of Solutions

• Solution - homogeneous mixture

• Solute - the substance in the mixture present in lesser quantity

• Solvent - the substance present in the largest quantity

• Aqueous solution - solution where the solvent is water

• Solutions can be liquids as well as solids and gases

Examples of Solutions

• Air - oxygen and several trace gases are dissolved in the gaseous solvent, nitrogen

• Alloys - brass and other homogeneous metal mixtures in the solid state

• Focus on liquid solutions as many important chemical reactions take place in liquid solutions

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• Clear, transparent, no visible particles

• May have color

• Electrolytes are formed from solutes that are soluble ionic compounds

• Nonelectrolytes do not dissociate

• Volumes of solute and solvent are not additive

– 1 L ethanol + 1 L water does not give 2 L of solution

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)(-Cl )(Na )NaCl( OH2 aqaqs

General Properties of Liquid Solutions

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• Colloidal suspension - contains solute particles which are not uniformly distributed– Due to larger size of particles (1nm - 200 nm)

– Appears identical to solution from the naked eye

– Smaller than 1 nm, have solution

– Larger than 1 nm, have a precipitate

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• Solubility - how much of a particular solute can dissolve in a certain solvent at a specified temperature

• Factors which affect solubility:1 Polarity of solute and solvent

• The more different they are, the lower the solubility2 Temperature

• Increase in temperature usually increases solubility3 Pressure

• Usually has no effect• If solubility is of gas in liquid, directly proportional

to applied pressure

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• Saturated solution - a solution that contains all the solute that can be dissolved at a particular temperature

• Supersaturated solution - contains more solute than can be dissolved at the current temperature

• How is this done? • Heat solvent, saturate it with solute then cool slowly

• Sometimes the excess will precipitate out

• If it doesn’t precipitate, the solution will be supersaturated

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• If excess solute is added to a solvent, some dissolves

• At first, rate of dissolution is large

• Later, reverse reaction – precipitation – occurs more quickly

• When equilibrium is reached the rates of dissolution and precipitation are equal, there is some dissolved and some undissolved solute

• A saturated solution is an example of a dynamic equilibrium

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• Henry’s law – the number of moles of a gas dissolved in a liquid at a given temperature is proportional to the partial pressure of the gas above the liquid

• Gas solubility in a liquid is directly proportional to the pressure of the gas in the atmosphere in contact with the liquid

• Gases are most soluble at low temperatures• Solubility decreases significantly at higher

temperatures– Carbonated beverages – CO2 solubility less when warm

– Respiration – facilitates O2 and CO2 exchange in lungs

6.2 Concentration Based on Mass

• Concentration - amount of solute dissolved in a given amount of solution

• Concentration of a solution has an effect on – Physical properties

• Melting and boiling points

– Chemical properties• Solution reactivity

6

• Amount of solute = mass of solute in grams

• Amount of solution = volume in milliliters

• Express concentration as a percentage by multiplying ratio by 100% = weight/volume percent or % (W/V)

%100solution of smilliliter

solute of grams

V

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solution ofamount

solute ofamount ion concentrat

Weight/Volume Percent6.

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PercentCalculate the percent composition or % (W/V) of 2.00 x 102 mL containing 20.0 g sodium chloride

20.0 g NaCl, mass of solute

2.00 x 102 mL, total volume of solution

% (W/V) = 20.0g NaCl / 2.00 x 102 mL x 100%

= 10.0% (W/V) sodium chloride

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Weight/Volume Percent

Calculate the number of grams of glucose in 7.50 x 102 mL of a 15.0% solution

15.0% (W/V) = Xg glucose/7.50 x 102 mL x 100%

Xg glucose x 100% = (15.0% W/V)(7.50 x 102 mL)

Xg glucose = 113 g glucose

%100solution of smilliliter

solute of grams

V

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%100solutions grams

solute grams

W

W%

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• Weight/weight percent is most useful for solutions of 2 solids whose masses are easily obtained

• Calculate % (W/W) of platinum in gold ring with 14.00 g Au and 4.500 g Pt[4.500 g Pt / (4.500 g Pt + 14.00 g Au)] x 100%

= 4.500 g / 18.50 g x 100% = 24.32% Pt

6.3 Concentration of Solutions: Moles and Equivalents

• Chemical equations represent the relative number of moles of reactants producing products

• Many chemical reactions occur in solution where it is most useful to represent concentrations on a molar basis

• The most common mole-based concentration unit is molarity

• Molarity– Symbolized M– Defined as the number of moles of solute per

liter of solution

Molarity6.

3 M

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solution L

solute molesM

• Calculate the molarity of 2.0 L of solution containing 5.0 mol NaOH

• Use the equation

• Substitute into the equation:

MNaOH = 5.0 mol solute

2.0 L solution

= 2.5 M

Calculating Molarity from Moles6.

3 M

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solution L

solute molesM

• If 5.00 g glucose are dissolved in 1.00 x 102 mL of solution, calculate molarity, M, of the glucose solution

• Convert from g glucose to moles glucose– Molar mass of glucose = 1.80 x 102 g/mol

5.00 g x 1 mol / 1.80 x 102 g = 2.78 x 10-2 mol glucose– Convert volume from mL to L

1.00 x 102 mL x 1 L / 103 mL = 1.00 x 10-1 L

• Substitute into the equation:

Mglucose = 2.78 x 10-2 mol glucose 1.00 x 10-1 L solution = 2.78 x 10-1 M

Calculating Molarity From Mass6.

3 M

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solution L

solute molesM

solution L

solute molesM6.

3 M

oles

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ents Dilution

Dilution is required to prepare a less concentrated solution from a more concentrated one

– M1 = molarity of solution before dilution

– M2 = molarity of solution after dilution

– V1 = volume of solution before dilution

– V2 = volume of solution after dilution

moles solute = (M)(L solution)

• In a dilution will the number of moles of solute change?

– No, only fewer per unit volume

• So,

• Knowing any three terms permits calculation of the fourth

M1V1 = M2V2

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• Calculate the molarity of a solution made by diluting 0.050 L of 0.10 M HCl solution to a volume of 1.0 L – M1 = 0.10 M molarity of solution before dilution

– M2 = X M molarity of solution after dilution

– V1 = 0.050 L volume of solution before dilution

– V2 = 1.0 L volume of solution after dilution

• Use dilution expression

• X M = (0.10 M) (0.050 L) / (1.0 L)

0.0050 M HCl OR 5.0 x 10-3 M HCl

M1V1 = M2V2

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ts Representation of Concentration of Ions in Solution

Two common ways of expressing concentration of ions in solution:

1. Moles per liter (molarity)

• Molarity emphasizes the number of individual ions

2. Equivalents per liter (eq/L)

• Emphasis on charge

ionon charges ofnumber

(g)ion of massmolar ionan of equivalent One 6.

3 M

oles

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Equivalents1 M Na3PO4

• What would the concentration of PO43- ions be?

• 1 M

• Equivalent is defined by the charge

• One Equivalent of an ion is the number of grams of the ion corresponding to Avogadro’s number of electrical charges

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ts Molarity vs. Equivalents – 1 M Na3PO4

• 1 mol Na+ = 1 equivalent Na+

• 1 mol PO43- = 3 equivalents PO4

3-

• Equivalents of Na+?

– 3 mol Na+ = 3 equivalents of Na+

• Equivalents of PO43-?

– 1 mol PO43- = 3 equivalents of PO4

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ts Calculating Ion Concentration

• Calculate eq/L of phosphate ion, PO43- in a

solution with 5.0 x 10-3 M phosphate

• Need to use two conversion factors:

– mol PO43- mol charge

– mol charge eq PO43

5.0 x 10-3 mol PO43- x 3 mol charge x 1 eq

1 L 1 mol PO43- 1mol charge

• 1.5 x 10-2 eq PO43- /L

6.4 Concentration-Dependent Solution Properties

• Colligative properties - properties of solutions that depend on the concentration of the solute particles, rather than the identity of the solute

• Four colligative properties of solutions1. vapor pressure lowering

2. boiling point elevation

3. freezing point depression

4. osmotic pressure

Vapor Pressure of a LiquidConsider Raoult’s law in molecular

terms• Vapor pressure of a solution

results from escape of solvent molecules from liquid to gas phase

• Partial pressure of gas phase solvent molecules increases until equilibrium vapor pressure is reached

• Presence of solute molecules hinders escape of solvent molecules, lowering equilibrium vapor pressure6.

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Vapor Pressure Lowering• Raoult’s law - when a nonvolatile solute is

added to a solvent, vapor pressure of the solvent decreases in proportion to the concentration of the solute

• Solute molecules (red below) serve as a barrier to the escape of solvent molecules resulting in a decrease in the vapor pressure

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Freezing Point Depression and Boiling Point Elevation

• Freezing point depression may be explained considering the equilibrium between solid and liquid states– Solute molecules interfere with the rate at which

liquid water molecules associate to form the solid state

• Boiling point elevation can be explained considering the definition as the temperature at which vapor pressure of the liquid equals the atmospheric pressure– If a solute is present, then the increase in boiling

temperature is necessary to raise the vapor pressure to atmospheric temperature

• Freezing point depression (Tf) - is proportional to the number of solute particles – Solute particles, not just solute

• How does an electrolyte behave?– Dissociate into ions

• An equal concentration of NaCl will affect the freezing point twice as much as glucose (a nonelectrolyte)

• Each solvent has a unique freezing point depression constant or proportionality factor

Tf=kf m6.4

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Freezing Point Depression

• Boiling point elevation (Tb) - is proportional to the number of solute particles

• An electrolyte will affect boiling point to a greater degree than a nonelectrolyte of the same concentration

• Each solvent has a unique boiling point elevation constant

Tb=kb m

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Boiling point elevation

Osmotic Pressure• Some types of membranes appear impervious

to matter, but actually have a network of small holes called pores

• These pores may be large enough to permit small solvent molecules to move from one side of the membrane to the other

• Solute molecules cannot cross the membrane as they are too large

• Semipermeable membrane - allows solvent but not solute to diffuse from one side to another

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Osmotic Pressure

• Osmosis - the movement of solvent from a dilute solution to a more concentrated solution through a semipermeable membrane

• Requires pressure to stop this flow

• Osmotic pressure ( - the amount of pressure required to stop the flow across a semipermeable membrane

• Osmolarity - the molarity of particles in solution

– Osmol, used for osmotic pressure calculation

=MRT

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Osmotic Pressure

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Tonicity and the Cell• Living cells contain aqueous solution and these cells

are also surrounded by aqueous solution• Cell function requires maintenance of the same osmotic

pressure inside and outside the cell• Solute concentration of fluid surrounding cells higher

than inside results in a hypertonic solution causing water to flow into the surroundings, causing collapse = crenation

• Solute concentration of fluid surrounding cells too low, results in a hypotonic solution causing water to flow into the cell, causing rupture = hemolysis

• Isotonic solutions have identical osmotic pressures and no osmotic pressure difference across the cell membrane

IsotonicHemolysisCrenation

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Tonicity and the Cell

Pickling Cucumber in Hypertonic Brine Due to Osmosis

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6.5 Water as a Solvent

• Water is often referred to as the “universal solvent”

• Excellent solvent for polar molecules • Most abundant liquid on earth• 60% of the human body is water

– transports ions, nutrients, and waste into and out of cells

– solvent for biochemical reactions in cells and digestive tract

– reactant or product in some biochemical processes