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Mark S. CracoliceEdward I. Peters

Mark S. Cracolice • The University of Montana

Chapter 16Solutions

Solution

Solution is a homogenous mixture. A solution can exist in any of the three states:

Solid solution: steel, brass, bronze

Liquid solution: alcohol in water, sugar water.

Gaseous solution: air, any mixture of gases

.

Solution Terminology

SoluteThe substance present in a relatively small amount in a

solution; the solid or gas when a substance in that state is dissolved in a liquid to make a solution.

SolventThe substance present in a relatively large amount in a

solution; the liquid when a solid or gas is dissolved to make a solution.

Solution Terminology

Concentrated Solutionhas a relatively large quantity of a specific solute

per unit amount of solution.

Dilute Solutionhas a relatively small amount of a specific solute

per unit amount of solution.

Solution TerminologySaturated solution

A solution that can exist in equilibrium with undissolved solute is a saturated solution.

SolubilityThe concentration of the saturated solution is called

the solubility of the solute.

UnsaturatedA solution is unsaturated if its concentration is

smaller than the solubility.

Solution Terminology

SupersaturatedA solution is said to be supersaturated if its concentration is greater than the solubility.

Supersaturated solution is not stable, a slight physical disturbance can start crystallization

The Formation of a SolutionThe water molecule is polar.

A polar molecule is one with an asymmetrical distribution of charge, resulting in positive and negative poles.

The Formation of a NaCl Solution

The Formation of a NaCl Solution

The formation of a sodium chloride solution from sodium chloride crystals is made possible by the interaction of water molecules with sodium and chloride ions.

NaCl (s) ↔ Na+ (aq) + Cl- (aq)

The sodium and chloride ions in solution are surrounded by polar water molecules, and are said to be hydrated.

The Formation of a Solution

The rate of dissolving depends on  1 The surface area. A finely divided solid dissolves

more rapidly.2 The diffusion of the solute from the surface. Stirring

increases the rate.3 Temperature. The higher the temperature, the

higher the rate. 

The Formation of a Solution

Crystallization rateThe rate per unit of surface area increases as the

solution concentration at the surface increases.

Equilibrium in a Saturated SolutionWhen the rates of dissolving and crystallization are the

same the solution is saturated.

The Formation of a Saturated Solution

When crystallization rate is equal to the dissolving rate the solution is saturated.

Factors That Determine Solubility

Solubility depends on three factors:

Intermolecular forces

Gas pressure (for a gas)

Temperature

Factors That Determine SolubilityIntermolecular Forces

Generally, if the forces between molecules A are about the same as the forces between molecules B, A and

B will probably dissolve in each other.

The rule of solubility is like dissolves like.

A nonpolar solvent can dissolve a nonpolar solute. A polar solvent can dissolve a polar or ionic solute

Factors That Determine Solubility

Ethylene glycol HO-CH2–CH2–OH, which is polar and can form hydrogen bond, is soluble in water. Motor oil is not soluble in water because hydrocarbon molecules are nonpolar.

Factors That Determine Solubility

Partial Pressure of Solute Gas Over Liquid Solution

• Pressure has little effect on the solubility of solids or liquids but has a pronounced effect on the solubility of gases.

• For an ideal solution, the solubility of a gas in a liquid is directly proportional to the partial pressure of the gas over the surface of the liquid.

Factors That Determine Solubility

Solubility of carbon dioxide decreases when partial pressure of carbon dioxide drops

Factors That Determine Solubility

Temperature

The solubility of most solids increases with rising temperature (but there are notable exceptions).

The solubilities of gases in liquids are generallylower at higher temperatures.

Factors That Determine Solubility

Factors That Determine Solubility

Percentage by Mass

Percentage by mass can be used as conversion factor between mass of solution and mass of solute.

100 solvent g + solute g

solute g = massby %

100 solution gsolute g = massby %

Percentage by Mass• When 125 grams of a solution is evaporated to

dryness, 42.3 grams of solute was recovered. What was the percentage of the solute?

100 solution g solute g = massby %

NaCl % 33.8 = 100 g 125g 42.3 =

MolarityMolarity, M

Moles of solute per liter of solution:

Molarity can be used as conversion factor between liters of solution and moles of solute.

Volume of solution x molarity = number of moles

MolarityExample:Calculate the molarity of a solution made by dissolving 15 g of

NaOH in water and diluting to 1.00 x 102 mL

Solution: First calculate the number of moles NaOH

NaOH mol38.0NaOH g 40.00

NaOH mol 1 x NaOH g 15

MolarityNext calculate the volume of solution in liters

1.00 x 102

Calculate the answer using the definition

L 100.0mL 1000

L 1x

NaOH M 3.8 solution L 0.100

NaOH mol 0.38 M

MolarityExample:• How many grams of silver nitrate must be dissolved to prepare

5.00 x102mL of 0.150 M AgNO3

• 5.00 x102mL x ( 1 L/ 1000 mL) x(0.150 mol AgNO3/L) x(169.9 g AgNO3/mol AgNO3)

= 12.7 g AgNO3

MolarityExampleFind the volume of a 1.40 M solution that contains 0.287 mole of

ammonia.Solution 

0.287 mole NH3 x ( 1L/ 1.40 mole NH3) = 0.205 L

MolarityTo prepare a solution of a specified molarity:

1. Weigh the appropriate amount of solute.2. Add less than the total volume of solvent.3. Mix to completely dissolve the solute.4. Add additional solvent until the total solution volume is

appropriate.

Preparation of 250 mL Solution KMnO4 0.0100M

250 mL x (1L/1000 mL) x (0.00100 mol KMnO4 /1L) x (158.04 g KMnO4/ 1 mol KMnO4) = 0.395 g KMnO4

MolalityMolality, m

The number of moles of solute dissolvedin one kilogram of solvent:

Molarity (mol/L) is temperature dependent;molality is temperature independent.

Normality: EquivalentEquivalent, eq

One equivalent of acid is the quantity that yieldsone mole of hydrogen ions in a chemical reaction.

Once equivalent of base is the quantity that reactswith one mole of hydrogen ions.

Normality: Equivalent eq /mol

H3PO4 + NaOH → NaH2PO4 + H2O 1

H3PO4 + 2 NaOH → Na2HPO4 + 2 H2O 2

H3PO4 + 3 NaOH → Na3PO4 + 3 H2O 3Number of equivalents of acid = Number of equivalents of base

Normality: Equivalent MassEquivalent mass: The number of grams per equivalent.

H3PO4 + NaOH → NaH2PO4 + H2O 97.99 g/eq

H3PO4 + 2 NaOH → Na2HPO4 + 2 H2O97.99 g/2 eq

H3PO4 + 3 NaOH → Na3PO4 + 3 H2O 97.99 g/3 eqEquivalent mass of H3PO4 for the three reactions are respectively: 97.99 g/eq, 49.00 g/eq, and 32.66 g/eq .

NormalityNormality, N

The number of equivalents, eq, per liter of solution:

NormalityExample:What is the normality of a solution made by dissolving 2.50

grams of sodium hydroxide in water to make 5.00 × 102 mL of solution?

Solution:By definition,

Therefore, we need to find equivalents of solute and liters of solution.

NormalityWhat is the normality of a solution made by dissolving 2.50

grams of sodium hydroxide in water to make 5.00 × 102 mL of solution?

Equivalents of solute:

GIVEN: 2.5 g NaOH WANTED: eq NaOH

NaOH eq 0.0625 = NaOH g 40.00

NaOH eq 1 NaOH g 2.5

NormalityWhat is the normality of a solution made by dissolving 2.50

grams of sodium hydroxide in water to make 5.00 × 102 mL of solution?

GIVEN: 5.00 × 102 mL WANTED: L

L 0.500 = mL 1000

L 1 mL 10 5.00 2

NormalityWhat is the normality of a solution made by dissolving 2.50

grams of sodium hydroxide in water to make 5.00 × 102 mL of solution?

We now have both parts of the normality fraction,0.0625 eq NaOH and 0.500 L solution:

NaOH N 0.125 = L 0.500NaOH eq 0.0625 =

Leq N

Solution Concentration: Summary

Dilution of Concentrated SolutionsConcentrated solutions are diluted by adding more solvent.

The amount of solute remains thesame before and after a dilution.

Mc x Vc = number of moles = Md x Vd

Mc x Vc = Md x Vd

n = mol = L L

mol = V M

moles ofNumber = Volume Molarity

Dilution of Concentrated SolutionsExample: How many milliliters of concentrated hydrochloric acid,

which is 11.6 M, should be used to prepare 5.50 liters of 0.500 M HCl ?

  Vc x Mc = Vd x Md

 Vc x 11.6 M = 5.50 L x 0.500 M

Vc = 5.50 L x 0.500 M /11.6 M = 0.237 L = 237 mL

Dilution of Concentrated SolutionsIf the volume of solution is increased 10 times, the concentration of diluted solution is 10 times smaller.

Solution StoichiometryHow many grams of lead(II) iodide will precipitate when excess potassium iodide solution is added to 50.0 mL of 0.811 M lead(II) nitrate.

Pb(NO3)2 (aq) + 2 KI(aq) PbI2(s) + 2 KNO3(aq)

Given: 50.0 mL of 0.811 M lead(II) nitrateWanted: gram of lead(II) iodide.

Need to convert volume of solution to moles of solute

Solution StoichiometryHow many grams of lead(II) iodide will precipitate when excess potassium iodide solution is added to 50.0 mL of 0.811 M lead(II) nitrate.

Pb(NO3)2 (aq) + 2 KI(aq) PbI2(s) + 2 KNO3(aq)

mL of Pb(NO3)2 liters of Pb(NO3)2moles of Pb(NO3)2

moles of PbI2gram of PbI2

50.0 mL of Pb(NO3)2 x (L of Pb(NO3)2/ 1000 mLof Pb(NO3)2)x (0.811 mol of Pb(NO3)2/L

Pb(NO3)2)x (1 mol PbI2 /1 mol Pb(NO3)2)x (461.0 g PbI2 /mol PbI2) = 18.7 g PbI2

Titration Using MolarityTitration: the very careful addition of one solution to

another by a burette.

Titration Using MolarityStandardize

Determination of the concentration of a solution to be used in a titration by titrating it against a primary standard.

Primary StandardA soluble solid of reasonable cost that is very stable and pure,

preferably with a high molar mass, that can be weighed accurately for use in a titration.

Titration Using Molarity1.18 g oxalic acid, H2C2O4. 2H2O (126.07 g/mol), are dissolved in water and the solution is titrated with a solution of NaOH of unknown concentration. 28.3 ml NaOH are required to neutralize the acid. Calculate the molarily of the NaOH solution.

  H2C2O4 + 2 NaOH → Na2C2O4 + 2 H2O

Given 1.18 g H2C2O4. 2H2O, 28.3 mL (0.0283L) NaOHWanted mol/L NaOH

g H2C2O4. 2H2O mol H2C2O4. 2H2Omol H2C2O4 mol NaOH molarity NaOH

Titration Using Molarity1.18 g oxalic acid, H2C2O4. 2H2O (126.07 g/mol), are dissolved in water and the solution is titrated with a solution of NaOH of unknown concentration. 28.3 ml NaOH are required to neutralize the acid. Calculate the molarily of the NaOH solution.

 H2C2O4 + 2 NaOH → Na2C2O4 + 2 H2O

1.18 g H2C2O4. 2H2O x (1 mol H2C2O4. 2H2O/ 126.07g H2C2O4. 2H2O )

x (1 mol H2C2O4. / 1 mol H2C2O4. 2H2O) x ( 2 mol NaOH/1 mol H2C2O4.) x ( 1/ 0.0283L NaOH) = 0.661 mol/L

Titration Using NormalityThe number of equivalents of all species

in a reaction is the same.

For an acid–base reaction,

equivalents of acid = equivalents of base

Titration Using Normality

A 25.0 mL sample of an electroplating solution is analyzed for its sulfuric acid concentration. It takes 46.8 mL of the 0.661 N NaOH to neutralize the sample, find the normality of the acid.

Na x Va = Nb x Vb

Na x 25.0 mL = 46.8 mL x 0.661 N

Na = (46.8 mL / 25.0 mL ) x 0.661 N = 1.24 N H2SO4

Colligative Properties of SolutionsA pure solvent has distinct physical properties.

Introducing a solute into the solvent affects these properties.

In dilute solutions of certain solutes, the change insome of these properties is proportional

to the molal concentration of the solute particles.

Colligative PropertySolution property that is determined only by the number of solute

particles dissolved in a fixed quantity of solvent and not by the identity of the solute particles.

Colligative Properties of SolutionsBoiling Point Elevation

The boiling point of a solution is higherthan the boiling point of the pure solvent.

Example:The normal boiling point of water is 100.0°C.

The normal boiling point of a 1 msolution of sugar water is 100.5°C.

Colligative Properties of SolutionsFreezing Point Depression

The freezing point of a solution is lowerthan the freezing point of the pure solvent.

Example:The normal freezing point of water is 0.0°C.

The normal boiling point of a 1 msolution of sugar water is –1.9°C.

Colligative Properties of Solutions

Colligative Properties of SolutionsBoiling point elevation and freezing point depression

are colligative properties:

They depend on the number of soluteparticles but not their identity.

∆Tb = change in boiling temperature

∆Tf = change in freezing temperature

Colligative Properties of Solutions

∆Tb = Kb × m ∆Tf = Kf × m

Kb = molal boiling-point elevation constant

Kf = molal freezing-point depression constant

Colligative Properties of SolutionsMolal Boiling Point Elevation Constant Values

Substance Boiling Point (°C) Kb (°C/m)Benzene 80 2.5Carbon disulfide 46 2.4Carbon tetrachloride 77 5.0Water 100 0.52

Molal Freezing Point Depression Constant ValuesSubstance Freezing Point (°C) Kf (°C/m)Benzene 6 5.1Carbon disulfide –112 3.8Carbon tetrachloride – 23 30Water 0 1.86

Colligative Properties of SolutionsWhat is the freezing point of a solution made by adding 12.0 g of urea, CO(NH2)2 , to 2.50 × 102 grams of water?

2222

2222 )CO(NH mol 0.200 =

)CO(NH g 60.06)CO(NH mol 1 )CO(NH g 12.0

OH g 10 2.50)CO(NH mol 0.200 =

solvent kgsolute mol m

22

22

222

2 )CO(NH m 0.800 = OH kg

OH g 1000

Colligative Properties of Solutions

∆Tf = Kf x m = 1.86 °C/m x 0.800 m = 1.49 0C

Tf = 0°C – 1.49°C = –1.49 °C

What is the freezing point of a solution made by adding 12.0 g of urea, CO(NH2)2 , to 2.50 × 102 grams of water?

Determination of molar mass Calculate the Molar Mass of a Solute from

Freezing Point Depression or Boiling Point Elevation Data

1. Calculate molality from m = ∆Tf/Kf or m = ∆Tb/Kb. Express as mol solute/kg solvent.

2. Using molality as a conversion factor between moles of solute and kilograms of solvent, find the number of moles of solute.

3. Use the defining equation for molar mass, MM = g/mol, to calculate the molar mass of the solute.

Determination of molar mass from ∆Tb

Example. Kb of benzene is 2.5 0C/m. A solution of 15.2 g of unknown solute in 91.0 g benzene boils at a temperature 2.1 C higher than the boiling point of pure benzene. Calculate the molar mass of the solute.

 Given: ΔTb = 2.1 0C, Kb = 2.5 0C/m Need: molar massFirst, calculate molality m : ∆Tb = Kb × m

Determination of molar mass from ∆Tb

Example. Kb of benzene is 2.5 0C/m. A solution of 15.2 g of unknown solute in 91.0 g benzene boils at a temperature 2.1 C higher than the boiling point of pure benzene. Calculate the molar mass of the solute.

  ∆Tb = Kb × mm = ΔTb / Kb = 2.1 0C/(2.5 0C/m) = 0.84 mol/kg Then calculate moles of solute and molar massMoles of solute in 91.0 g of benzene = 0.84 mol/kg x

( 1kg/1000g) x 91.0 g = 0.077 mol Molar mass = g/ mol = 15.2 g/ 0.077 mol

= 197.40 g/mol = 200 g/mol

Homework• Homework: 33, 43, 45, 49, 61, 71, 85, 89, 93, 103, 111, 121,

125, 127, 132.