Acids, Bases, and Salts Module 6: Characteristics of Acids and Bases Strong and Weak Acids and Bases...
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Transcript of Acids, Bases, and Salts Module 6: Characteristics of Acids and Bases Strong and Weak Acids and Bases...
![Page 1: Acids, Bases, and Salts Module 6: Characteristics of Acids and Bases Strong and Weak Acids and Bases Ionization of Water pH and pOH Neutralization Reactions.](https://reader035.fdocuments.net/reader035/viewer/2022070400/56649f145503460f94c2846e/html5/thumbnails/1.jpg)
Acids, Bases, and SaltsModule 6:
Characteristics of Acids and Bases
Strong and Weak Acids and Bases
Ionization of Water
pH and pOH
Neutralization Reactions
Buffers
How Buffers Maintain pH Constant
Acidosis and Alkalosis
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Properties of Acids
Produce hydrogen (H+) or hydronium (H3O+) ions when dissolved in water
Taste: Sour Are electrolytes. (Electrolytes are substances that conduct
electricity.) Turns blue litmus paper red Neutralizes bases Has a pH level ranging from -1 to 7
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Properties of Bases
Produce OH- in aqueous solutions. Taste: Bitter Turn red litmus paper blue Are electrolytes. e) React with acids. f) Feel slippery or soapy Has a pH level ranging from 7 to 14
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Acids and Bases
H2O
Acid Base
MOH
hydroxide ion written at end
metal or ammonium ion
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Strong and Weak Acids and Bases
Strong acids and bases dissociate completely, so they are considered strong electrolytes
Compounds that dissociates less than 50% are considered weak electrolytes (weak acids and bases).
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Strong Acids
Strong Acid: are strong electrolytes.
Strong acids ionize completely.
A strong acid examples are: HCl, HBr, HI, HClO4, HNO3, H2SO4
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Strong Bases
Strong Base: are also strong electrolytes
All salts/bases dissociate 100%
A strong base consists of:a [Group 1A metal] + [Hydroxides] and Ca(OH)2 , Sr(OH)2 , Ba(OH)2
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Acids and Bases
Substance Strong Electrolyte
Weak Electrolyte
HBr
Ni(OH)2
KOH
HCN
Tap water
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Acids and Bases
Substance Strong Electrolyte
Weak Electrolyte
HBr
Ni(OH)2
KOH
HCN
Tap water
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Ionization of Water
The above equation is often simplified as follows:
H+ + OH- H2O
H2O + H2O H3O+ + OH-
hydronium hydroxide
The double arrow indicates that this is an equilibrium
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http://www.chem.umass.edu/genchem/whelan/class_images/110_pH_Scale.jpg
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pH and pOH
Notice that pH + pOH = 14
0123456
141312111098
100
10-1
10-2
10-3
10-4
10-5
10-6
10-14
10-13
10-12
10-11
10-10
10-9
10-8
7 7 10-7 10-7 neutral
891011121314
6543210
10-8
10-9
10-10
10-11
10-12
10-13
10-14
10-6
10-5
10-4
10-3
10-2
10-1
100
pH pOH [H+] [OH-]
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pH and pOH
pH pOH [H+] [OH-] acidic, basic, or neutral
11
10-9
7
10-4
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pH and pOH
pH pOH [H+] [OH-] acidic, basic, or neutral
3 11 10-3 10-11 acidic
9 5 10-9 10-5 basic
7 7 10-7 10-7 neutral
10 4 10-10 10-4 basic
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Neutralization Reactions
the salt is formed when the positive ion from the base combines
with the negative ion from acid
water is formed when H+
combines with OH-
NaCl + HOHHCl + NaOH
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Neutralization Reactions
NiCl3 + LiOH Ni(OH)3 + LiCl
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Neutralization Reactions
NiCl3 + LiOH Ni(OH)3 + LiCl
NiCl3 + LiOH Ni(OH)3 + 3 LiCl
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Neutralization Reactions
NiCl3 + LiOH Ni(OH)3 + LiCl
NiCl3 + LiOH Ni(OH)3 + 3 LiCl
NiCl3 + 3 LiOH Ni(OH)3 + 3 LiCl
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Buffers
•A buffer solution is a solution whose pH remains relatively constant when either acids or bases are added to it.
•Buffers do not have an unlimited ability to resist pH changes. This is the buffering capacity.
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Buffers
•A buffer solution is a solution whose pH remains relatively constant when either acids or bases are added to it.
•Buffers do not have an unlimited ability to resist pH changes. This is the buffering capacity.
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Buffers•Buffers must be able to react with excess acid or excess base
•It has two components•one to react with the acid and •one to react with the base
•Most common buffers consist of:•a weak acid and its salt or •a weak base and its salt
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Example Calculate the mass of NaCl found in 200
mL of 6.0% (m/v) NaCl solution.
iongNaClsolutmLsolution
gramsNaClmLsolution
12100
06200
.
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Example What volume of a solution that is 5.0% (m/v)
copper (II) nitrate contains 50.0g of Cu(NO3) 2?
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Example What volume of a solution that is 5.0% copper
(II) nitrate contains 50.0g of Cu(NO3) 2?
mLsolution
NOgCuNOCu
100
0505 23
23
)(.)(%.
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Example What volume of a solution that is 5.0% copper
(II) nitrate contains 50.0g of Cu(NO3) 2?
23
23 05
100050
)(.)(.
NOgCu
mLsolutionNOgCu
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Example What volume of a solution that is 5.0% copper
(II) nitrate contains 50.0g of Cu(NO3) 2?
mLsolution
NOgCu
mLsolutionNOgCu
1000
05
100050
2323
)(.)(.
![Page 27: Acids, Bases, and Salts Module 6: Characteristics of Acids and Bases Strong and Weak Acids and Bases Ionization of Water pH and pOH Neutralization Reactions.](https://reader035.fdocuments.net/reader035/viewer/2022070400/56649f145503460f94c2846e/html5/thumbnails/27.jpg)
Example
Calculate the percent v/v of a solution prepared by dissolving 25 ml of ethanol in 50mL water (assume the volumes are additive).
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Example Calculate the percent v/v of a solution prepared
by dissolving 25 ml of ethanol in 50mL water (assume the volumes are additive).
First, we need to determine the total volume of the solution:
25mL ethanol + 50 mL water = 75 mL solution
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Example
Calculate the percent v/v of a solution prepared by dissolving 25 ml of ethanol in 50mL water (assume the volumes are additive).
First, we need to determine the total volume of the solution:
25mL ethanol + 50 mL water = 75 mL solution
mLsolution
mLethanol
75
25
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Example
Calculate the percent v/v of a solution prepared by dissolving 25 ml of ethanol in 50mL water (assume the volumes are additive).
First, we need to determine the total volume of the solution:
25mL ethanol + 50 mL water = 75 mL solution
10075
25
mLsolution
mLethanol
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Example
Calculate the percent v/v of a solution prepared by dissolving 25 ml of ethanol in 50mL water (assume the volumes are additive).
First, we need to determine the total volume of the solution:
25mL ethanol + 50 mL water = 75 mL solution
solution ethanol %3310075
25
mLsolution
mLethanol
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Molarity
Molarity
Moles solute = M Liter of solution
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Example What is the molarity (M) of a solution that
contains 30.0g copper (II) nitrate in 250mL solution?
![Page 34: Acids, Bases, and Salts Module 6: Characteristics of Acids and Bases Strong and Weak Acids and Bases Ionization of Water pH and pOH Neutralization Reactions.](https://reader035.fdocuments.net/reader035/viewer/2022070400/56649f145503460f94c2846e/html5/thumbnails/34.jpg)
Example What is the molarity (M) of a solution that
contains 30.0g copper (II) nitrate in 250mL solution?
30 01
187 53 2
3 2
3 2
. ( )( )
. ( )gCu NO
moleCu NO
gCu NO
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Example What is the molarity (M) of a solution that
contains 30.0g copper (II) nitrate in 250mL solution?
30 01
187 5
0160
3 2
3 2
3 2
3 2
. ( )( )
. ( )
. ( )
gCu NOmoleCu NO
gCu NO
moleCu NO
![Page 36: Acids, Bases, and Salts Module 6: Characteristics of Acids and Bases Strong and Weak Acids and Bases Ionization of Water pH and pOH Neutralization Reactions.](https://reader035.fdocuments.net/reader035/viewer/2022070400/56649f145503460f94c2846e/html5/thumbnails/36.jpg)
Example What is the molarity (M) of a solution that
contains 30.0g copper (II) nitrate in 250mL solution?
2501
1000mLsolution
Lsolution
mLsolution
![Page 37: Acids, Bases, and Salts Module 6: Characteristics of Acids and Bases Strong and Weak Acids and Bases Ionization of Water pH and pOH Neutralization Reactions.](https://reader035.fdocuments.net/reader035/viewer/2022070400/56649f145503460f94c2846e/html5/thumbnails/37.jpg)
Example What is the molarity (M) of a solution that
contains 30.0g copper (II) nitrate in 250mL solution?
2501
10000 250
mLsolutionLsolution
mLsolutionLsolution
.
![Page 38: Acids, Bases, and Salts Module 6: Characteristics of Acids and Bases Strong and Weak Acids and Bases Ionization of Water pH and pOH Neutralization Reactions.](https://reader035.fdocuments.net/reader035/viewer/2022070400/56649f145503460f94c2846e/html5/thumbnails/38.jpg)
Example What is the molarity (M) of a solution that
contains 30.0g copper (II) nitrate in 250mL solution?
0160
0 2503 2. ( )
.
moleCu NO
Lsolution
![Page 39: Acids, Bases, and Salts Module 6: Characteristics of Acids and Bases Strong and Weak Acids and Bases Ionization of Water pH and pOH Neutralization Reactions.](https://reader035.fdocuments.net/reader035/viewer/2022070400/56649f145503460f94c2846e/html5/thumbnails/39.jpg)
Example What is the molarity (M) of a solution that
contains 30.0g copper (II) nitrate in 250mL solution?
0160
0 2500 640
3 2
3 2
. ( )
.. ( )
moleCu NO
LsolutionMCu NO
![Page 40: Acids, Bases, and Salts Module 6: Characteristics of Acids and Bases Strong and Weak Acids and Bases Ionization of Water pH and pOH Neutralization Reactions.](https://reader035.fdocuments.net/reader035/viewer/2022070400/56649f145503460f94c2846e/html5/thumbnails/40.jpg)
Example Calculate the mass of NaOH needed to
prepare 150mL of a 0.220M solution of sodium hydroxide.
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Example Calculate the mass of NaOH needed to
prepare 150mL of a 0.220M solution of sodium hydroxide.
Lsolution
moleNaOHMsolution
1
22002200
..
![Page 42: Acids, Bases, and Salts Module 6: Characteristics of Acids and Bases Strong and Weak Acids and Bases Ionization of Water pH and pOH Neutralization Reactions.](https://reader035.fdocuments.net/reader035/viewer/2022070400/56649f145503460f94c2846e/html5/thumbnails/42.jpg)
Example Calculate the mass of NaOH needed to
prepare 150mL of a 0.220M solution of sodium hydroxide.
mLsolution
LsolutionmLsolution
1000
10150 .
![Page 43: Acids, Bases, and Salts Module 6: Characteristics of Acids and Bases Strong and Weak Acids and Bases Ionization of Water pH and pOH Neutralization Reactions.](https://reader035.fdocuments.net/reader035/viewer/2022070400/56649f145503460f94c2846e/html5/thumbnails/43.jpg)
Example Calculate the mass of NaOH needed to
prepare 150mL of a 0.220M solution of sodium hydroxide.
Lsolution
mLsolution
LsolutionmLsolution
1500
1000
10150
.
.
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Example Calculate the mass of NaOH needed to
prepare 150mL of a 0.220M solution of sodium hydroxide.
moleNaOH
gNaOH
Lsolution
moleNaOHLsolution
1
040
1
22001500
...
![Page 45: Acids, Bases, and Salts Module 6: Characteristics of Acids and Bases Strong and Weak Acids and Bases Ionization of Water pH and pOH Neutralization Reactions.](https://reader035.fdocuments.net/reader035/viewer/2022070400/56649f145503460f94c2846e/html5/thumbnails/45.jpg)
Example Calculate the mass of NaOH needed to
prepare 150mL of a 0.220M solution of sodium hydroxide.
gNaOH
moleNaOH
gNaOH
Lsolution
moleNaOHLsolution
321
1
040
1
22001500
.
...
![Page 46: Acids, Bases, and Salts Module 6: Characteristics of Acids and Bases Strong and Weak Acids and Bases Ionization of Water pH and pOH Neutralization Reactions.](https://reader035.fdocuments.net/reader035/viewer/2022070400/56649f145503460f94c2846e/html5/thumbnails/46.jpg)
Dilutions
V1 X C1 = V2 X C2
where C is concentration, it may be Molarity, mass%, or other concentration units
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Example Calculate the volume of 6.0MKCl needed to prepare
250mL of a 0.300M solution of potassium chloride.
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Example Calculate the volume of 6.0MKCl needed to prepare
250mL of a 0.300M solution of potassium chloride. V1 X C1 = V2 X C2
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Example Calculate the volume of 6.0MKCl needed to prepare
250mL of a 0.300M solution of potassium chloride. V1 X C1 = V2 X C2
(? )( . )( )( . . )
(? )
mL MKClsolutionmL Msol
mL
6 0250 0 300
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Example Calculate the volume of 6.0MKCl needed to prepare
250mL of a 0.300M solution of potassium chloride. V1 X C1 = V2 X C2
(? )( . )( )( . . )
(? )
mL MKClsolutionmL Msol
mL
6 0250 0 300
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Solutions, Colloids and Suspensions
SOLUTION COLLOIDS SUSPENSIONS
Solute Particles: small particles(ions or molecules)
larger particles very large particles(macromolecules)
Examples:
salted water, beer starch and water, fog, smoke, sprays
calamine lotion
Characteristics:
Transparent.Particles cannot be separated by filters or semipermeablemembranes.
Tyndall Effect.Particles can be separated by semi-permeablemembranes.
Opaque.Particles can be separated by filters.
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Osmosis
The flow of a solvent through a semipermeable membrane into a solution of higher solute concentration.
A semipermeable membrane, such as cellophane, contains tiny holes far too small to be seen but large enough to let solvent molecules pass through, but NOT large solute particles.
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Osmosis
• Compartment A – pure water
• Compartment B – glucose solution
dissolved
A B
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Osmosis
dissolved
A B osmotic pressure
Osmosis reaches an equilibrium and the level of compartment B is higher than A. These levels can be made equal again by applying an external pressure to compartment B. The amount of pressure necessary to equalize both levels is called osmotic pressure.
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Osmotic PressureDepends on the number of particles in the solution. In the blood,
Solution Osmotic pressure
Examples
hypotonic Lower than red blood cells
Pure water, solutions LESS than 5% glucose, 0.9% NaCl solution
isotonic Same as red blood cells
Plasma, 5% glucose, 0.9% NaCl solution
hypertonic Higher than red blood cells
Solutions GREATER than 5% glucose, 0.9% NaCl solution
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Dialysis
Uses a semipermeable membrane to separate dissolved ions and molecules from larger colloidal particles dispersed in the solution.
In kidneys the waste products of the blood dialyze out through a semipermeable membrane.
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Acid, Bases and Salts
THE END