Chapter 11 Solutions

Copyright © 2005 by Pearson Education, Inc.Publishing as Benjamin Cummings

Solute and SolventSolutions• Are homogeneous mixtures of two or more substances.• Consist of a solvent and one or more solutes.

Solutes• Spread evenly

throughout the solution.

• Cannot be separated by filtration.

• Can be separated by evaporation.

• Are not visible, but can give a color to the solution.

Nature of Solutes in Solutions

Copyright © 2005 by Pearson Education, Inc.Publishing as Benjamin Cummings

Examples of Solutions

WaterWater• Is the most common solvent.• Is a polar molecule.• Forms hydrogen bonds between the hydrogen atom in one molecule and the oxygen atom in a different water molecule.

Copyright © 2005 by Pearson Education, Inc.Publishing as Benjamin Cummings

(Intermolecular Forces)

Formation of a Solution

Na+ and Cl- ions,• On the surface of a NaCl

crystal are attracted to polar water molecules.

• In solution are hydrated as several H2O molecules surround each.

When NaCl(s) dissolves in water, the reaction can be written as:

H2O

NaCl(s) Na+(aq) + Cl- (aq)

solid separation of ions

Equations for Solution Formation

Solid LiCl is added to water. It dissolves because:A. The Li+ ions are attracted to the 1) oxygen atom ( -) of water.

2) hydrogen atom (+) of water.

B. The Cl- ions are attracted to the 1) oxygen atom ( -) of water. 2) hydrogen atom (+) of water.

Learning Check

Two substances form a solution:

• When there is an attraction between the particles of the solute and solvent.

• When a polar solvent, such as water, dissolves polar solutes such as sugar and ionic solutes such as NaCl.

• When a nonpolar solvent, such as hexane (C6H14),

dissolves nonpolar solutes such as oil or grease.

Like Dissolves Like

Which of the following solutes will dissolve in water? Why?

1) Na2SO4

2) gasoline (nonpolar)

3) I2

4) HCl

Learning Check

In water, • Strong electrolytes produce ions and conduct an

electric current. • Weak electrolytes produce a few ions. • Nonelectrolytes do not produce ions.

Solutes and Ionic Charge

Copyright © 2005 by Pearson Education, Inc.Publishing as Benjamin Cummings

Strong electrolytes, • Dissociate in water producing positive and negative ions.• Produce an electric current in water.• In equations show the formation of ions in aqueous (aq)

solutions.

H2O 100% ions

NaCl(s) Na+(aq) + Cl− (aq)

H2O

CaBr2(s) Ca2+(aq) + 2Br− (aq)

Strong Electrolytes

A weak electrolyte,• Dissociates only slightly in water.• In water forms a solution of only a few ions and

mostly undissociated molecules.

HF(g) + H2O(l) H3O+(aq) + F- (aq)

NH3(g) + H2O(l) NH4+(aq) + OH- (aq)

Weak Electrolytes

Solubility • Is the maximum amount of solute that dissolves in a

specific amount of solvent. • Can be expressed as grams of solute in 100 grams of

solvent, usually water.

g of solute

100 g water

Solubility

Unsaturated solutions • Contain less than the maximum

amount of solute. • Can dissolve more solute.

Saturated solutions • Contain the maximum amount of solute that can dissolve. • Have undissolved solute at the bottom of the container.

Soluble and Insoluble SaltsIonic compounds that• Dissolve in water are

soluble salts.• Do NOT dissolve in water

are insoluble salts.

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Equations for Forming SolidsA molecular equation shows the formulas of the

compounds.

Pb(NO3)(aq) + 2NaCl(aq) PbCl2(s) + 2NaNO3(aq)

An ionic equation shows the ions of the compounds.

Pb2+(aq) + 2NO3−(aq) + 2Na+(aq) + 2Cl−(aq)

PbCl2(s) + 2Na+(aq) + 2NO3−(aq)

NOTE: the “spectator ions”

A net ionic equation shows only the ions that form a solid. Pb2+(aq) + 2Cl−(aq) PbCl2(s)

Learning Check A precipitate forms in the following reaction.

Write the molecular, ionic and net ionic equations for the reaction.

BaCl2(aq) + Na2SO4(aq)

The mass percent (%m/m)

• Concentration is the percent by mass of solute in a solution.

mass percent (%m/m)

= g of solute x 100

g of solute + g of solvent

• Is the g of solute in 100 g of solution.

mass percent = g of solute x 100

100 g of solution

Mass Percent

Mass percent (%m/m) is calculated from the grams of solute (g KCl) and the grams of solution (g KCl solution).

g of KCl = 8.00 g

g of solvent (water) = 42.00 g

g of KCl solution = 50.00 g

8.00 g KCl (solute) x 100 = 16.0% (m/m)

50.00 g KCl solution

Calculating Mass Percent

The volume percent (%v/v) is:

• Percent volume (mL) of solute (liquid) to volume (mL) of solution.

volume % (v/v) = mL of solute x 100 mL of solution

• Solute (mL) in 100 mL of solution.

volume % (v/v) = mL of solute 100 mL of solution

Volume Percent

Percent Conversion Factors

• Two conversion factors can be written for each type of % value.

Write two conversion factors for each solutions:

A. 8.50%(m/m) NaOH

B. 5.75%(v/v) ethanol

Learning Check

How many grams of NaCl are needed to prepare225 g of a 10.0% (m/m) NaCl solution?STEP 1 Given: 225 g solution; 10.0% (m/m) NaCl

Need: g of NaClSTEP 2 g solution g NaClSTEP 3 Write the 10.0% (m/m) as conversion factors.

10.0 g NaCl and 100 g solution 100 g solution 10.0 g NaCl

STEP 4 Set up using the factor that cancels g solution.225 g solution x 10.0 g NaCl = 22.5 g NaCl

100 g solution

Using Percent Factors

Molarity (M)

Molarity (M) is:

• A concentration term for solutions.

• Gives the moles of solute in 1 L solution.

• moles of soluteliter of solution

Preparing a 1.0 Molar Solution

A 1.00 M NaCl solution is prepared:• By weighing out 58.5 g NaCl (1.00 mol) and• Adding water to make 1.00 liter of solution.

What is the molarity of 325 mL of a solution containing 46.8 g of NaHCO3?

1) 0.557 M

2) 1.44 M

3) 1.71 M

Learning Check

Molarity Conversion FactorsThe units of molarity are used as conversion factors in calculations with solutions.

How many grams of AlCl3 are needed to prepare

125 mL of a 0.150 M solution?

1) 20.0 g AlCl3

2) 16.7g AlCl3

3) 2.50 g AlCl3

Learning Check

How many milliliters of 2.00 M HNO3 contain 24.0 g HNO3?

1) 12.0 mL

2) 83.3 mL

3) 190. mL

Learning Check

DilutionIn a dilution,

• Water is added

• Volume increases

• Concentration decreases

M1V1 = M2V2

initial diluted

Learning Check

What is the final volume (mL) of 15.0 mL of a 1.80 M

KOH diluted to give a 0.300 M solution?

1) 27.0 mL

2) 60.0 mL

3) 90.0 mL

Using Molarity of Reactants

How many mL of 3.00 M HCl are needed to completely

react with 4.85 g CaCO3?

2HCl(aq) + CaCO3(s) CaCl2(aq) + CO2(g) + H2O(l)

STEP 1 Given: 3.00 M HCl; 4.85 g CaCO3

Need: volume in mL

STEP 2 Plan:

g CaCO3 mol CaCO3 mol HCl mL HCl

Using Molarity of Reactants (cont.)

2HCl(aq) + CaCO3(s) CaCl2(aq) + CO2(g) + H2O(l)

STEP 3 Equalities

1 mol CaCO3 = 100.09 g; 1 mol CaCO3 = 2 mol HCl

1000 mL HCl = 3.00 mol HCl

STEP 4 Setup

4.85 g CaCO3 x 1 mol CaCO3 x 2 mol HCl x 1000 mL HCl

100.09 g CaCO3 1 mol CaCO3 3.00 mol HCl

= 32.3 mL HCl required

Learning Check

If 22.8 mL of 0.100 M MgCl2 is needed to completely

react 15.0 mL of AgNO3 solution, what is the molarity of

the AgNO3 solution?

MgCl2(aq) + 2AgNO3(aq) 2AgCl(s) + Mg(NO3)2(aq)

1) 0.0760 M

2) 0.152 M

3) 0.304 M