Tuesday, August 31, 2010
make up Quiz ep 4
find me video links on Van Der Waals Forces, pi sigma bonding, hybridization, or metallic bonding
Sunday, August 29, 2010
Week Aug 30, 2010 Topics
Ep 4 Exam topics--chapter 6-9--periodic law from ch 6, covalent, ionic bonds, polyatomics ions, and naming--more details forthcoming.
Ep 5 Exam Topics: Ch 15 and 16--aqueous solutions, moles revisited, molarity, molality, dilutions, percent problems, and colligative properties.
Ep 6 Exam topics: Ch 19-- Buffers, neutralization, titrations, dilutions, and pH.
This week will be about the problems from this topic. Next week, we will start the review sheets for the exam.
Exam format--vocab--pick 10 out of 20 to define, and then, short answers that will comprise of problems, questions about application of chemical theories, and explaining certain theories to demonstrate understanding.
Mr. Gary leaves Sept 4 until Sept 10 for the Debate Tournament in the Phillipines.
Ep 5 Exam Topics: Ch 15 and 16--aqueous solutions, moles revisited, molarity, molality, dilutions, percent problems, and colligative properties.
Ep 6 Exam topics: Ch 19-- Buffers, neutralization, titrations, dilutions, and pH.
This week will be about the problems from this topic. Next week, we will start the review sheets for the exam.
Exam format--vocab--pick 10 out of 20 to define, and then, short answers that will comprise of problems, questions about application of chemical theories, and explaining certain theories to demonstrate understanding.
Mr. Gary leaves Sept 4 until Sept 10 for the Debate Tournament in the Phillipines.
Saturday, August 28, 2010
Wednesday, August 25, 2010
Make up quiz 1 ep 4, 5, and 6
I want a video link for an interesting 3-7 minute video on the following:
ep 4--stoichemistry, limiting reagents, moles
ep 5 solutions, colloids, tyndall effect, brownian motion
ep 6 acid bases, oxidation/reduction, or electrochemistry.
if i look at the video, and i'm bored in one minute, you get half bonus for your quiz.
also include a summation of the top three points in the quiz--what's going on, what's interesting, and a description proving you watched the video--example:
solutions video--link www.blahblah.com
The video starts out with a definition of solution formation. It has an interesting discussion of solute and solvents and shows a supersaturation of acetate. Students will like how the acetate solution freezes over after the temperature drops.
ep 4--stoichemistry, limiting reagents, moles
ep 5 solutions, colloids, tyndall effect, brownian motion
ep 6 acid bases, oxidation/reduction, or electrochemistry.
if i look at the video, and i'm bored in one minute, you get half bonus for your quiz.
also include a summation of the top three points in the quiz--what's going on, what's interesting, and a description proving you watched the video--example:
solutions video--link www.blahblah.com
The video starts out with a definition of solution formation. It has an interesting discussion of solute and solvents and shows a supersaturation of acetate. Students will like how the acetate solution freezes over after the temperature drops.
Tuesday, August 24, 2010
Ch 9.3 Polyatomics
MEMORIZE table 9.3 on page 257 for quiz
Sometimes the same two or three elements combine in different ratios to form different polyatomic ions. You can see examples in Table 9.3. Look for pairs of ions for which there is both an -ite and an -ate ending, for example, sulfite and sulfate. In the list below, examine the charge on each ion in the pair. Note the number of oxygen atoms and the endings on each name. You should be able to discern a pattern in the naming convention.
The charge on each polyatomic ion in a given pair is the same. The -ite ending indicates one less oxygen atom than the -ate ending. However, the ending does not tell you the actual number of oxygen atoms in the ion. For example, the nitrite ion has two oxygen atoms and the sulfite ion has three oxygen atoms. All anions with names ending in -ite or -ate contain oxygen.
Sometimes the same two or three elements combine in different ratios to form different polyatomic ions. You can see examples in Table 9.3. Look for pairs of ions for which there is both an -ite and an -ate ending, for example, sulfite and sulfate. In the list below, examine the charge on each ion in the pair. Note the number of oxygen atoms and the endings on each name. You should be able to discern a pattern in the naming convention.
The charge on each polyatomic ion in a given pair is the same. The -ite ending indicates one less oxygen atom than the -ate ending. However, the ending does not tell you the actual number of oxygen atoms in the ion. For example, the nitrite ion has two oxygen atoms and the sulfite ion has three oxygen atoms. All anions with names ending in -ite or -ate contain oxygen.
EP 4 ASSIGNMENTS: LAST THREE
assignment 1: p. 256 #1, 2 page 258 # 3, 4 page 263 # 10-11 skip b
assignment 2: p. 258 # 8, 9 p. 265 #12, 13 p. 266 #17-19
Vocab section 8.3 and 7.3
assignment 2: p. 258 # 8, 9 p. 265 #12, 13 p. 266 #17-19
Vocab section 8.3 and 7.3
Monday, August 23, 2010
Ch 19: Problems
(40) Key Concept What type of salt produces an acidic solution? A basic solution?Hint
(41) Key Concept What substances are combined to make a buffer? Hint
(42)Which of these salts would form an acidic aqueous solution?
a. KC2H3O2
b. LiCl
c. NaHCO3
d. (NH4)2SO4
(43)Using equations, show what happens when acid is added to an ammonium ion–ammonia buffer. What happens when base is added?
(41) Key Concept What substances are combined to make a buffer? Hint
(42)Which of these salts would form an acidic aqueous solution?
a. KC2H3O2
b. LiCl
c. NaHCO3
d. (NH4)2SO4
(43)Using equations, show what happens when acid is added to an ammonium ion–ammonia buffer. What happens when base is added?
Word Origins Buffer
Word Origins
Buffer comes from the Old English word buff, meaning “firmly or sturdily.” A buffer solution resists changes in its pH even when acidic or basic solutions are added to it. What does it mean if authorities set up a buffer zone around a fire investigation site?
Buffer comes from the Old English word buff, meaning “firmly or sturdily.” A buffer solution resists changes in its pH even when acidic or basic solutions are added to it. What does it mean if authorities set up a buffer zone around a fire investigation site?
Ch 19: Intro to Buffers 2
Buffers
The addition of 10 mL of 0.10M sodium hydroxide to 1 L of pure water increases the pH by 4.0 pH units (from 7.0 to 11.0). A solution containing 0.20 mol/L each of ethanoic acid and sodium ethanoate has a pH of 4.76. When moderate amounts of either acid or base are added to this solution, however, the pH changes little. The addition of 10 mL of 0.10M sodium hydroxide to 1 L of this solution, for example, increases the pH by only 0.01 pH unit, from 4.76 to 4.77. Figure 19.27 shows what happens when 1.0 mL of 0.01M HCl solution is added to an unbuffered solution.
Figure 19.27 A buffer is a solution in which the pH remains relatively constant. a. The indicator shows that the buffered solution on the left and the unbuffered solution on the right are basic—pH about 8. b. After the addition of 1.0 mL of 0.01M HCl solution, the pH of the buffered solution shows no visible change. The pH of the unbuffered solution, however, is now about 3—the solution is acidic.
The solution of ethanoic acid and sodium ethanoate is an example of a typical buffer. A buffer is a solution in which the pH remains relatively constant when small amounts of acid or base are added. A buffer is a solution of a weak acid and one of its salts, or a solution of a weak base and one of its salts.
A buffer solution is better able to resist drastic changes in pH than is pure water. Figure 19.28 illustrates how a buffer works. Ethanoic acid (CH3COOH) and its anion (CH3COO−) act as reservoirs of neutralizing power. They react with any hydroxide ions or hydrogen ions added to the solution. For example, consider the buffer solution in which the sodium ethanoate (CH3COONa) is completely ionized.
The addition of 10 mL of 0.10M sodium hydroxide to 1 L of pure water increases the pH by 4.0 pH units (from 7.0 to 11.0). A solution containing 0.20 mol/L each of ethanoic acid and sodium ethanoate has a pH of 4.76. When moderate amounts of either acid or base are added to this solution, however, the pH changes little. The addition of 10 mL of 0.10M sodium hydroxide to 1 L of this solution, for example, increases the pH by only 0.01 pH unit, from 4.76 to 4.77. Figure 19.27 shows what happens when 1.0 mL of 0.01M HCl solution is added to an unbuffered solution.
Figure 19.27 A buffer is a solution in which the pH remains relatively constant. a. The indicator shows that the buffered solution on the left and the unbuffered solution on the right are basic—pH about 8. b. After the addition of 1.0 mL of 0.01M HCl solution, the pH of the buffered solution shows no visible change. The pH of the unbuffered solution, however, is now about 3—the solution is acidic.
The solution of ethanoic acid and sodium ethanoate is an example of a typical buffer. A buffer is a solution in which the pH remains relatively constant when small amounts of acid or base are added. A buffer is a solution of a weak acid and one of its salts, or a solution of a weak base and one of its salts.
A buffer solution is better able to resist drastic changes in pH than is pure water. Figure 19.28 illustrates how a buffer works. Ethanoic acid (CH3COOH) and its anion (CH3COO−) act as reservoirs of neutralizing power. They react with any hydroxide ions or hydrogen ions added to the solution. For example, consider the buffer solution in which the sodium ethanoate (CH3COONa) is completely ionized.
Ch 19: Salt Hydrolysis
The ethanoate ion is a Brønsted-Lowry base, which means it is a hydrogen-ion acceptor. It establishes an equilibrium with water, forming electrically neutral ethanoic acid and negative hydroxide ions.
This process is called hydrolysis because it splits a hydrogen ion off a water molecule. The resulting solution contains a hydroxide-ion concentration greater than the hydrogen-ion concentration. Thus the solution is basic.
Ammonium chloride (NH4Cl) is the salt of a strong acid (hydrochloric acid, HCl) and a weak base (ammonia, NH3). It is completely ionized in solution.
NH4Cl(aq) → NH4+(aq) + Cl−(aq)
Figure 19.25 Vapors of the strong acid HCl(aq) and the weak base NH3(aq) combine to form the acidic white salt ammonium chloride (NH4Cl).
The ammonium ion (NH4+) is a strong enough acid to donate a hydrogen ion to a water molecule, although the equilibrium is strongly to the left.
This process is also called hydrolysis. It results in the formation of unionized ammonia and hydronium (hydrogen) ions. The [H3O+] is greater than the [OH−]. Thus, a solution of ammonium chloride is acidic. To determine if a salt solution is acidic or basic, remember the following rules:
Strong acid + Strong base → Neutral solution
Strong acid + Weak base → Acidic solution
Weak acid + Strong base → Basic solution
This process is called hydrolysis because it splits a hydrogen ion off a water molecule. The resulting solution contains a hydroxide-ion concentration greater than the hydrogen-ion concentration. Thus the solution is basic.
Ammonium chloride (NH4Cl) is the salt of a strong acid (hydrochloric acid, HCl) and a weak base (ammonia, NH3). It is completely ionized in solution.
NH4Cl(aq) → NH4+(aq) + Cl−(aq)
Figure 19.25 Vapors of the strong acid HCl(aq) and the weak base NH3(aq) combine to form the acidic white salt ammonium chloride (NH4Cl).
The ammonium ion (NH4+) is a strong enough acid to donate a hydrogen ion to a water molecule, although the equilibrium is strongly to the left.
This process is also called hydrolysis. It results in the formation of unionized ammonia and hydronium (hydrogen) ions. The [H3O+] is greater than the [OH−]. Thus, a solution of ammonium chloride is acidic. To determine if a salt solution is acidic or basic, remember the following rules:
Strong acid + Strong base → Neutral solution
Strong acid + Weak base → Acidic solution
Weak acid + Strong base → Basic solution
Ch 19: Intro to Buffers
Salt Hydrolysis
A salt consists of an anion from an acid and a cation from a base. It forms as a result of a neutralization reaction. Although solutions of many salts are neutral, some are acidic and others are basic. Solutions of sodium chloride and of potassium sulfate are neutral. A solution of ammonium chloride is acidic. A solution of sodium ethanoate (sodium acetate) is basic. Figure 19.24 shows a titration curve obtained by adding a solution of sodium hydroxide, a strong base, to a solution of ethanoic (acetic) acid, a weak acid. An aqueous solution of sodium ethanoate exists at the equivalence point.
The pH at the equivalence point is 8.7—basic.
For a strong acid–strong base titration, the pH at the equivalence point is 7, or neutral. This difference exists because some salts promote hydrolysis. In salt hydrolysis, the cations or anions of a dissociated salt remove hydrogen ions from or donate hydrogen ions to water. Depending on the direction of the hydrogen-ion transfer, solutions containing hydrolyzing salts may be either acidic or basic. Hydrolyzing salts are usually derived from a strong acid and a weak base, or from a weak acid and a strong base. Sodium carbonate, washing soda, is the salt of the strong base sodium hydroxide and carbonic acid, a weak acid. Ammonium nitrate, used in fertilizers, is the salt of the weak base ammonia and nitric acid, a strong acid. Soap is the salt of a strong base, usually sodium hydroxide, and stearic acid, a weak acid present in fats. In general, salts that produce acidic solutions contain positive ions that release protons to water. Salts that produce basic solutions contain negative ions that attract protons from water.
Sodium ethanoate (CH3COONa) is the salt of a weak acid (ethanoic acid, CH3COOH) and a strong base (sodium hydroxide, NaOH). In solution, the salt is completely ionized.
A salt consists of an anion from an acid and a cation from a base. It forms as a result of a neutralization reaction. Although solutions of many salts are neutral, some are acidic and others are basic. Solutions of sodium chloride and of potassium sulfate are neutral. A solution of ammonium chloride is acidic. A solution of sodium ethanoate (sodium acetate) is basic. Figure 19.24 shows a titration curve obtained by adding a solution of sodium hydroxide, a strong base, to a solution of ethanoic (acetic) acid, a weak acid. An aqueous solution of sodium ethanoate exists at the equivalence point.
The pH at the equivalence point is 8.7—basic.
For a strong acid–strong base titration, the pH at the equivalence point is 7, or neutral. This difference exists because some salts promote hydrolysis. In salt hydrolysis, the cations or anions of a dissociated salt remove hydrogen ions from or donate hydrogen ions to water. Depending on the direction of the hydrogen-ion transfer, solutions containing hydrolyzing salts may be either acidic or basic. Hydrolyzing salts are usually derived from a strong acid and a weak base, or from a weak acid and a strong base. Sodium carbonate, washing soda, is the salt of the strong base sodium hydroxide and carbonic acid, a weak acid. Ammonium nitrate, used in fertilizers, is the salt of the weak base ammonia and nitric acid, a strong acid. Soap is the salt of a strong base, usually sodium hydroxide, and stearic acid, a weak acid present in fats. In general, salts that produce acidic solutions contain positive ions that release protons to water. Salts that produce basic solutions contain negative ions that attract protons from water.
Sodium ethanoate (CH3COONa) is the salt of a weak acid (ethanoic acid, CH3COOH) and a strong base (sodium hydroxide, NaOH). In solution, the salt is completely ionized.
Ch 19: Acids/Bases 19.5 Buffers
Ch 19
Key Concepts
• When is the solution of a salt acidic or basic?
• What are the components of a buffer?
Vocabulary
• salt hydrolysis
• buffer
• buffer capacity
Key Concepts
• When is the solution of a salt acidic or basic?
• What are the components of a buffer?
Vocabulary
• salt hydrolysis
• buffer
• buffer capacity
Ch 8: Problems
23) Key Concept How are atomic and molecular orbitals related?Hint
(24) Key Concept Explain how the VSEPR theory can be used to predict the shapes of molecules.Hint
(25) Key Concept How is orbital hybrization useful in describing molecules?Hint
(26)What shape would you expect a simple carbon-containing compound to have if the carbon atom has the following hybridizations?
a. sp2
b. sp3
c. sp
(27)What is a sigma bond? Describe, with the aid of a diagram, how the overlap of two half-filled 1s orbitals produces a sigma bond?
(28)How many sigma and how many pi bonds are in an ethyne molecule (C2H2)?
(29)The BF3 molecule is planar. The attachment of a fluoride ion to the boron in BF3, through a coordinate covalent bond, creates the BF4− ion. What is the geometric shape of this ion?
(24) Key Concept Explain how the VSEPR theory can be used to predict the shapes of molecules.Hint
(25) Key Concept How is orbital hybrization useful in describing molecules?Hint
(26)What shape would you expect a simple carbon-containing compound to have if the carbon atom has the following hybridizations?
a. sp2
b. sp3
c. sp
(27)What is a sigma bond? Describe, with the aid of a diagram, how the overlap of two half-filled 1s orbitals produces a sigma bond?
(28)How many sigma and how many pi bonds are in an ethyne molecule (C2H2)?
(29)The BF3 molecule is planar. The attachment of a fluoride ion to the boron in BF3, through a coordinate covalent bond, creates the BF4− ion. What is the geometric shape of this ion?
Ch 8: Hybridization Orbitals
Hybrid Orbitals
The VSEPR theory works well when accounting for molecular shapes, but it does not help much in describing the types of bonds formed. Orbital hybridization provides information about both molecular bonding and molecular shape. In hybridization, several atomic orbitals mix to form the same total number of equivalent hybrid orbitals.
Hybridization Involving Single Bonds
Recall that the carbon atom’s outer electron configuration is 2s2 2p2, but one of the 2s electrons is promoted to a 2p orbital to give one 2s electron and three 2p electrons, allowing it to bond to four hydrogen atoms in methane. You might suspect that one bond would be different from the other three. In fact, all the bonds are identical. This is explained by orbital hybridization.
The one 2s orbital and three 2p orbitals of a carbon atom mix to form four sp3 hybrid orbitals. These are at the tetrahedral angle of 109.5°. As you can see in Figure 8.19, the four sp3 orbitals of carbon overlap with the 1s orbitals of the four hydrogen atoms. The sp3 orbitals extend farther into space than either s or p orbitals, allowing a great deal of overlap with the hydrogen 1s orbitals. The eight available valence electrons fill the molecular orbitals to form four C —H sigma bonds. The extent of overlap results in unusually strong covalent bonds.
Figure 8.19
Hybridization Involving Double Bonds
Hybridization is also useful in describing double covalent bonds. Ethene is a relatively simple molecule that has one carbon–carbon double bond and four carbon–hydrogen single bonds.
Experimental evidence indicates that the H—C—H bond angles in ethene are about 120°. In ethene, sp2 hybrid orbitals form from the combination of one 2s and two 2p atomic orbitals of carbon. As you can see in Figure 8.20, each hybrid orbital is separated from the other two by 120°. Two sp2 hybrid orbitals of each carbon form sigma-bonding molecular orbitals with the four available hydrogen 1s orbitals. The third sp2 orbitals of each of the two carbons overlap to form a carbon–carbon sigma-bonding orbital. The nonhybridized 2p carbon orbitals overlap side-by-side to form a pi-bonding orbital. A total of twelve electrons fill six bonding molecular orbitals. Thus five sigma bonds and one pi bond hold the ethene molecule together. The sigma bonds and the pi bond are two-electron covalent bonds. Although they are drawn alike in structural formulas, pi bonds are weaker than sigma bonds. In chemical reactions that involve breaking one bond of a carbon–carbon double bond, the pi bond is more likely to break than the sigma bond.
Hybridization Involving Triple Bonds
A third type of covalent bond is a triple bond, such as is found in ethyne (C2H2), also called acetylene.
As with other molecules, the hybrid orbital description of ethyne is guided by an understanding of the properties of the molecule. Ethyne is a linear molecule. The best hybrid orbital description is obtained if a 2s atomic orbital of carbon mixes with only one of the three 2p atomic orbitals. The result is two sp hybrid orbitals for each carbon.
The carbon–carbon sigma-bonding molecular orbital of the ethyne molecule in Figure 8.21 forms from the overlap of one sp orbital from each carbon. The other sp orbital of each carbon overlaps with the 1s orbital of each hydrogen, also forming sigma-bonding molecular orbitals. The remaining pair of p atomic orbitals on each carbon overlap side-by-side. They form two pi-bonding molecular orbitals that surround the central carbons. The ten available electrons completely fill five bonding molecular orbitals. The bonding of ethyne consists of three sigma bonds and two pi bonds
The VSEPR theory works well when accounting for molecular shapes, but it does not help much in describing the types of bonds formed. Orbital hybridization provides information about both molecular bonding and molecular shape. In hybridization, several atomic orbitals mix to form the same total number of equivalent hybrid orbitals.
Hybridization Involving Single Bonds
Recall that the carbon atom’s outer electron configuration is 2s2 2p2, but one of the 2s electrons is promoted to a 2p orbital to give one 2s electron and three 2p electrons, allowing it to bond to four hydrogen atoms in methane. You might suspect that one bond would be different from the other three. In fact, all the bonds are identical. This is explained by orbital hybridization.
The one 2s orbital and three 2p orbitals of a carbon atom mix to form four sp3 hybrid orbitals. These are at the tetrahedral angle of 109.5°. As you can see in Figure 8.19, the four sp3 orbitals of carbon overlap with the 1s orbitals of the four hydrogen atoms. The sp3 orbitals extend farther into space than either s or p orbitals, allowing a great deal of overlap with the hydrogen 1s orbitals. The eight available valence electrons fill the molecular orbitals to form four C —H sigma bonds. The extent of overlap results in unusually strong covalent bonds.
Figure 8.19
Hybridization Involving Double Bonds
Hybridization is also useful in describing double covalent bonds. Ethene is a relatively simple molecule that has one carbon–carbon double bond and four carbon–hydrogen single bonds.
Experimental evidence indicates that the H—C—H bond angles in ethene are about 120°. In ethene, sp2 hybrid orbitals form from the combination of one 2s and two 2p atomic orbitals of carbon. As you can see in Figure 8.20, each hybrid orbital is separated from the other two by 120°. Two sp2 hybrid orbitals of each carbon form sigma-bonding molecular orbitals with the four available hydrogen 1s orbitals. The third sp2 orbitals of each of the two carbons overlap to form a carbon–carbon sigma-bonding orbital. The nonhybridized 2p carbon orbitals overlap side-by-side to form a pi-bonding orbital. A total of twelve electrons fill six bonding molecular orbitals. Thus five sigma bonds and one pi bond hold the ethene molecule together. The sigma bonds and the pi bond are two-electron covalent bonds. Although they are drawn alike in structural formulas, pi bonds are weaker than sigma bonds. In chemical reactions that involve breaking one bond of a carbon–carbon double bond, the pi bond is more likely to break than the sigma bond.
Hybridization Involving Triple Bonds
A third type of covalent bond is a triple bond, such as is found in ethyne (C2H2), also called acetylene.
As with other molecules, the hybrid orbital description of ethyne is guided by an understanding of the properties of the molecule. Ethyne is a linear molecule. The best hybrid orbital description is obtained if a 2s atomic orbital of carbon mixes with only one of the three 2p atomic orbitals. The result is two sp hybrid orbitals for each carbon.
The carbon–carbon sigma-bonding molecular orbital of the ethyne molecule in Figure 8.21 forms from the overlap of one sp orbital from each carbon. The other sp orbital of each carbon overlaps with the 1s orbital of each hydrogen, also forming sigma-bonding molecular orbitals. The remaining pair of p atomic orbitals on each carbon overlap side-by-side. They form two pi-bonding molecular orbitals that surround the central carbons. The ten available electrons completely fill five bonding molecular orbitals. The bonding of ethyne consists of three sigma bonds and two pi bonds
Ch 8: Molecular Orbitals
Molecular Orbitals
The model for covalent bonding you have been using assumes that the orbitals are those of the individual atoms. There is a quantum mechanical model of bonding, however, that describes the electrons in molecules using orbitals that exist only for groupings of atoms. When two atoms combine, this model assumes that their atomic orbitals overlap to produce molecular orbitals, or orbitals that apply to the entire molecule.
In some ways, atomic orbitals and molecular orbitals are similar. Just as an atomic orbital belongs to a particular atom, a molecular orbital belongs to a molecule as a whole. Each atomic orbital is filled if it contains two electrons. Similarly, two electrons are required to fill a molecular orbital. A molecular orbital that can be occupied by two electrons of a covalent bond is called a bonding orbital.
Sigma Bonds
When two atomic orbitals combine to form a molecular orbital that is symmetrical around the axis connecting two atomic nuclei, a sigma bond is formed, as illustrated in Figure 8.13. The symbol for this bond is the Greek letter sigma (σ).
In general, covalent bonding results from an imbalance between the attractions and repulsions of the nuclei and electrons involved. Because their charges have opposite signs, the nuclei and electrons attract each other. Because their charges have the same sign, nuclei repel other nuclei and electrons repel other electrons. In a hydrogen molecule, the nuclei repel each other, as do the electrons. In a bonding molecular orbital of hydrogen, however, the attractions between the hydrogen nuclei and the electrons are stronger than the repulsions. The balance of all the interactions between the hydrogen atoms is thus tipped in favor of holding the atoms together. The result is a stable diatomic molecule of H2.
Atomic p orbitals can also overlap to form molecular orbitals. A fluorine atom, for example, has a half-filled 2p orbital. When two fluorine atoms combine, as shown in Figure 8.14, the p orbitals overlap to produce a bonding molecular orbital. There is a high probability of finding a pair of electrons between the positively charged nuclei of the two fluorines. The fluorine nuclei are attracted to this region of high electron density. This attraction holds the atoms together in the fluorine molecule (F2). The overlap of the 2p orbitals produces a bonding molecular orbital that is symmetrical when viewed around the F —F bond axis connecting the nuclei. Therefore, the F —F bond is a sigma bond.
Pi Bonds
In the sigma bond of the fluorine molecule, the p atomic orbitals overlap end-to-end. In some molecules, however, orbitals can overlap side-by-side. As shown in Figure 8.15, the side-by-side overlap of atomic p orbitals produces what are called pi molecular orbitals. When a pi molecular orbital is filled with two electrons, a pi bond results. In a pi bond (symbolized by the Greek letter Ï€), the bonding electrons are most likely to be found in sausage-shaped regions above and below the bond axis of the bonded atoms. It is not symmetrical around the F- —F bond axis. Atomic orbitals in pi bonding overlap less than in sigma bonding. Therefore, pi bonds tend to be weaker than sigma bonds.
VESPR—not very small elephants playing in the rain…
The valence-shell electron-pair repulsion theory, or VSEPR theory, explains the three-dimensional shape of methane. According to VSEPR theory, the repulsion between electron pairs causes molecular shapes to adjust so that the valence-electron pairs stay as far apart as possible. The methane molecule has four bonding electron pairs and no unshared pairs. The bonding pairs are farthest apart when the angle between the central carbon and its attached hydrogens is 109.5°. This is the H— C —H bond angle found experimentally.
Unshared pairs of electrons are also important in predicting the shapes of molecules. The nitrogen in ammonia (NH3) is surrounded by four pairs of valence electrons, so you might predict the tetrahedral angle of 109.5° for the H—N—H bond angle. However, one of the valence-electron pairs shown in Figure 8.16b is an unshared pair. No bonding atom is vying for these unshared electrons. Thus they are held closer to the nitrogen than are the bonding pairs. The unshared pair strongly repels the bonding pairs, pushing them together. The measured H—N—H bond angle is only 107°.
The model for covalent bonding you have been using assumes that the orbitals are those of the individual atoms. There is a quantum mechanical model of bonding, however, that describes the electrons in molecules using orbitals that exist only for groupings of atoms. When two atoms combine, this model assumes that their atomic orbitals overlap to produce molecular orbitals, or orbitals that apply to the entire molecule.
In some ways, atomic orbitals and molecular orbitals are similar. Just as an atomic orbital belongs to a particular atom, a molecular orbital belongs to a molecule as a whole. Each atomic orbital is filled if it contains two electrons. Similarly, two electrons are required to fill a molecular orbital. A molecular orbital that can be occupied by two electrons of a covalent bond is called a bonding orbital.
Sigma Bonds
When two atomic orbitals combine to form a molecular orbital that is symmetrical around the axis connecting two atomic nuclei, a sigma bond is formed, as illustrated in Figure 8.13. The symbol for this bond is the Greek letter sigma (σ).
In general, covalent bonding results from an imbalance between the attractions and repulsions of the nuclei and electrons involved. Because their charges have opposite signs, the nuclei and electrons attract each other. Because their charges have the same sign, nuclei repel other nuclei and electrons repel other electrons. In a hydrogen molecule, the nuclei repel each other, as do the electrons. In a bonding molecular orbital of hydrogen, however, the attractions between the hydrogen nuclei and the electrons are stronger than the repulsions. The balance of all the interactions between the hydrogen atoms is thus tipped in favor of holding the atoms together. The result is a stable diatomic molecule of H2.
Atomic p orbitals can also overlap to form molecular orbitals. A fluorine atom, for example, has a half-filled 2p orbital. When two fluorine atoms combine, as shown in Figure 8.14, the p orbitals overlap to produce a bonding molecular orbital. There is a high probability of finding a pair of electrons between the positively charged nuclei of the two fluorines. The fluorine nuclei are attracted to this region of high electron density. This attraction holds the atoms together in the fluorine molecule (F2). The overlap of the 2p orbitals produces a bonding molecular orbital that is symmetrical when viewed around the F —F bond axis connecting the nuclei. Therefore, the F —F bond is a sigma bond.
Pi Bonds
In the sigma bond of the fluorine molecule, the p atomic orbitals overlap end-to-end. In some molecules, however, orbitals can overlap side-by-side. As shown in Figure 8.15, the side-by-side overlap of atomic p orbitals produces what are called pi molecular orbitals. When a pi molecular orbital is filled with two electrons, a pi bond results. In a pi bond (symbolized by the Greek letter Ï€), the bonding electrons are most likely to be found in sausage-shaped regions above and below the bond axis of the bonded atoms. It is not symmetrical around the F- —F bond axis. Atomic orbitals in pi bonding overlap less than in sigma bonding. Therefore, pi bonds tend to be weaker than sigma bonds.
VESPR—not very small elephants playing in the rain…
The valence-shell electron-pair repulsion theory, or VSEPR theory, explains the three-dimensional shape of methane. According to VSEPR theory, the repulsion between electron pairs causes molecular shapes to adjust so that the valence-electron pairs stay as far apart as possible. The methane molecule has four bonding electron pairs and no unshared pairs. The bonding pairs are farthest apart when the angle between the central carbon and its attached hydrogens is 109.5°. This is the H— C —H bond angle found experimentally.
Unshared pairs of electrons are also important in predicting the shapes of molecules. The nitrogen in ammonia (NH3) is surrounded by four pairs of valence electrons, so you might predict the tetrahedral angle of 109.5° for the H—N—H bond angle. However, one of the valence-electron pairs shown in Figure 8.16b is an unshared pair. No bonding atom is vying for these unshared electrons. Thus they are held closer to the nitrogen than are the bonding pairs. The unshared pair strongly repels the bonding pairs, pushing them together. The measured H—N—H bond angle is only 107°.
Ch 8: Molecular Orbitals
Molecular Orbitals
The model for covalent bonding you have been using assumes that the orbitals are those of the individual atoms. There is a quantum mechanical model of bonding, however, that describes the electrons in molecules using orbitals that exist only for groupings of atoms. When two atoms combine, this model assumes that their atomic orbitals overlap to produce molecular orbitals, or orbitals that apply to the entire molecule.
In some ways, atomic orbitals and molecular orbitals are similar. Just as an atomic orbital belongs to a particular atom, a molecular orbital belongs to a molecule as a whole. Each atomic orbital is filled if it contains two electrons. Similarly, two electrons are required to fill a molecular orbital. A molecular orbital that can be occupied by two electrons of a covalent bond is called a bonding orbital.
Sigma Bonds
When two atomic orbitals combine to form a molecular orbital that is symmetrical around the axis connecting two atomic nuclei, a sigma bond is formed, as illustrated in Figure 8.13. The symbol for this bond is the Greek letter sigma (σ).
In general, covalent bonding results from an imbalance between the attractions and repulsions of the nuclei and electrons involved. Because their charges have opposite signs, the nuclei and electrons attract each other. Because their charges have the same sign, nuclei repel other nuclei and electrons repel other electrons. In a hydrogen molecule, the nuclei repel each other, as do the electrons. In a bonding molecular orbital of hydrogen, however, the attractions between the hydrogen nuclei and the electrons are stronger than the repulsions. The balance of all the interactions between the hydrogen atoms is thus tipped in favor of holding the atoms together. The result is a stable diatomic molecule of H2.
Atomic p orbitals can also overlap to form molecular orbitals. A fluorine atom, for example, has a half-filled 2p orbital. When two fluorine atoms combine, as shown in Figure 8.14, the p orbitals overlap to produce a bonding molecular orbital. There is a high probability of finding a pair of electrons between the positively charged nuclei of the two fluorines. The fluorine nuclei are attracted to this region of high electron density. This attraction holds the atoms together in the fluorine molecule (F2). The overlap of the 2p orbitals produces a bonding molecular orbital that is symmetrical when viewed around the F —F bond axis connecting the nuclei. Therefore, the F —F bond is a sigma bond.
Pi Bonds
In the sigma bond of the fluorine molecule, the p atomic orbitals overlap end-to-end. In some molecules, however, orbitals can overlap side-by-side. As shown in Figure 8.15, the side-by-side overlap of atomic p orbitals produces what are called pi molecular orbitals. When a pi molecular orbital is filled with two electrons, a pi bond results. In a pi bond (symbolized by the Greek letter Ï€), the bonding electrons are most likely to be found in sausage-shaped regions above and below the bond axis of the bonded atoms. It is not symmetrical around the F- —F bond axis. Atomic orbitals in pi bonding overlap less than in sigma bonding. Therefore, pi bonds tend to be weaker than sigma bonds.
VESPR—not very small elephants playing in the rain…
The valence-shell electron-pair repulsion theory, or VSEPR theory, explains the three-dimensional shape of methane. According to VSEPR theory, the repulsion between electron pairs causes molecular shapes to adjust so that the valence-electron pairs stay as far apart as possible. The methane molecule has four bonding electron pairs and no unshared pairs. The bonding pairs are farthest apart when the angle between the central carbon and its attached hydrogens is 109.5°. This is the H— C —H bond angle found experimentally.
Unshared pairs of electrons are also important in predicting the shapes of molecules. The nitrogen in ammonia (NH3) is surrounded by four pairs of valence electrons, so you might predict the tetrahedral angle of 109.5° for the H—N—H bond angle. However, one of the valence-electron pairs shown in Figure 8.16b is an unshared pair. No bonding atom is vying for these unshared electrons. Thus they are held closer to the nitrogen than are the bonding pairs. The unshared pair strongly repels the bonding pairs, pushing them together. The measured H—N—H bond angle is only 107°.
The model for covalent bonding you have been using assumes that the orbitals are those of the individual atoms. There is a quantum mechanical model of bonding, however, that describes the electrons in molecules using orbitals that exist only for groupings of atoms. When two atoms combine, this model assumes that their atomic orbitals overlap to produce molecular orbitals, or orbitals that apply to the entire molecule.
In some ways, atomic orbitals and molecular orbitals are similar. Just as an atomic orbital belongs to a particular atom, a molecular orbital belongs to a molecule as a whole. Each atomic orbital is filled if it contains two electrons. Similarly, two electrons are required to fill a molecular orbital. A molecular orbital that can be occupied by two electrons of a covalent bond is called a bonding orbital.
Sigma Bonds
When two atomic orbitals combine to form a molecular orbital that is symmetrical around the axis connecting two atomic nuclei, a sigma bond is formed, as illustrated in Figure 8.13. The symbol for this bond is the Greek letter sigma (σ).
In general, covalent bonding results from an imbalance between the attractions and repulsions of the nuclei and electrons involved. Because their charges have opposite signs, the nuclei and electrons attract each other. Because their charges have the same sign, nuclei repel other nuclei and electrons repel other electrons. In a hydrogen molecule, the nuclei repel each other, as do the electrons. In a bonding molecular orbital of hydrogen, however, the attractions between the hydrogen nuclei and the electrons are stronger than the repulsions. The balance of all the interactions between the hydrogen atoms is thus tipped in favor of holding the atoms together. The result is a stable diatomic molecule of H2.
Atomic p orbitals can also overlap to form molecular orbitals. A fluorine atom, for example, has a half-filled 2p orbital. When two fluorine atoms combine, as shown in Figure 8.14, the p orbitals overlap to produce a bonding molecular orbital. There is a high probability of finding a pair of electrons between the positively charged nuclei of the two fluorines. The fluorine nuclei are attracted to this region of high electron density. This attraction holds the atoms together in the fluorine molecule (F2). The overlap of the 2p orbitals produces a bonding molecular orbital that is symmetrical when viewed around the F —F bond axis connecting the nuclei. Therefore, the F —F bond is a sigma bond.
Pi Bonds
In the sigma bond of the fluorine molecule, the p atomic orbitals overlap end-to-end. In some molecules, however, orbitals can overlap side-by-side. As shown in Figure 8.15, the side-by-side overlap of atomic p orbitals produces what are called pi molecular orbitals. When a pi molecular orbital is filled with two electrons, a pi bond results. In a pi bond (symbolized by the Greek letter Ï€), the bonding electrons are most likely to be found in sausage-shaped regions above and below the bond axis of the bonded atoms. It is not symmetrical around the F- —F bond axis. Atomic orbitals in pi bonding overlap less than in sigma bonding. Therefore, pi bonds tend to be weaker than sigma bonds.
VESPR—not very small elephants playing in the rain…
The valence-shell electron-pair repulsion theory, or VSEPR theory, explains the three-dimensional shape of methane. According to VSEPR theory, the repulsion between electron pairs causes molecular shapes to adjust so that the valence-electron pairs stay as far apart as possible. The methane molecule has four bonding electron pairs and no unshared pairs. The bonding pairs are farthest apart when the angle between the central carbon and its attached hydrogens is 109.5°. This is the H— C —H bond angle found experimentally.
Unshared pairs of electrons are also important in predicting the shapes of molecules. The nitrogen in ammonia (NH3) is surrounded by four pairs of valence electrons, so you might predict the tetrahedral angle of 109.5° for the H—N—H bond angle. However, one of the valence-electron pairs shown in Figure 8.16b is an unshared pair. No bonding atom is vying for these unshared electrons. Thus they are held closer to the nitrogen than are the bonding pairs. The unshared pair strongly repels the bonding pairs, pushing them together. The measured H—N—H bond angle is only 107°.
Ch 8 Orbitals
Ch 8
Key Concepts
• How are atomic and molecular orbitals related?
• How does VSEPR theory help predict the shapes of molecules?
• In what ways is orbital hybridization useful in describing molecules?
Vocabulary
• molecular orbitals
• bonding orbital
• sigma bond
• pi bond
• tetrahedral angle
• VSEPR theory
• hybridization
Key Concepts
• How are atomic and molecular orbitals related?
• How does VSEPR theory help predict the shapes of molecules?
• In what ways is orbital hybridization useful in describing molecules?
Vocabulary
• molecular orbitals
• bonding orbital
• sigma bond
• pi bond
• tetrahedral angle
• VSEPR theory
• hybridization
Problems Ch 16
(16) Key Concept How do you calculate the molarity of a solution? Hint
(17) Key Concept Compare the number of moles of solute before dilution with the number of moles of solute after dilution. Hint
(18) Key Concept What are two ways of expressing the concentration of a solution as a percent? Hint
(19)Calculate the molarity of a solution containing 400 g CuSO4 in 4.00 L of solution.
(20)How many moles of solute are present in 50.0 mL of 0.20M KNO3?
(21)How many milliliters of a stock solution of 2.00M KNO3 would you need to prepare 100.0 mL of 0.150M KNO3?
(22)What is the concentration, in percent (v/v), of a solution containing 50 mL of diethyl ether (C4H10O) in 2.5 L of solution?
(23)How many grams of K2SO4 would you need to prepare 1500 g of 5.0% K2SO4 (m/m) solution?
(17) Key Concept Compare the number of moles of solute before dilution with the number of moles of solute after dilution. Hint
(18) Key Concept What are two ways of expressing the concentration of a solution as a percent? Hint
(19)Calculate the molarity of a solution containing 400 g CuSO4 in 4.00 L of solution.
(20)How many moles of solute are present in 50.0 mL of 0.20M KNO3?
(21)How many milliliters of a stock solution of 2.00M KNO3 would you need to prepare 100.0 mL of 0.150M KNO3?
(22)What is the concentration, in percent (v/v), of a solution containing 50 mL of diethyl ether (C4H10O) in 2.5 L of solution?
(23)How many grams of K2SO4 would you need to prepare 1500 g of 5.0% K2SO4 (m/m) solution?
Ch 16: Percent Mass
Concentration in Percent (Mass/Mass)
Another way to express the concentration of a solution is as a percent (mass/mass), which is the number of grams of solute in 100 grams of solution. Percent by mass is sometimes a convenient unit of concentration when the solute is a solid. For example, a solution containing 7 g of sodium chloride in 100 grams of solution is 7 percent (mass/mass), or 7% (m/m).
Suppose you want to make 2000 g of a solution of glucose in water that has a 2.8% (m/m) concentration of glucose. How much glucose should you use? In a 2.8 percent solution, each 100 g of solution contains 2.8 g of solute. Thus, to find the grams of solute, you can multiply the mass of the solution by the ratio of grams of solute to grams of solution.
How much solvent (water) should be used? The mass of the solvent equals the mass of the solution minus the mass of the solute or 1944 g (2000 g − 56 g). Thus a 2.8% (m/m) glucose solution contains 56 g of glucose dissolved in 1944 g of water.
Information is often expressed as percent composition on food labels. For example, the label on a fruit drink or on maple-flavored pancake syrup should show the percent of fruit juice or maple syrup contained in the product. Such information can be misleading unless the units are given. When you use percentages to express concentration, be sure to state the units (v/v) or (m/m).
Another way to express the concentration of a solution is as a percent (mass/mass), which is the number of grams of solute in 100 grams of solution. Percent by mass is sometimes a convenient unit of concentration when the solute is a solid. For example, a solution containing 7 g of sodium chloride in 100 grams of solution is 7 percent (mass/mass), or 7% (m/m).
Suppose you want to make 2000 g of a solution of glucose in water that has a 2.8% (m/m) concentration of glucose. How much glucose should you use? In a 2.8 percent solution, each 100 g of solution contains 2.8 g of solute. Thus, to find the grams of solute, you can multiply the mass of the solution by the ratio of grams of solute to grams of solution.
How much solvent (water) should be used? The mass of the solvent equals the mass of the solution minus the mass of the solute or 1944 g (2000 g − 56 g). Thus a 2.8% (m/m) glucose solution contains 56 g of glucose dissolved in 1944 g of water.
Information is often expressed as percent composition on food labels. For example, the label on a fruit drink or on maple-flavored pancake syrup should show the percent of fruit juice or maple syrup contained in the product. Such information can be misleading unless the units are given. When you use percentages to express concentration, be sure to state the units (v/v) or (m/m).
CH 16: Percent Solutions
Percent Solutions
Another way to describe the concentration of a solution is by the percent of a solute in the solvent. The concentration of a solution in percent can be expressed in two ways: as the ratio of the volume of the solute to the volume of the solution or as the ratio of the mass of the solute to the mass of the solution.
Concentration in Percent (Volume/Volume)
If both the solute and the solvent are liquids, a convenient way to make a solution is to measure the volumes of the solute and the solution. The concentration of the solute is then expressed as a percent of the solution by volume. For example, isopropyl alcohol (2-propanol) is sold as a 91% solution, as shown in Figure 16.12. This solution consists of 91 mL of isopropyl alcohol mixed with enough water to make 100 mL of solution. The concentration can be expressed as 91 percent (volume/volume), or 91% (v/v). The relationship between percent by volume and the volumes of solute and solution is
Figure 16.12 The label clearly distinguishes this solution of isopropyl alcohol from rubbing alcohol which is a 70% solution of isopropyl alcohol. Applying Concepts How many milliliters of isopropyl alcohol are in 100 mL of 91% alcohol?
Another way to describe the concentration of a solution is by the percent of a solute in the solvent. The concentration of a solution in percent can be expressed in two ways: as the ratio of the volume of the solute to the volume of the solution or as the ratio of the mass of the solute to the mass of the solution.
Concentration in Percent (Volume/Volume)
If both the solute and the solvent are liquids, a convenient way to make a solution is to measure the volumes of the solute and the solution. The concentration of the solute is then expressed as a percent of the solution by volume. For example, isopropyl alcohol (2-propanol) is sold as a 91% solution, as shown in Figure 16.12. This solution consists of 91 mL of isopropyl alcohol mixed with enough water to make 100 mL of solution. The concentration can be expressed as 91 percent (volume/volume), or 91% (v/v). The relationship between percent by volume and the volumes of solute and solution is
Figure 16.12 The label clearly distinguishes this solution of isopropyl alcohol from rubbing alcohol which is a 70% solution of isopropyl alcohol. Applying Concepts How many milliliters of isopropyl alcohol are in 100 mL of 91% alcohol?
Ch 16: Dilutions
Figure 16.10 The student is preparing 100 mL of 0.40M MgSO4 from a stock solution of 2.0M MgSO4. She measures 20 mL of the stock solution with a 20-mL pipet. She transfers the 20 mL to a 100-mL volumetric flask. She carefully adds water to the mark to make 100 mL of solution. Inferring How many significant figures does the new molarity have?
Moles of solute before dilution = moles of solute after dilution
Recall the definition of molarity.
Rearranging the equation gives an expression for moles of solute.
Moles of solute = molarity (M) × liters of solution (V)
The total number of moles of solute remains unchanged upon dilution, so you can write this equation.
Moles of solute = M1 × V1 = M2 × V2
M1 and V1 are the molarity and volume of the initial solution, and M2 and V2 are the molarity and volume of the diluted solution. Volumes can be in liters or milliliters, as long as the same units are used for both V1 and V2.
Moles of solute before dilution = moles of solute after dilution
Recall the definition of molarity.
Rearranging the equation gives an expression for moles of solute.
Moles of solute = molarity (M) × liters of solution (V)
The total number of moles of solute remains unchanged upon dilution, so you can write this equation.
Moles of solute = M1 × V1 = M2 × V2
M1 and V1 are the molarity and volume of the initial solution, and M2 and V2 are the molarity and volume of the diluted solution. Volumes can be in liters or milliliters, as long as the same units are used for both V1 and V2.
Ch 16: Molarity Calculations
Molarity revisited
Sometimes, you may need to determine the number of moles of solute dissolved in a given volume of solution. You can do this if the molarity of the solution is known. For example, how many moles are in 2.00 L of 2.5M lithium chloride (LiCl)? Rearrange the formula for molarity to solve for the number of moles.
Moles of solute = molarity (M) × liters of solution (V)
Thus 2.00 L of 2.5 M lithium chloride solution contains 5.0 mol LiCl.
Sometimes, you may need to determine the number of moles of solute dissolved in a given volume of solution. You can do this if the molarity of the solution is known. For example, how many moles are in 2.00 L of 2.5M lithium chloride (LiCl)? Rearrange the formula for molarity to solve for the number of moles.
Moles of solute = molarity (M) × liters of solution (V)
Thus 2.00 L of 2.5 M lithium chloride solution contains 5.0 mol LiCl.
Ch 16: Molarity Calculations
Key Concepts
• How do you calculate the molarity of a solution?
• What effect does dilution have on the total moles of solute in solution?
• What are two ways to express the percent concentration of a solution?
Vocabulary
• concentration
• dilute solution
• concentrated solution
• molarity (M)
Reading Strategy
Summarizing When you summarize, you restate the key ideas in your own words. As you read about molarity, making dilutions, and percent solutions, summarize the main ideas in the text. In your summary, be sure to include all the key terms and the sentences in boldfaced type.
• How do you calculate the molarity of a solution?
• What effect does dilution have on the total moles of solute in solution?
• What are two ways to express the percent concentration of a solution?
Vocabulary
• concentration
• dilute solution
• concentrated solution
• molarity (M)
Reading Strategy
Summarizing When you summarize, you restate the key ideas in your own words. As you read about molarity, making dilutions, and percent solutions, summarize the main ideas in the text. In your summary, be sure to include all the key terms and the sentences in boldfaced type.
Ch 16. Solubility factors: Pressure
Pressure
Changes in pressure have little affect on the solubility of solids and liquids, but pressure strongly influences the solubility of gases. Gas solubility increases as the partial pressure of the gas above the solution increases. Carbonated beverages are a good example. These drinks contain large amounts of carbon dioxide (CO2) dissolved in water. Dissolved CO2 makes the liquid fizz and your mouth tingle. The drinks are bottled under a high pressure of CO2 gas, which forces large amounts of the gas into solution. When a carbonated-beverage container is opened, the partial pressure of CO2 above the liquid decreases. Immediately, bubbles of CO2 form in the liquid and escape from the open bottle,
Changes in pressure have little affect on the solubility of solids and liquids, but pressure strongly influences the solubility of gases. Gas solubility increases as the partial pressure of the gas above the solution increases. Carbonated beverages are a good example. These drinks contain large amounts of carbon dioxide (CO2) dissolved in water. Dissolved CO2 makes the liquid fizz and your mouth tingle. The drinks are bottled under a high pressure of CO2 gas, which forces large amounts of the gas into solution. When a carbonated-beverage container is opened, the partial pressure of CO2 above the liquid decreases. Immediately, bubbles of CO2 form in the liquid and escape from the open bottle,
Ch 16. Solubility factors: Pressure
Pressure
Changes in pressure have little affect on the solubility of solids and liquids, but pressure strongly influences the solubility of gases. Gas solubility increases as the partial pressure of the gas above the solution increases. Carbonated beverages are a good example. These drinks contain large amounts of carbon dioxide (CO2) dissolved in water. Dissolved CO2 makes the liquid fizz and your mouth tingle. The drinks are bottled under a high pressure of CO2 gas, which forces large amounts of the gas into solution. When a carbonated-beverage container is opened, the partial pressure of CO2 above the liquid decreases. Immediately, bubbles of CO2 form in the liquid and escape from the open bottle,
Changes in pressure have little affect on the solubility of solids and liquids, but pressure strongly influences the solubility of gases. Gas solubility increases as the partial pressure of the gas above the solution increases. Carbonated beverages are a good example. These drinks contain large amounts of carbon dioxide (CO2) dissolved in water. Dissolved CO2 makes the liquid fizz and your mouth tingle. The drinks are bottled under a high pressure of CO2 gas, which forces large amounts of the gas into solution. When a carbonated-beverage container is opened, the partial pressure of CO2 above the liquid decreases. Immediately, bubbles of CO2 form in the liquid and escape from the open bottle,
Ch 16. Solubility factors: Temperature on Gases
The effect of temperature on the solubility of gases in liquid solvents is opposite that of solids. The solubilities of most gases are greater in cold water than in hot. For example, Table 16.1 shows that the most important component of air for living beings—oxygen—becomes less soluble in water as the temperature of the solution rises. This fact has some important consequences. When an industrial plant takes water from a lake to use for cooling and then dumps the resulting heated water back into the lake, the temperature of the entire lake increases. Such a change in temperature is known as thermal pollution. Aquatic animal and plant life can be severely affected because the increase in temperature lowers the concentration of dissolved oxygen in the lake water.
Ch 16: Super saturation
Suppose you make a saturated solution of sodium ethanoate (sodium acetate) at 30°C and let the solution stand undisturbed as it cools to 25°C. Because the solubility of this compound is greater at 30°C than at 25°C, you expect that solid sodium ethanoate will crystallize from the solution as the temperature drops. But no crystals form. You have made a supersaturated solution. A supersaturated solution contains more solute than it can theoretically hold at a given temperature. The crystallization of a supersaturated solution can be initiated if a very small crystal, called a seed crystal, of the solute is added. The rate at which excess solute deposits upon the surface of a seed crystal can be very rapid, as shown in Figure 16.6. Crystallization can also occur if the inside of the container is scratched.
Figure 16.6 A supersaturated solution crystallizes rapidly when disturbed. The solution is clear before a seed crystal is added. Crystals begin to form in the solution immediately after the addition of a seed crystal. Excess solute crystallizes rapidly. Applying Concepts When the crystallization has ceased, will the solution be saturated or unsaturated?
Another example of crystallization in a supersaturated solution is the production of rock candy. A solution is supersaturated with sugar. Seed crystals cause the sugar to crystallize out of solution onto a string for you to enjoy!
Figure 16.6 A supersaturated solution crystallizes rapidly when disturbed. The solution is clear before a seed crystal is added. Crystals begin to form in the solution immediately after the addition of a seed crystal. Excess solute crystallizes rapidly. Applying Concepts When the crystallization has ceased, will the solution be saturated or unsaturated?
Another example of crystallization in a supersaturated solution is the production of rock candy. A solution is supersaturated with sugar. Seed crystals cause the sugar to crystallize out of solution onto a string for you to enjoy!
Ch 16. Solubility factors
Factors Affecting Solubility
You have read that solubility is defined as the mass of solute that dissolves in a given mass of a solvent at a specified temperature. Temperature affects the solubility of solid, liquid, and gaseous solutes in a solvent; both temperature and pressure affect the solubility of gaseous solutes.
Temperature
The solubility of most solid substances increases as the temperature of the solvent increases. Figure 16.4 shows how the solubility of several substances changes as temperature increases. The mineral deposits around hot springs, such as the one shown in Figure 16.5, result from the cooling of the hot, saturated solution of minerals emerging from the spring. As the solution cools in air, it cannot contain the same concentration of minerals as it did at a higher temperature, so some of the minerals precipitate.
Figure 16.5 Mineral deposits form around the edges of this hot spring because the hot water is saturated with minerals. As the water cools, some of the minerals crystallize because they are less soluble at the lower temperature.
For a few substances, solubility decreases with temperature. For example, the solubility of ytterbium sulfate (Yb2(SO4)3) in water drops from 44.2 g per 100 g of water at 0°C to 5.8 g per 100 g of water at 90°C. Table 16.1 lists the solubilities of some common substances at various temperatures.
You have read that solubility is defined as the mass of solute that dissolves in a given mass of a solvent at a specified temperature. Temperature affects the solubility of solid, liquid, and gaseous solutes in a solvent; both temperature and pressure affect the solubility of gaseous solutes.
Temperature
The solubility of most solid substances increases as the temperature of the solvent increases. Figure 16.4 shows how the solubility of several substances changes as temperature increases. The mineral deposits around hot springs, such as the one shown in Figure 16.5, result from the cooling of the hot, saturated solution of minerals emerging from the spring. As the solution cools in air, it cannot contain the same concentration of minerals as it did at a higher temperature, so some of the minerals precipitate.
Figure 16.5 Mineral deposits form around the edges of this hot spring because the hot water is saturated with minerals. As the water cools, some of the minerals crystallize because they are less soluble at the lower temperature.
For a few substances, solubility decreases with temperature. For example, the solubility of ytterbium sulfate (Yb2(SO4)3) in water drops from 44.2 g per 100 g of water at 0°C to 5.8 g per 100 g of water at 90°C. Table 16.1 lists the solubilities of some common substances at various temperatures.
Word Origins Miscible
Word Origins
Miscible comes from the Latin word miscere, meaning “to mix.” Completely miscible liquids dissolve in each other in all proportions. If the prefix im- means “not,” what would you call two liquids that are insoluble in each other?
Miscible comes from the Latin word miscere, meaning “to mix.” Completely miscible liquids dissolve in each other in all proportions. If the prefix im- means “not,” what would you call two liquids that are insoluble in each other?
Ch 16.1 Solution Formation
1. Stirring and Solution Formation
If a teaspoon of granulated sugar (sucrose) is placed in a glass of tea, the crystals dissolve slowly. If the contents of the glass are stirred, however, the crystals dissolve more quickly. The dissolving process occurs at the surface of the sugar crystals. Stirring speeds up the process because fresh solvent (the water in tea) is continually brought into contact with the surface of the solute (sugar). It's important to realize, however, that agitation (stirring or shaking) affects only the rate at which a solid solute dissolves. It does not influence the amount of solute that will dissolve. An insoluble substance remains undissolved regardless of how vigorously or for how long the solvent/solute system is agitated.
Temperature and Solution Formation
Temperature also influences the rate at which a solute dissolves. Sugar dissolves much more rapidly in hot tea than in iced tea. At higher temperatures, the kinetic energy of water molecules is greater than at lower temperatures so they move faster. The more rapid motion of the solvent molecules leads to an increase in the frequency and the force of the collisions between water molecules and the surfaces of the sugar crystals.
Particle Size and Solution Formation
The rate at which a solute dissolves also depends upon the size of the solute particles. A spoonful of granulated sugar dissolves more quickly than a sugar cube because the smaller particles in granulated sugar expose a much greater surface area to the colliding water molecules. Remember, the dissolving process is a surface phenomenon. The more surface of the solute that is exposed, the faster the rate of dissolving.
If a teaspoon of granulated sugar (sucrose) is placed in a glass of tea, the crystals dissolve slowly. If the contents of the glass are stirred, however, the crystals dissolve more quickly. The dissolving process occurs at the surface of the sugar crystals. Stirring speeds up the process because fresh solvent (the water in tea) is continually brought into contact with the surface of the solute (sugar). It's important to realize, however, that agitation (stirring or shaking) affects only the rate at which a solid solute dissolves. It does not influence the amount of solute that will dissolve. An insoluble substance remains undissolved regardless of how vigorously or for how long the solvent/solute system is agitated.
Temperature and Solution Formation
Temperature also influences the rate at which a solute dissolves. Sugar dissolves much more rapidly in hot tea than in iced tea. At higher temperatures, the kinetic energy of water molecules is greater than at lower temperatures so they move faster. The more rapid motion of the solvent molecules leads to an increase in the frequency and the force of the collisions between water molecules and the surfaces of the sugar crystals.
Particle Size and Solution Formation
The rate at which a solute dissolves also depends upon the size of the solute particles. A spoonful of granulated sugar dissolves more quickly than a sugar cube because the smaller particles in granulated sugar expose a much greater surface area to the colliding water molecules. Remember, the dissolving process is a surface phenomenon. The more surface of the solute that is exposed, the faster the rate of dissolving.
Ch 16: Solutions Key Terms/Questions
16.1
Key Concepts
1. What factors determine the rate at which a substance dissolves?
2. How is solubility usually expressed?
3. What conditions determine the amount of solute that will dissolve in a given solvent?
Vocabulary
• saturated solution
• solubility
• unsaturated solution
• miscible
• immiscible
• supersaturated solution
• Henry’s law
Key Concepts
1. What factors determine the rate at which a substance dissolves?
2. How is solubility usually expressed?
3. What conditions determine the amount of solute that will dissolve in a given solvent?
Vocabulary
• saturated solution
• solubility
• unsaturated solution
• miscible
• immiscible
• supersaturated solution
• Henry’s law
Changes to AC Chemistry
Mr. Gary is implementing a new program in lieu of our action research plan for the academic year 2010. The impact of multimedia and online assistance for class supplementation of work, quizzes, and laboratory assignments will be tested as well as an extensive assessment of student understanding, providing tiered lessons for both the advanced and remedial student. Pragmatically, what does this mean? More structure, more testing, and more accountability combined with more opportunities for bonus and enrichment activities.
Step 1. www.acchemistry.blogspot.com will be updated weekly to include all assignments for the week for each section—from reading, notes, questions to be answered, and quiz topics.
Step 2. Every double period will involve a quiz on the vocabulary of that week’s section, pertinent problems in the notes, and past problems.
Step 3. Notebooks will be checked every two weeks along with progress reports for any deficient students being sent home to parents.
Step 4. The website will have multimedia sections, further notes, and problem sets that can be done for improvement as well as an opportunity to raise quiz scores with a quiz make up—up to one grade.
Step 5. Grades will be tracked and students surveyed at the end of sessions comparing past performance with current performance to see if there are any improvements. Hits will be tracked on the website to see if activity increases over time. Finally, notebook samples will be checked to see any improvement.
Step 1. www.acchemistry.blogspot.com will be updated weekly to include all assignments for the week for each section—from reading, notes, questions to be answered, and quiz topics.
Step 2. Every double period will involve a quiz on the vocabulary of that week’s section, pertinent problems in the notes, and past problems.
Step 3. Notebooks will be checked every two weeks along with progress reports for any deficient students being sent home to parents.
Step 4. The website will have multimedia sections, further notes, and problem sets that can be done for improvement as well as an opportunity to raise quiz scores with a quiz make up—up to one grade.
Step 5. Grades will be tracked and students surveyed at the end of sessions comparing past performance with current performance to see if there are any improvements. Hits will be tracked on the website to see if activity increases over time. Finally, notebook samples will be checked to see any improvement.
Wednesday, August 18, 2010
Debate goes to PDS Tuesday Aug 24, 2010
The Debate Team will be going to PDS on Tuesday Aug 24, 2010 for a crosstraining from 4:30 p.m. to 7:30 p.m.
Retest EP 4
do this hand written in your notebook:
page 123 #58, 61, 62, 70, 73,
page 124 #85
page 125 #1-9
page 123 #58, 61, 62, 70, 73,
page 124 #85
page 125 #1-9
Subscribe to:
Posts (Atom)
About Me
- Gary Hi10spro Sakuma
- I have played for 25 years and coached for the last 17 years--certified United States Professional Tennis Association Professional One--worked for Punahou Schools-voted the #1 Sports School in the United States, as a Program Supervisor, in charge of coaching the High Performance Players as well as coordinating programs for K-12 and Tennis Pro Education.