Chapter 9 Solutions

1Copyright © 2004 Pearson Education Inc., publishing as Benjamin Cummings.

9.1 Properties of Water

9.2 Solutions

9.3 Electrolytes and Nonelectrolytes

9.6 Percent Concentration

9.7 Molarity

Chapter 9 Solutions


Water is the most common solvent. The water molecule is polar. Hydrogen bonds form between the hydrogen

atom in one molecule and the oxygen atom in a different water molecule.

9.1 Water


Water


Water for the body is obtained from fluids as well as foods.

Some foods have a high percentage of water.

Water in Foods


Water molecules At the surface form hydrogen bonds with

molecules on or below the surface, which pulls them closer.

At the surface behave like a thin, elastic membrane, or “skin.”

Cannot hydrogen bond when compounds called surfactants are added.

Surface Tension


Solutions Are homogeneous

mixtures of two or more substances.

Consist of a solvent and one or more solutes.

9.2 Solutions: Solute and Solvent


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


Examples of Solutions

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


Identify the solute and the solvent in each.

A. brass: 20 g zinc + 50 g copper solute = 1) zinc 2) copper solvent = 1) zinc 2) copper

B. 100 g H2O + 5 g KCl

solute = 1) KCl 2) H2O

solvent = 1) KCl 2) H2O

Learning Check


Identify the solute and the solvent in each.A. brass: 20 g zinc + 50 g copper

solute = 1) zinc solvent = 2) copper

B. 100 g H2O + 5 g KCl solute = 1) KCl solvent = 2) H2O

Solution


Identify the solute in each of the following solutions:

A. 2 g sugar (1) and 100 mL water (2)B. 60.0 mL of ethyl alcohol (1) and 30.0 mL of

methyl alcohol (2)

C. 55.0 mL water (1) and 1.50 g NaCl (2)

D. Air: 200 mL O2 (1) and 800 mL N2 (2)

Learning Check


Identify the solute in each of the following solutions:

A. 2 g sugar (1)

B. 30.0 mL of methyl alcohol (2)

C. 1.5 g NaCl (2)

D. 200 mL O2 (1)

Solution


A solution forms when there is an attraction between the particles of the solute and solvent.

A polar solvent such as water dissolves polar solutes such as sugar and ionic solutes such as NaCl.

A nonpolar solvent such as hexane (C6H14)

dissolves nonpolar solutes such as oil or grease.

Like Dissolves Like


Examples of Like Dissolves Like

Solvents Solutes

Water (polar) Ni(NO3)2

(ionic)

CH2Cl2 (nonpolar)

I2 (nonpolar)


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

1) Na2SO4

2) gasoline (nonpolar)

3) I2

4) HCl

Learning Check


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

1) Na2SO4 Yes, ionic

2) gasoline No, nonnpolar

3) I2 No, nonpolar

4) HCl Yes, polar

Most polar and ionic solutes dissolve in water

because water is a polar solvent.

Solution


Formation of a Solution Na+ and Cl- ions on

the surface of a NaCl crystal are attracted to polar water molecules.

In solution, the ions 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 becauseA. 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


Solid LiCl is added to water. It dissolves because

A. The Li+ ions are attracted to the

1) oxygen atom ( –) of water.

B. The Cl- ions are attracted to the

2) hydrogen atom ( +) of water.

Solution


Hydrates Hydrates are solid

compounds that contain water molecules as part of the crystal structure.

Heating a hydrate releases the water of hydration to give the anhydrate salt.


Examples of Hydrates


Learning Check

Write the equation for the dehydration of AlCl3 • 6H2O.


Solution

Write the equation for the dehydration of AlCl3 • 6H2O.

AlCl3 • 6H2O AlCl3 + 6H2O


Electrolytes Produce positive (+) and negative (-) ions

when they dissolve in water. In water conduct an electric current.

9.3 Electrolytes


Strong electrolytes ionize 100% in solution. Equations for the dissociation of strong

electrolytes 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


Complete each of the following dissociation equations for strong electrolytes dissolved in water: H2O

A. CaCl2 (s) 1) CaCl2

2) Ca2+ + Cl2-

3) Ca2+ + 2Cl-

H2O

B. K3PO4 (s) 1) 3K+ + PO43-

2) K3PO4

3) K3+ + P3- + O4

-

Learning Check


Complete each of the following dissociation equations for strong electrolytes dissolved in water:

H2O

A. 3) CaCl2 (s) Ca2+ + 2Cl-

H2O

B. 1) K3PO4 (s) 3K+ + PO43-

Solution


A weak electrolyte Dissolves mostly as molecules in solution. Produces only a few ions in aqueous

solutions. Has an equilibrium that favors the

reactants.HF + H2O H3O+(aq) + F- (aq)

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

Weak Electrolytes


Nonelectrolytes Form only

molecules in water. Do not produce ions

in water. Do not conduct an

electric current.

Nonelectrolytes


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

amount of soluteamount of solution

The percent concentration describes the amount of solute that is dissolved in 100 parts of solution.

amount of solute100 parts solution

9.6 Percent Concentration


The mass percent (%m/m) Concentration is the percent by mass of solute in a

solution.mass percent = g of solute x 100%

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

mass percent = g of solute 100 g of solution

Mass Percent


grams of solute + grams of solvent

50.0 g KCl solution

Mass of Solution


Mass percent (%m/m) is calculated from the grams of solute (g KCl) and the grams of solution (g KCl solution).g of KCl = 8.0 gg of solvent (water) = 42.0 gg of KCl solution = 50.0 g

8.0 g KCl (solute) x 100 = 16% (m/m) KCl 50.0 g KCl solution

Calculating Mass Percent


A solution is prepared by mixing 15 g Na2CO3 and 235 g of H2O. Calculate the mass percent (%m/m) of the solution.

1) 15% (m/m) Na2CO3

2) 6.4% (m/m) Na2CO3

3) 6.0% (m/m) Na2CO3

Learning Check


3) 6.0% (m/m) Na2CO3

mass solute = 15 g Na2CO3

mass solution = 15 g + 235 g = 250 g

mass %(m/m) = 15 g Na2CO3 x 100

250 g solution

= 6.0% Na2CO3 solution

Solution


The mass/volume percent (%m/v) Concentration is the ratio of the mass in grams (g)

of solute in a volume (mL) of solution.

mass/volume % = g of solute x 100% mL of solution

Is the g of solute in 100 mL of solution.

mass/volume % = g of solute 100 mL of solution

Mass/Volume Percent


Preparing a Solution with a Mass/Volume % Concentration

A percent mass/volume solution is prepared by weighing out the grams of solute (g) and adding water to give the final volume of the solution.


Calculation of Mass/Volume Percent

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

g of KI = 5.0 g KI

mL of KI solution = 250.0 mL

5.0 g KI (solute) x 100 = 2.0%(m/v) KI

250.0 mL KI solution


A 500. mL samples of an IV glucose solution contains 25 g glucose (C6H12O6) in water.

What is the mass/volume % (%m/v) of glucose of the IV solution?1) 5.0% 2) 20.% 3) 50.%

Learning Check


1) 5.0% Mass/volume %(m/v)

= 25 g glucose x 100

500. mL solution

= 5.0 % (m/v) glucose solution

Solution


The volume percent (%v/v) Concentration is the percent volume (mL) of

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

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

Is the mL of solute in 100 mL of solution.

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

Volume Percent


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

Percent Conversion Factors


Write two conversion factors for each solutions:

A. 8%(m/v) NaOH

B. 12%(v/v) ethyl alcohol

Learning Check


A. 8%(m/v) NaOH

8 g NaOH and 100 mL solution 100 mL solution 8 g NaOH

B. 12%(v/v) ethyl alcohol

12 mL alcohol and 100 mL solution

100 mL solution 12 mL alcohol

Solution


How many grams of NaCl are needed to prepare250 g of a 10.0% (m/m) NaCl solution?

1. Write the 10.0 % (m/m) as conversion factors.

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

2. Use the factor that cancels given (g solution).

250 g solution x 10.0 g NaCl = 25 g NaCl 100 g solution

Using Percent Factors


How many grams of NaOH are needed to prepare 2.0 L of a 12%(m/v) NaOH solution?1) 24 g NaOH

2) 240 g NaOH

3) 2400 g NaOH

Learning Check


2) 240 g NaOH

2.0 L x 1000 mL = 2000 mL 1 L

2000 mL x 12 g NaOH = 240 g NaOH 100 mL

12 % (m/v) factor

Solution


How many milliliters of 5% (m/v) glucose solution are given if a patient receives 150 g of glucose?

1) 30 mL

2) 3000 mL

3) 7500 mL

Learning Check


2) 3000 mL

150 g glucose x 100 mL = 3000 mL 5 g glucose 5% m/v factor (inverted)

Solution


Molarity is a concentration unit for the moles of solute in the liters (L) of solution.

Molarity (M) = moles of solute = moles liter of solution L

Examples:2.0 M HCl = 2.0 moles HCl

1 L 6.0 M HCl = 6.0 moles HCl

1 L

9.7 Molarity (M)


Preparing a 1.0 Molar Solution A 1.0 M NaCl solution is prepared by weighing out

58.5 g NaCl ( 1.0 mole) and adding water to make 1.0 liter of solution.


What is the molarity of a NaOH solution prepared by adding 4.0 g of solid NaOH to water to make 0.50 L of solution ?

1. Determine the moles of solute. 4.0 g NaOH x 1 mole NaOH = 0.10 mole 40.0 g NaOH

2. Calculate molarity. 0.10 mole = 0.20 mole = 0.20 M NaOH

0.50 L 1 L

Calculation of Molarity


Calculate the molarity of an NaHCO3 solution prepared by dissolving 36 g of solid NaHCO3 in water to give a solution volume of 240 mL.

1) 0.43 M

2) 1.8 M

3) 15 M

Learning Check


2) 1.8 M

36 g x 1 mole NaHCO3 = 0.43 mole NaHCO3

84 g

0.43 mole NaHCO3 = 1.8 M NaHCO3

0.240 L

Solution


A glucose solution with a volume of 2.0 L contains 72 g glucose (C6H12O6). If glucose has a molar mass of 180. g/mole, what is the molarity of the glucose solution?1) 0.20 M2) 5.0 M3) 36 M

Learning Check


1) 0.20 M

72 g x 1 mole x 1 = 0.20 moles

180. g 2.0 L 1 L

= 0.20 M

Solution


The units in molarity can be used to write conversion factors.

Molarity Conversion Factors


Stomach acid is 0.10 M HCl solution. How many moles of HCl are present in 1500 mL of stomach acid?

1) 15 moles HCl

2) 1.5 moles HCl

3) 0.15 mole HCl

Learning Check


3) 0.15 mole HCl 1500 mL x 1 L = 1.5 L

1000 mL

1.5 L x 0.10 mole HCl = 0.15 mole HCl 1 L

Molarity factor

Solution


Calculate the grams of KCl that must be dissolved in water to prepare 0.25 L of a 2.0 M KCl solution.

1) 150 g KCl

2) 37 g KCl

3) 19 g KCl

Learning Check


3) 37 g KCl

Determine the number of moles of KCl.0.25 L x 2.0 mole KCl = 0.50 moles KCl

1 L Convert the moles to grams of KCl.0.50 moles KCl x 74.6 g KCl = 37 g KCl

1 mole KCl

molar mass of KCl

Solution


How many milliliters of 6.0 M HNO3 contain

0.15 mole of HNO3?

1) 25 mL

2) 90 mL

3) 400 mL

Learning Check


1) 25 mL

0.15 mole HNO3 x 1 L x 1000 mL

6.0 moles HNO3 1 L

Molarity factor inverted

= 25 mL HNO3

Solution

Chapter 9 Solutions

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