Chapter 7 SolutionsChapter 7 Solutions

7.1 Solutions

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Solute and SolventSolute and Solvent

Solutions

• are homogeneous mixtures of two or more substances.

• consist of a solvent and one or more solutes.

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Nature of Solutes in SolutionsNature of Solutes in Solutions

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.

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Examples of SolutionsExamples of Solutions

The solute and solvent in a solution can be a solid, liquid, and/or a gas.

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ExampleExample

Identify the solute in each of the following solutions.

A. 2 g of sugar and 100 mL of water

B. 60.0 mL of ethyl alcohol and 30.0 mL of methyl alcohol

C. 55.0 mL of water and 1.50 g of NaCl

D. Air: 200 mL of O2 and 800 mL of N2

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WaterWaterWater• 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.

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Formation of a SolutionFormation of a Solution

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

crystal are attracted to polar water molecules.

• are hydrated in solution with many H2O molecules surrounding each ion.

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

H2O

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

solid separation of ions

ExampleExample

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.

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Like Dissolves LikeLike Dissolves LikeTwo 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.

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Like Dissolves LikeLike Dissolves Like

Solvents Solutes

Water (polar)

Ni(NO3)2

CH2Cl2(nonpolar) (polar)

I2 (nonpolar)

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ExampleExample

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

1) Na2SO4

2) gasoline (nonpolar)

3) I2

4) HCl

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7.27.2 Electrolytes and NonelectrolytesElectrolytes and Nonelectrolytes

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In water, • strong electrolytes produce ions and conduct an

electric current. • weak electrolytes produce a few ions. • nonelectrolytes do not produce ions.

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7.27.2 Electrolytes and Electrolytes and NonelectrolytesNonelectrolytes

Strong ElectrolytesStrong Electrolytes

Strong electrolytes • dissociate in water, producing positive and negative

ions.• conduct 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)

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ExamplesExamples

Complete each for strong electrolytes in water. H2O

A. CaCl2(s)

1) CaCl2(s)

2) Ca2+(aq) + Cl2−(aq)

3) Ca2+(aq) + 2Cl−(aq)

H2O

B. K3PO4(s)

1) 3K+(aq) + PO43−(aq)

2) K3PO4(s)

3) K3+(aq) + P3−(aq) + O4

−(aq)

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Weak ElectrolytesWeak Electrolytes

A weak electrolyte• dissociates only slightly in water.• in water forms a solution of a few ions and mostly

undissociated molecules.

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

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

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NonelectrolytesNonelectrolytes

Nonelectrolytes • dissolve as molecules in

water. • do not produce ions in

water.• do not conduct an

electric current.

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ExamplesExamples Classify each of the solute represented in the following

equations as a strong, weak, or nonelectrolyte

a. NH4OH(aq) NH4+(aq) + -OH(aq)

b.HCl (aq) H+(aq) + Cl-(aq)

EquivalentsEquivalents

An equivalent (Eq) is the amount of an electrolyte or an ion that provides 1 mole of electrical charge (+ or -).

1 mole of Na+ = 1 equivalent

1 mole of Cl− = 1 equivalent

1 mole of Ca2+ = 2 equivalents

1 mole of Fe3+ = 3 equivalents

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Electrolytes in Body FluidsElectrolytes in Body Fluids

In replacement solutions for body fluids, the electrolytes

are given in milliequivalents per liter (mEq/L).

Ringer’s Solution

Na+ 147 mEq/L

Cl− 155 mEq/L

K+ 4 mEq/L

Ca2+ 4 mEq/L

The milliequivalents per liter of cations must equal the milliequivalents per liter of anions.

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ExamplesExamples

A. In 1 mole of Fe3+, there are

1) 1 Eq. 2) 2 Eq. 3) 3 Eq.

B. In 2.5 moles of SO42−, there are

1) 2.5 Eq. 2) 5.0 Eq. 3) 1.0 Eq.

C. An IV bottle contains NaCl. If the Na+ is

34 mEq/L, the Cl− is

1) 34 mEq/L. 2) 0 mEq/L. 3) 68 mEq/L.

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7.3 Solubility7.3 Solubility

Solubility is

• the maximum amount of solute that dissolves in a specific amount of solvent.

• expressed as grams of solute in 100 grams of solvent, usually water.

g of solute100 g water

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Unsaturated SolutionsUnsaturated Solutions

Unsaturated solutions

• contain less than the maximum amount of solute.

• can dissolve more solute.

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Saturated SolutionsSaturated Solutions

Saturated solutions

• contain the maximum amount of solute that can dissolve.

• have undissolved solute at the bottom of the container.

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ExamplesExamples

At 40 C, the solubility of KBr is 80 g/100 g of H2O.Identify the following solutions as either

1) saturated or 2) unsaturated. Explain.

A. 60 g KBr added to 100 g of water at 40 C.

B. 200 g KBr added to 200 g of water at 40 C.

C. 25 g KBr added to 50 g of water at 40 C.

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ExamplesExamples The solubility of NaNO3 in water at 50oC is 110/100mL.

In a laboratory, a student use 50.0 g of NaNO3 with 200 g of water at the same temperature

◦ How many grams of NaNO3 will dissovle?

◦ Is the solution saturated or unsaturated?

◦ What is the mass, in grams, of any solid NaNO3 on the bottom of the container?

Effect of Temperature on Effect of Temperature on SolubilitySolubility

Solubility• depends on temperature.• of most solids increases as temperature increases.

• Hot tea dissolves more sugar than does cold tea because the solubility of sugar is much greater in higher temperature

• When a saturated solution is carefully cooled, it becomes a supersaturated solution because it contains more solute than the solubility allows.

• of gases decreases as temperature increases.• At higher temperature, more gas molecules have the energy to

escape from the solution

• Leaving higher pressure inside the container

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Solubility and PressureSolubility and Pressure

Henry’s law states • the solubility of a gas

in a liquid is directly related to the pressure of that gas above the liquid.

• at higher pressures, more gas molecules dissolve in the liquid.

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A. Why could a bottle of carbonated drink possibly burst (explode) when it is left out in the hot sun?

B. Why do fish die in water that is too warm?

ExamplesExamples

Soluble and Insoluble SaltsSoluble and Insoluble Salts Ionic compounds that dissolve in water => soluble

salts

Ionic compounds that do not separate into ions in water => insoluble salt

How do we know which ionic compounds is soluble or insoluble?

Precipitation Reactions and Precipitation Reactions and Solubility GuidelinesSolubility Guidelines

ExamplesExamplesPredict whether the following

ionic salt is soluble in water◦FeCl3

◦Na2CO3

◦AgCl

◦BaSO4

7.4 Percent Concentration7.4 Percent Concentration

The concentration of a solution

is the amount of solute dissolved in a specific amount of solution.

amount of soluteamount of solution

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Mass PercentMass Percent

Mass percent (% m/m) is the• concentration by mass of solute in a solution.

mass percent = g of solute x 100 g of solute + g of solvent

• amount in g of solute in 100 g of solution.

mass percent = g of solute a

100 g of solution

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Calculating Mass PercentCalculating Mass PercentThe calculation of mass percent (% m/m) requires the• grams of solute (g KCl) and• grams of solution (g KCl solution).

g of KCl = 8.00 gg of solvent (water) = 42.00 gg of KCl solution = 50.00 g

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ExampleExample

A solution is prepared by mixing 15.0 g of Na2CO3 and 235 g of H2O. Calculate the mass percent (% m/m) of the solution.

1) 15.0% (m/m) Na2CO3

2) 6.38% (m/m) Na2CO3

3) 6.00% (m/m) Na2CO3

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Volume PercentVolume PercentThe 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

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Mass/Volume PercentMass/Volume Percent

The mass/volume percent (% m/v) is

• percent mass (g) of solute to volume (mL) of solution.

mass/volume % (m/v) = g of solute x 100 mL of

solution

• solute (g) in 100 mL of solution.

mass/volume % (m/v) = g of solute

100 mL of solution

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Percent Conversion FactorsPercent Conversion FactorsTwo conversion factors can be written for each type of % value.

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Copyright © 2009 by Pearson Education, Inc.

Using Percent Concentration Using Percent Concentration (m/m) (m/m) as Conversion Factorsas Conversion Factors

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 NaCl

STEP 2: g solution g NaCl

STEP 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.

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ExampleExample

How many milliliters of a 5.75% (v/v) ethanol solution can be prepared from 2.25 mL of ethanol?

1) 2.56 mL

2) 12.9 mL

3) 39.1 mL

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Using Percent Concentration (m/v) Using Percent Concentration (m/v)

as Conversion Factorsas Conversion FactorsHow many mL of a 4.20% (m/v) will contain 3.15 g of KCl?

STEP 1: Given: 3.15 g of KCl(solute); 4.20% (m/v) KCl Need: mL of KCl solution

STEP 2: Plan: g of KCl mL of KCl solution

STEP 3: Write conversion factors. 4.20 g KCl and 100 mL solution

100 mL solution 4.20 g KCl

STEP 4: Set up the problem

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ExampleExampleHow many grams of NaOH are needed to prepare 125 mL

of a 8.80% (m/v) NaOH solution?

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ExampleExample What is the % m/v of NaOH in a solution prepared by

dissovling 12.0 g of NaOH in enough water to make 220.0 mL solution?