Chapter 7 Solutions

Chapter 7

Solutions

Solute and Solvent

7.1 Solutions

Solutions • are homogeneous mixtures of two or more substances. • consist of a solvent and one or more solutes.

Copyright © 2009 by Pearson Education, Inc.

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Nature of Solutes in Solutions

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Examples of 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.

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



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Learning Check

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Solution Identify the solute in each of the following solutions.

Identify the solute in each of the following solutions.

A. 2 g of sugar

A. 2 g of sugar and 100 mL of water

B. 30.0 mL of methyl alcohol

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

C. 1.5 g of NaCl D. 200 mL of O2

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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Water

Formation of a Solution

Water • 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.

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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Equations for Solution Formation

Learning Check Solid LiCl is added to water. It dissolves because

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

H2 O

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A. the Li+ ions are attracted to the 1) oxygen atom (δ -) of water. 2) hydrogen atom (δ+) of water.

Na+(aq) + Cl-(aq) separation of ions

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

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Solution

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Like Dissolves Like

Solid LiCl is added to water. It dissolves because

Two substances form a solution

Li+

A. the ions are attracted to the 1) oxygen atom (δ -) of water.

• 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.

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

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Water and a Polar Solute

Like Dissolves Like

Solvents

Solutes

Water (polar)

Ni(NO3)2 (polar)

CH2Cl2(nonpolar)

I2 (nonpolar)



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Learning Check

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Solution Which of the following solutes will dissolve in water? Why? yes, ionic 1) Na2SO4 2) gasoline no, nonpolar 3) I2 no, nonpolar 4) HCl yes, polar

Which of the following solutes will dissolve in water? Why? 1) Na2SO4 2) gasoline (nonpolar) 3) I2 4) HCl

Most polar and ionic solutes dissolve in water because water is a polar solvent.

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Chapter 7

Solutions

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Solutes and Ionic Charge In water, • strong electrolytes produce ions and conduct an electric current. • weak electrolytes produce a few ions. • nonelectrolytes do not produce ions.

7.2 Electrolytes and Nonelectrolytes

Copyright © 2009 by Pearson Education, Inc. Copyright © 2009 by Pearson Education, Inc.

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Strong Electrolytes

Learning Check Complete each for strong electrolytes in water.

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. 100% ions H2O NaCl(s) Na+(aq) + Cl−(aq) CaBr2(s)

H2O

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)

Ca2+(aq) + 2Br−(aq)

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Solution

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

Complete each for strong electrolytes in water: H2 O A. CaCl2(s) 3) Ca2+(aq) + 2Cl−(aq) H2 O B. K3PO4 (s) 1) 3K+(aq) + PO43−(aq)

HF(g) +

H2O(l)

NH3(g) + H2O(l)

H3O+(aq) + F-(aq) NH4+(aq) + OH-(aq)

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Nonelectrolytes

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Equivalents

Nonelectrolytes • dissolve as molecules in water. • do not produce ions in water. • do not conduct an electric current.

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 equivalent

1 mole of Ca2+

=

2 equivalents

Fe3+

=

3 equivalents

1 mole of

1 mole of

Cl−


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Electrolytes in Body Fluids

Electrolytes in Body Fluids In replacement solutions for body fluids, the electrolytes are given in milliequivalents per liter (mEq/L). Ringer’s Solution Na+ K+ Ca2+

147 mEq/L 4 mEq/L 4 mEq/L

Cl−

155 mEq/L

The milliequivalents per liter of cations must equal the milliequivalents per liter of anions. Copyright © 2009 by Pearson Education, Inc.

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Learning Check A.

B.

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Solution

In 1 mole of Fe3+, there are 1) 1 Eq. 2) 2 Eq.

3) 3 Eq.

In 2.5 moles of SO42−, there are 1) 2.5 Eq. 2) 5.0 Eq.

3) 1.0 Eq.

A. 3) 3 Eq. B. 2) 5.0 Eq. 2.5 mole SO42− x 2 Eq = 5.0 Eq 1 mole SO42− C. 1) 34 mEq/L

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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Chapter 7

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Solutions

Solubility

7.3 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 solute 100 g water


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Unsaturated Solutions

Saturated Solutions

Unsaturated solutions

Saturated solutions

• contain less than the maximum amount of solute.

• contain the maximum amount of solute that can dissolve.

• can dissolve more solute.

• have undissolved solute at the bottom of the container.



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Learning Check

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Solution

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. 2

Amount of 60 g KBr/100 g of water is less than the solubility of 80 g KBr/100 g of water.

B. 1

In 100 g of water, 100 g of KBr exceeds the solubility of 80 g KBr /100 g of water at 40 °C.

C. 2

This is the same as 50 g KBr in 100 g of water, which is less than the solubility of 80 g KBr/ 100 g of water at 40 °C.

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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Effect of Temperature on Solubility

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Learning Check

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

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?


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Solution

Solubility 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.

A. The pressure in a bottle increases as the gas leaves solution when it becomes less soluble at higher temperatures. As pressure increases, the bottle could burst. B. Because O2 gas is less soluble in warm water, fish cannot obtain the amount of O2 required for their survival.


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Chapter 7

Solutions

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Percent Concentration The concentration of a solution

7.4 Percent Concentration

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


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Mass Percent

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Mass of Solution

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

x 100 g of solute g of solute + g of solvent

Add water to give 50.00 g of solution

8.00 g KCl

• amount in g of solute in 100 g of solution. mass percent = g of solute a 100 g of solution

50.00 g of KCl solution Copyright © 2009 by Pearson Education, Inc.

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Guide to Calculating Solution Concentrations

Calculating Mass Percent The 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 g g of solvent (water) = 42.00 g g of KCl solution = 50.00 g 8.00 g KCl (solute) 50.00 g KCl solution

x 100 = 16.0% (m/m)


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Solution

Learning Check

3) 6.00% (m/m) Na2CO3 STEP 1: mass solute = 15.0 g of Na2CO3 mass solution = 15.0 g + 235 g = 250. g STEP 2: Use g solute/g solution ratio

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

STEP 3: mass %(m/m) = g solute x 100 g solution STEP 4: Set up problem mass %(m/m) = 15.0 g Na2CO3 x 100 = 6.00% Na2CO3 250. g solution

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Volume Percent

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Mass/Volume Percent

The volume percent (% v/v) is

The mass/volume percent (% m/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

• 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 Factors

Learning Check

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 C. 4.8% (m/v) HCl


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Using Percent Concentration (m/m) as Conversion Factors

Solution A. 8.50 g NaOH 100 g solution

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and

How many grams of NaCl are needed to prepare 225 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. and 100 g solution 10.0 g NaCl 100 g solution 10.0 g NaCl STEP 4: Set up using the factor that cancels g solution. = 22.5 g NaCl 225 g solution x 10.0 g NaCl 100 g solution

100 g solution 8.50 g NaOH

B. 5.75 mL alcohol and 100 mL solution

100 mL solution 5.75 mL alcohol

and C. 4.8 g HCl 100 mL solution

100 mL HCl 4.8 g HCl

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Learning Check

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Solution

How many grams of NaOH are needed to prepare 75.0 g of 14.0% (m/m) NaOH solution?

1) 10.5 g NaOH 75.0 g solution x 14.0 g NaOH = 10.5 g NaOH 100 g solution

1) 10.5 g of NaOH 2) 75.0 g of NaOH 3) 536 g of NaOH

14.0% (m/m) factor

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Learning Check

Solution

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

3) 39.1 mL 2.25 mL ethanol x 100 mL solution 5.75 mL ethanol

1) 2.56 mL 2) 12.9 mL 3) 39.1 mL

5.75% (v/v) inverted

= 39.1 mL of solution

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Using Percent Concentration (m/v) as Conversion Factors

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Learning Check

How 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. and 100 mL solution 4.20 g KCl 100 mL solution 4.20 g KCl STEP 4: Set up the problem 3.15 g KCl x 100 mL KCl solution = 75.0 mL of KCl 4.20 g KCl

How many grams of NaOH are needed to prepare 125 mL of a 8.80% (m/v) NaOH solution?

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Solution

Chapter 7

How many grams of NaOH are needed to prepare 125 mL of a 8.80% (m/v) NaOH solution? 125 mL solution x 8.80 g NaOH = 100 mL solution

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Solutions 7.5 Molarity and Dilution

11.0 g of NaOH


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Molarity (M)

Preparing a 1.0 Molar Solution

Molarity (M)

A 1.00 M NaCl solution is prepared • by weighing out 58.5 g of NaCl (1.00 mole) and • adding water to make 1.00 liter of solution.

• is a concentration term for solutions. • gives the moles of solute in 1 L of solution. • moles of solute liter of solution


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Calculation of Molarity

Calculation of Molarity (continued)

What is the molarity of 0.500 L of NaOH solution if it contains 6.00 g of NaOH?

STEP 3: Conversion factors 1 mole of NaOH = 40.0 g of NaOH 1 mole NaOH and 40.0 g NaOH 40.0 g NaOH 1 mole NaOH

STEP 1: Given 6.00 g of NaOH in 0.500 L of solution Need molarity (mole/L) STEP 2: Plan g NaOH

mole NaOH

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STEP 4: Calculate molarity. 6.00 g NaOH x 1 mole NaOH = 0.150 mole 40.0 g NaOH 0.150 mole = 0.300 mole = 0.300 M NaOH 0.500 L 1L

molarity

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Learning Check

Solution

What is the molarity of 325 mL of a solution containing 46.8 g of NaHCO3? 1) 2) 3)

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3) 1.71 M 46.8 g NaHCO3 x 1 mole NaHCO3 = 0.557 mole NaHCO3 84.0 g NaHCO3

0.557 M 1.44 M 1.71 M

0.557 mole of NaHCO3 = 1.71 M NaHCO3 0.325 L

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Learning Check

Solution 2) 1.53 M 34.8 g KNO3 x 1 mole KNO3 = 0.344 mole of KNO3 101.1 g KNO3

What is the molarity of 225 mL of a KNO3 solution containing 34.8 g of KNO3? 1) 0.344 M 2) 1.53 M 3) 15.5 M

M = mole = 0.344 mole KNO3 = 1.53 M L 0.225 L In one setup: 34.8 g KNO3 x 1 mole KNO3 x 1 = 1.53 M 101.1 g KNO3 0.225 L

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Molarity Conversion Factors

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Calculations Using Molarity

The units of molarity are used as conversion factors in calculations with solutions.

How many grams of KCl are needed to prepare 125 mL of a 0.720 M KCl solution?

Molarity 3.5 M HCl

STEP 1: Given 125 mL (0.125 L) of 0.720 M KCl Need g of KCl

Equality 1 L = 3.5 moles of HCl

STEP 2: Plan

Written as Conversion Factors and 1L 3.5 moles HCl 1L 3.5 moles HCl

L KCl

moles KCl

g KCl

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Calculations Using Molarity

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Learning Check

STEP 3: Conversion factors 1 mole of KCl = 74.6 g 1 mole KCl and 74.6 g KCl 74.6 g KCl 1 mole KCl

How many grams of AlCl3 are needed to prepare 125 mL of a 0.150 M solution? 1)

1 L KCl = 0.720 mole of KCl and 0.720 mole KCl 1L 0.720 mole KCl 1L

20.0 g of AlCl3

2)

16.7 g of AlCl3

3)

2.50 g of AlCl3

STEP 4: Calculate grams. 0.125 L x 0.720 mole KCl x 74.6 g KCl = 6.71 g of KCl 1L 1 mole KCl 71

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Solution 3)

Learning Check

2.50 g AlCl3

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

0.125 L x 0.150 mole x 133.5 g = 2.50 g of AlCl3 1L 1 mole

1) 12.0 mL 2) 83.3 mL 3) 190. mL

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Dilution

Solution 24.0 g HNO3 x 1 mole HNO3 x 63.0 g HNO3

In a dilution • water is added. • volume increases. • concentration decreases.

1000 mL 2.00 mole HNO3

Molarity factor inverted

= 190. mL of HNO3


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Guide to Calculating Dilution Quantities

Comparing Initial and Diluted Solutions In the initial and diluted solution, • the moles of solute are the same. • the concentrations and volumes are related by the following equations: For percent concentration: C1V1 = C2V2 initial

diluted

For molarity: M1V1 = M2V2 initial

diluted Copyright © 2009 by Pearson Education, Inc.

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Dilution Calculations with Percent

Learning Check

What volume of a 2.00% (m/v) HCl solution can be prepared by diluting 25.0 mL of 14.0% (m/v) HCl solution? Prepare a table: C1= 14.0% (m/v) V1 = 25.0 mL V2 = ? C2= 2.00% (m/v) Solve dilution equation for unknown and enter values: C1V1 = C2V2 V2

= V1C1 C2

What is the percent (% m/v) of a solution prepared by diluting 10.0 mL of 9.00% NaOH to 60.0 mL?

= (25.0 mL)(14.0%) = 175 mL 2.00%

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Solution

Dilution Calculations with Molarity

What is the percent (% m/v) of a solution prepared by diluting 10.0 mL of 9.00% NaOH to 60.0 mL? Prepare a table: V1 = 10.0 mL C1= 9.00 %(m/v) V2 = 60.0 mL C2= ? Solve dilution equation for unknown and enter values: C1V1 = C2V2 C2

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What is the molarity (M) of a solution prepared by diluting 0.180 L of 0.600 M HNO3 to 0.540 L? Prepare a table: V1 = 0.180 L M1= 0.600 M V2 = 0.540 L M2= ? Solve dilution equation for unknown and enter values: M1V1 = M2V2

= C1 V1 = (10.0 mL)(9.00%) = 1.50% (m/v) 60.0 mL V2

M2

= M1V1 V2

= (0.600 M)(0.180 L) = 0.200 M 0.540 L

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Learning Check

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Solution What is the final volume (mL) of 15.0 mL of a 1.80 M KOH diluted to give a 0.300 M solution? Prepare a table: V1 = 15.0 mL M1= 1.80 M V2 = ? M2= 0.300M Solve dilution equation for V2 and enter values: M1V1 = M2V2

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

V2

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= M1V1 M2

= (1.80 M)(15.0 mL) = 90.0 mL 0.300 M

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Chapter 7

Solutions

Molarity in Chemical Reactions In a chemical reaction, • the volume and molarity of a solution are used to determine the moles of a reactant or product.

7.6 Solutions in Chemical Reactions

molarity ( mole ) x volume (L) = moles 1L • if molarity (mole/L) and moles are given, the volume (L) can be determined. = volume (L) moles x 1L moles


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Using Molarity of Reactants

Using Molarity of Reactants (cont.)

How many mL of 3.00 M HCl are needed to completely react with 4.85 g of CaCO3? 2 HCl(aq) + CaCO3(s)

2HCl(aq) + CaCO3(s) STEP 3: Equalities 1 mole of CaCO3 1 mole of CaCO3 1000 mL of HCl

CaCl2(aq) + CO2(g) + H2O(l)

STEP 1: Given 3.00 M HCl; 4.85 g of CaCO3 Need volume in mL STEP 2: Plan mole CaCO3 mole HCl g CaCO3

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CaCl2(aq) + CO2(g) + H2O(l)

= 100.1 g of CaCO3 = 2 moles of HCl = 3.00 moles of HCl

STEP 4: Set Up mL HCl

4.85 g CaCO3 x 1 mole CaCO3 x 2 mole HCl x 1000 mL HCl 100.1 g CaCO3 1 mole CaCO3 3.00 mole HCl = 32.3 mL of HCl required 87

Learning Check

Solution

How many mL of a 0.150 M Na2S solution are needed to completely react 18.5 mL of 0.225 M NiCl2 solution? NiCl2(aq) + Na2S(aq)

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3) 27.0 mL 0.0185 L x 0.225 mole NiCl2 x 1 mole Na2S x 1000 mL 1L 1 mole NiCl2 0.150 mole

NiS(s) + 2NaCl(aq)

= 27.0 mL of Na2S solution

1) 4.16 mL 2) 6.24 mL 3) 27.0 mL

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Learning Check

Solution

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)

3) 0.304 M AgNO3 0.0228 L x 0.100 mole MgCl2 x 2 mole AgNO3 x 1 P 1L 1 mole MgCl2 0.0150 L

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

= 0.304 mole/L = 0.304 M AgNO3

1) 0.0760 M 2) 0.152 M 3) 0.304 M

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Learning Check

Solution

How many liters of H2 gas at STP are produced when Zn reacts with 125 mL of 6.00 M HCl? Zn(s) + 2HCl(aq)

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2) 8.40 L H2 gas 0.125 L x 6.00 mole HCl 1L

ZnCl2 (aq) + H2(g)

1) 4.20 L of H2 2) 8.40 L of H2 3) 16.8 L of H2

x 1 mole H2 x 22.4 L 2 mole HCl 1 mole

= 8.40 L of H2 gas

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Chapter 7 Solutions

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Solutions

7.7 Properties of Solutions

Solutions • contain small particles (ions or molecules). • are transparent. • do not separate. • cannot be filtered.


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Examples of Colloids

Colloids

Examples of colloids include

Colloids

• fog

• have medium-size particles.

• whipped cream

• cannot be filtered.

• milk

• can be separated by semipermeable membranes.

• cheese • blood plasma • pearls


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Solutions, Colloids, and Suspensions

Suspensions Suspensions • have very large particles. • settle out. • can be filtered. • must be stirred to stay suspended. Examples include: blood platelets, muddy water, and calamine lotion.


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Learning Check

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Solution

A mixture that has solute particles that do not settle out, but are too large to pass through a semipermeable membrane is called a

A mixture that has solute particles that do not settle out, but are too large to pass through a semipermeable membrane is called a

1) solution. 2) colloid. 3) suspension.

2) colloid.

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Osmosis

Osmosis

In osmosis, • water (solvent) flows from the lower solute concentration into the higher solute concentration. • the level of the solution with the higher solute concentration rises. • the concentrations of the two solutions become equal with time.

Suppose a semipermeable membrane separates a 4% starch solution from a 10% starch solution. Starch is a colloid and cannot pass through the membrane, but water can. What happens? semipermeable membrane

4% starch

10% starch

H2O Copyright © 2009 by Pearson Education, Inc.

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Osmotic Pressure

Water Flow Equalizes • The 10% starch solution is diluted by the flow of water out of the 4% and its volume increases. • The 4% solution loses water and its volume decreases. • Eventually, the water flow between the two becomes equal.

Osmotic pressure is • produced by the solute particles dissolved in a solution. • equal to the pressure that would prevent the flow of additional water into the more concentrated solution. • greater as the number of dissolved particles in the solution increases.

7% starch 7% starch H O 2

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Learning Check

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Solution

A semipermeable membrane separates a 10% sucrose solution A from a 5% sucrose solution B. If sucrose is a colloid, fill in the blanks in the statements below.

A semipermeable membrane separates a 10% sucrose solution A from a 5% sucrose solution B. If sucrose is a colloid, fill in the blanks in the statements below.

1. Solution ____ has the greater osmotic pressure. 2. Water initially flows from ___ into ___. 3. The level of solution ____will be lower.

1. Solution A has the greater osmotic pressure. 2. Water initially flows from B into A. 3. The level of solution B will be lower.

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Isotonic Solutions

Osmotic Pressure of the Blood

An isotonic solution • exerts the same osmotic pressure as red blood cells. • is known as a “physiological solution.” • of 5.0% glucose or 0.90% NaCl is used medically because each has a solute concentration equal to the osmotic pressure equal to red blood cells.

Red blood cells • have cell walls that are semipermeable membranes. • maintain an osmotic pressure that cannot change or damage occurs. • must maintain an equal flow of water between the red blood cell and its surrounding environment.

H2O


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Hypotonic Solutions A hypotonic solution • has a lower osmotic pressure than red blood cells. • has a lower concentration than physiological solutions. • causes water to flow into red blood cells. • causes hemolysis: RBCs swell and may burst.

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Hypertonic Solutions A hypertonic solution • has a higher osmotic pressure than RBCs. • has a higher concentration than physiological solutions. • causes water to flow out of RBCs. • cause crenation: RBCs shrink in size.

H2O

H2O



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Dialysis

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Learning Check

In dialysis, • solvent and small solute particles pass through an artificial membrane.

Indicate if each of the following solutions is 1) isotonic, 2) hypotonic, or 3) hypertonic. A.____ 2% NaCl solution B.____ 1% glucose solution C.____ 0.5% NaCl solution D.____ 5% glucose solution

• large particles are retained inside. • waste particles such as urea from blood are removed using hemodialysis (artificial kidney).

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Solution

Learning Check When placed in each of the following, indicate if a red blood cell will 1) not change, 2) hemolyze, or 3) crenate.

Indicate if each of the following solutions is 1) isotonic, 2) hypotonic, or 3) hypertonic. A. 3 2% NaCl solution B. 2 1% glucose solution C. 2 0.5% NaCl solution D. 1 5% glucose solution

A.____ 5% glucose solution B.____ 1% glucose solution C.____ 0.5% NaCl solution D.____ 2% NaCl solution

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Solution

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Learning Check Each of the following mixtures is placed in a dialyzing bag and immersed in pure water. Which substance, if any, will be found in the water outside the bag?

When placed in each of the following, indicate if a red blood cell will 1) not change, 2) hemolyze, or 3) crenate. A._1_ 5% glucose solution B._2_ 1% glucose solution C._2_ 0.5% NaCl solution D._3_ 2% NaCl solution

A. 10% KCl solution B. 5% starch solution C. 5% NaCl and 5% starch solutions

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Solution Each of the following mixtures is placed in a dialyzing bag and immersed in pure water. Which substance, if any, will be found in the water outside the bag? A. 10% KCl solution KCl ( K+, Cl−) B. 5% starch solution None; starch is retained. C. 5% NaCl and 5% starch solutions NaCl (Na+, Cl−), but starch is retained.

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