SSSSSSSSSSSSSSSSSSSSSSSAcids and Bases Nice.ppt
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About This Presentation
SSSSSSSSSSSSSSSSSSSSSSSAcids and Bases Nice.pptSSSSSSSSSSSSSSSSSSSSSSSAcids and Bases Nice.pptSSSSSSSSSSSSSSSSSSSSSSSAcids and Bases Nice.pptSSSSSSSSSSSSSSSSSSSSSSSAcids and Bases Nice.pptSSSSSSSSSSSSSSSSSSSSSSSAcids and Bases Nice.pptSSSSSSSSSSSSSSSSSSSSSSSAcids and Bases Nice.pptSSSSSSSSSSSSSSSSSS...
Slide Content
11
The Chemistry of The Chemistry of
Acids and BasesAcids and Bases Chemistry I – Chapter 19Chemistry I – Chapter 19
Chemistry I HD – Chapter 16Chemistry I HD – Chapter 16
ICP – Chapter 23ICP – Chapter 23
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print "Handouts" instead of
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The Chemistry of The Chemistry of
Acids and BasesAcids and Bases Chemistry I – Chapter 19Chemistry I – Chapter 19
Chemistry I HD – Chapter 16Chemistry I HD – Chapter 16
ICP – Chapter 23ICP – Chapter 23
SAVE PAPER AND INK!!! When you
print out the notes on PowerPoint,
print "Handouts" instead of
"Slides" in the print setup. Also,
turn off the backgrounds
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22
Acid and BasesAcid and Bases
Acid and BasesAcid and Bases
33
Acid and BasesAcid and Bases
Acid and BasesAcid and Bases
44
Acid and BasesAcid and Bases
Acid and BasesAcid and Bases
55
Acids
Have a sour taste. Vinegar is a solution of acetic acid. CitrusHave a sour taste. Vinegar is a solution of acetic acid. Citrus
fruits contain citric acid.fruits contain citric acid.
React with certain metals to produce hydrogen gasReact with certain metals to produce hydrogen gas..
React with carbonates and bicarbonates to produce carbon React with carbonates and bicarbonates to produce carbon
dioxide gasdioxide gas
Have a bitter taste.Have a bitter taste.
Feel slippery. Many soaps contain bases.Feel slippery. Many soaps contain bases.
Bases
Acids
Have a sour taste. Vinegar is a solution of acetic acid. CitrusHave a sour taste. Vinegar is a solution of acetic acid. Citrus
fruits contain citric acid.fruits contain citric acid.
React with certain metals to produce hydrogen gasReact with certain metals to produce hydrogen gas..
React with carbonates and bicarbonates to produce carbon React with carbonates and bicarbonates to produce carbon
dioxide gasdioxide gas
Have a bitter taste.Have a bitter taste.
Feel slippery. Many soaps contain bases.Feel slippery. Many soaps contain bases.
Bases
66
Some Properties of Acids
Produce H
+
(as H
3O
+
) ions in water (the hydronium ion is a
hydrogen ion attached to a water molecule)
Taste sour
Corrode metals
Electrolytes
React with bases to form a salt and water
pH is less than 7
Turns blue litmus paper to red “Blue to Red A-CID”
Some Properties of Acids
Produce H
+
(as H
3O
+
) ions in water (the hydronium ion is a
hydrogen ion attached to a water molecule)
Taste sour
Corrode metals
Electrolytes
React with bases to form a salt and water
pH is less than 7
Turns blue litmus paper to red “Blue to Red A-CID”
77
Anion
Ending Acid Name
-ide hydro-(stem)-ic acid
-ate (stem)-ic acid
-ite (stem)-ous acid
Acid Nomenclature Review
No OxygenNo Oxygen
w/Oxygen w/Oxygen
An easy way to remember which goes with which…An easy way to remember which goes with which…
““In the cafeteria, you In the cafeteria, you ATEATE something something ICICky”ky”
Anion
Ending Acid Name
-ide hydro-(stem)-ic acid
-ate (stem)-ic acid
-ite (stem)-ous acid
Acid Nomenclature Review
No OxygenNo Oxygen
w/Oxygen w/Oxygen
An easy way to remember which goes with which…An easy way to remember which goes with which…
““In the cafeteria, you In the cafeteria, you ATEATE something something ICICky”ky”
88
•HBr HBr
(aq)(aq)
•HH
22COCO
33
•HH
22SOSO
33
hydrohydrobromicbromic acidacid
carboncarbonicic acidacid
sulfursulfurousous acidacid
Acid Nomenclature Review
•HBr HBr
(aq)(aq)
•HH
22COCO
33
•HH
22SOSO
33
hydrohydrobromicbromic acidacid
carboncarbonicic acidacid
sulfursulfurousous acidacid
Acid Nomenclature Review
99
Name ‘Em!
•HI HI
(aq)(aq)
•HCl HCl
(aq)(aq)
•HH
22SOSO
33
•HNOHNO
33
•HIOHIO
44
Name ‘Em!
•HI HI
(aq)(aq)
•HCl HCl
(aq)(aq)
•HH
22SOSO
33
•HNOHNO
33
•HIOHIO
44
1010
Some Properties of Bases
Produce OHProduce OH
--
ions in water ions in water
Taste bitter, chalkyTaste bitter, chalky
Are electrolytesAre electrolytes
Feel soapy, slipperyFeel soapy, slippery
React with acids to form salts and waterReact with acids to form salts and water
pH greater than 7pH greater than 7
Turns red litmus paper to blue “Turns red litmus paper to blue “BBasic asic BBlue”lue”
Some Properties of Bases
Produce OHProduce OH
--
ions in water ions in water
Taste bitter, chalkyTaste bitter, chalky
Are electrolytesAre electrolytes
Feel soapy, slipperyFeel soapy, slippery
React with acids to form salts and waterReact with acids to form salts and water
pH greater than 7pH greater than 7
Turns red litmus paper to blue “Turns red litmus paper to blue “BBasic asic BBlue”lue”
1111
Some Common Bases
NaOHNaOHsodium hydroxidesodium hydroxide lyelye
KOHKOHpotassium hydroxidepotassium hydroxideliquid soapliquid soap
Ba(OH)Ba(OH)
22 barium hydroxidebarium hydroxide stabilizer for plasticsstabilizer for plastics
Mg(OH)Mg(OH)
22 magnesium hydroxidemagnesium hydroxide“MOM” Milk of magnesia“MOM” Milk of magnesia
Al(OH)Al(OH)
33aluminum hydroxidealuminum hydroxideMaalox (antacid)Maalox (antacid)
Some Common Bases
NaOHNaOHsodium hydroxidesodium hydroxide lyelye
KOHKOHpotassium hydroxidepotassium hydroxideliquid soapliquid soap
Ba(OH)Ba(OH)
22 barium hydroxidebarium hydroxide stabilizer for plasticsstabilizer for plastics
Mg(OH)Mg(OH)
22 magnesium hydroxidemagnesium hydroxide“MOM” Milk of magnesia“MOM” Milk of magnesia
Al(OH)Al(OH)
33aluminum hydroxidealuminum hydroxideMaalox (antacid)Maalox (antacid)
1212
Copyright McGraw-Hill 2009 12
All the other acids and bases are weak
electrolytes
Copyright McGraw-Hill 2009 12
All the other acids and bases are weak
electrolytes
1313
1.T/F. Acids pH 1-7. F 1-6.9
2.T/F. Bases taste bitter T
3.HF Hydrofluoric acid/ Hydroflouric
4.T/F. Strong bases, OH paired with elements
in Grp 3 – F Grp 1-2
5.NaOH – Sodium Hydroxide
6.H2SO4 – Sulfuric Acid
7.What is your basis of a strong acid? –
Complete dissociation, pH close to 1
8.HNO3, is it a strong acid? (Yes/No) - Yes
9.T/F Per if the no. oxygen increases T
10. Give one significance of acid or base?
1.T/F. Acids pH 1-7. F 1-6.9
2.T/F. Bases taste bitter T
3.HF Hydrofluoric acid/ Hydroflouric
4.T/F. Strong bases, OH paired with elements
in Grp 3 – F Grp 1-2
5.NaOH – Sodium Hydroxide
6.H2SO4 – Sulfuric Acid
7.What is your basis of a strong acid? –
Complete dissociation, pH close to 1
8.HNO3, is it a strong acid? (Yes/No) - Yes
9.T/F Per if the no. oxygen increases T
10. Give one significance of acid or base?
1414
Copyright McGraw-Hill 2009 14
•Types of acids
–Monoprotic: one ionizable hydrogen
HCl + H
2
O H
3
O
+
+ Cl
–Diprotic: two ionizable hydrogens
H
2
SO
4
+ H
2
O H
3
O
+
+ HSO
4
HSO
4
+ H
2
O H
3
O
+
+ SO
4
2
Copyright McGraw-Hill 2009 14
•Types of acids
–Monoprotic: one ionizable hydrogen
HCl + H
2
O H
3
O
+
+ Cl
–Diprotic: two ionizable hydrogens
H
2
SO
4
+ H
2
O H
3
O
+
+ HSO
4
HSO
4
+ H
2
O H
3
O
+
+ SO
4
2
1515
Copyright McGraw-Hill 2009 15
–Triprotic: three ionizable hydrogens
H
3
PO
4
+ H
2
O H
3
O
+
+ H
2
PO
4
H
2PO
4
+ H
2O H
3O
+
+ HPO
4
2
HPO
4
2
+ H
2
O H
3
O
+
+ PO
4
3
–Polyprotic: generic term meaning
more than one ionizable hydrogen
Copyright McGraw-Hill 2009 15
–Triprotic: three ionizable hydrogens
H
3
PO
4
+ H
2
O H
3
O
+
+ H
2
PO
4
H
2PO
4
+ H
2O H
3O
+
+ HPO
4
2
HPO
4
2
+ H
2
O H
3
O
+
+ PO
4
3
–Polyprotic: generic term meaning
more than one ionizable hydrogen
1616
Copyright McGraw-Hill 2009 16
•Types of bases
–Monobasic: One OH
group
KOH K
+
+ OH
–Dibasic: Two OH
groups
Ba(OH)
2
Ba
2+
+ 2OH
Copyright McGraw-Hill 2009 16
•Types of bases
–Monobasic: One OH
group
KOH K
+
+ OH
–Dibasic: Two OH
groups
Ba(OH)
2
Ba
2+
+ 2OH
1717
Acid/Base definitions
•Definition #1: Arrhenius (traditional)
Acids – produce H
+
ions (or hydronium ions
H
3O
+
)
Bases – produce OH
-
ions
(problem: some bases don’t have hydroxide
ions!)
Acid/Base definitions
•Definition #1: Arrhenius (traditional)
Acids – produce H
+
ions (or hydronium ions
H
3O
+
)
Bases – produce OH
-
ions
(problem: some bases don’t have hydroxide
ions!)
1818
Arrhenius acid is a substance that produces H
+
(H
3
O
+
) in water
Arrhenius base is a substance that produces OH
-
in water
Arrhenius acid is a substance that produces H
+
(H
3
O
+
) in water
Arrhenius base is a substance that produces OH
-
in water
1919
Acid/Base Definitions
•Definition #2: Brønsted – Lowry
Acids – proton donor
Bases – proton acceptor
A “proton” is really just a hydrogen
atom that has lost it’s electron!
Acid/Base Definitions
•Definition #2: Brønsted – Lowry
Acids – proton donor
Bases – proton acceptor
A “proton” is really just a hydrogen
atom that has lost it’s electron!
2020
A Brønsted-Lowry acid is a proton donor
A Brønsted-Lowry base is a proton acceptor
acid
conjugate
base
base
conjugate
acid
A Brønsted-Lowry acid is a proton donor
A Brønsted-Lowry base is a proton acceptor
acid
conjugate
base
base
conjugate
acid
2121
2222
2323
ACID-BASE THEORIESACID-BASE THEORIES
The Brønsted definition means NHThe Brønsted definition means NH
33 is is
aa BASEBASE in water — and water is in water — and water is
itself anitself an ACIDACID
BaseAcidAcidBase
NH
4
+
+ OH
-
NH
3 + H
2O
ACID-BASE THEORIESACID-BASE THEORIES
The Brønsted definition means NHThe Brønsted definition means NH
33 is is
aa BASEBASE in water — and water is in water — and water is
itself anitself an ACIDACID
BaseAcidAcidBase
NH
4
+
+ OH
-
NH
3 + H
2O
2424
Conjugate PairsConjugate Pairs
Conjugate PairsConjugate Pairs
2525
How to determine a conjugate acid
and conjugate base
How to determine a conjugate acid
and conjugate base
2626
2727
2828
2929
3030
3131
3232
3333
3434
Acids & Base DefinitionsAcids & Base Definitions
Lewis acid - a Lewis acid - a
substance that substance that
accepts an electron accepts an electron
pairpair
Lewis base - a Lewis base - a
substance that substance that
donates an electron donates an electron
pairpair
Definition #3 – Lewis
Acids & Base DefinitionsAcids & Base Definitions
Lewis acid - a Lewis acid - a
substance that substance that
accepts an electron accepts an electron
pairpair
Lewis base - a Lewis base - a
substance that substance that
donates an electron donates an electron
pairpair
Definition #3 – Lewis
3535
Formation ofFormation of hydronium ion hydronium ion is also an is also an
excellent example.excellent example.
Lewis Acids & BasesLewis Acids & Bases
•Electron pair of the new O-H bond Electron pair of the new O-H bond
originates on the Lewis base.originates on the Lewis base.
H
H
H
BASE
••
•
•
••
O—H
O—H
H
+
ACID
Formation ofFormation of hydronium ion hydronium ion is also an is also an
excellent example.excellent example.
Lewis Acids & BasesLewis Acids & Bases
•Electron pair of the new O-H bond Electron pair of the new O-H bond
originates on the Lewis base.originates on the Lewis base.
H
H
H
BASE
••
•
•
••
O—H
O—H
H
+
ACID
3636
Lewis Acid/Base ReactionLewis Acid/Base Reaction
Lewis Acid/Base ReactionLewis Acid/Base Reaction
3737
Lewis Acid-Base Interactions Lewis Acid-Base Interactions
in Biologyin Biology
•The heme group The heme group
in hemoglobin can in hemoglobin can
interact with Ointeract with O
22
and CO.and CO.
•The Fe ion in The Fe ion in
hemoglobin is a hemoglobin is a
Lewis acidLewis acid
•OO
22 and CO can act and CO can act
as Lewis basesas Lewis bases
Heme group
Lewis Acid-Base Interactions Lewis Acid-Base Interactions
in Biologyin Biology
•The heme group The heme group
in hemoglobin can in hemoglobin can
interact with Ointeract with O
22
and CO.and CO.
•The Fe ion in The Fe ion in
hemoglobin is a hemoglobin is a
Lewis acidLewis acid
•OO
22 and CO can act and CO can act
as Lewis basesas Lewis bases
Heme group
3838
The The pH scalepH scale is a way of is a way of
expressing the strength expressing the strength
of acids and bases. of acids and bases.
Instead of using very Instead of using very
small numbers, we just small numbers, we just
use the NEGATIVE use the NEGATIVE
power of 10 on the power of 10 on the
Molarity of the HMolarity of the H
++
(or (or
OHOH
--
) ion.) ion.
Under 7 = acidUnder 7 = acid
7 = neutral 7 = neutral
Over 7 = baseOver 7 = base
The The pH scalepH scale is a way of is a way of
expressing the strength expressing the strength
of acids and bases. of acids and bases.
Instead of using very Instead of using very
small numbers, we just small numbers, we just
use the NEGATIVE use the NEGATIVE
power of 10 on the power of 10 on the
Molarity of the HMolarity of the H
++
(or (or
OHOH
--
) ion.) ion.
Under 7 = acidUnder 7 = acid
7 = neutral 7 = neutral
Over 7 = baseOver 7 = base
3939
pH of Common pH of Common
SubstancesSubstances
pH of Common pH of Common
SubstancesSubstances
4040
Calculating the pH
pH = - log [H+]
(Remember that the [ ] mean Molarity)
Example: If [H
+
] = 1 X 10
-10
pH = - log 1 X 10
-10
pH = - (- 10)
pH = 10
Example: If [H
+
] = 1.8 X 10
-5
pH = - log 1.8 X 10
-5
pH = - (- 4.74)
pH = 4.74
Calculating the pH
pH = - log [H+]
(Remember that the [ ] mean Molarity)
Example: If [H
+
] = 1 X 10
-10
pH = - log 1 X 10
-10
pH = - (- 10)
pH = 10
Example: If [H
+
] = 1.8 X 10
-5
pH = - log 1.8 X 10
-5
pH = - (- 4.74)
pH = 4.74
4141
Try These!Try These!
Find the pH of Find the pH of
these:these:
1) A 0.15 M solution 1) A 0.15 M solution
of Hydrochloric of Hydrochloric
acidacid
2) A 3.00 X 102) A 3.00 X 10
-7-7
M M
solution of Nitric solution of Nitric
acidacid
Try These!Try These!
Find the pH of Find the pH of
these:these:
1) A 0.15 M solution 1) A 0.15 M solution
of Hydrochloric of Hydrochloric
acidacid
2) A 3.00 X 102) A 3.00 X 10
-7-7
M M
solution of Nitric solution of Nitric
acidacid
4242
pH calculations – Solving for H+pH calculations – Solving for H+
If the pH of Coke is 3.12, [HIf the pH of Coke is 3.12, [H
++
] = ???] = ???
Because pH = - log [HBecause pH = - log [H
++
] then] then
- pH = log [H- pH = log [H
++
]]
Take antilog (10Take antilog (10
xx
) of both) of both
sides and get sides and get
1010
-pH -pH
==
[H[H
++
]]
[H[H
++
] = 10] = 10
-3.12-3.12
= 7.6 x 10 = 7.6 x 10
-4-4
M M
*** to find antilog on your calculator, look for “Shift” or “2*** to find antilog on your calculator, look for “Shift” or “2
nd nd
function” and then the log buttonfunction” and then the log button
pH calculations – Solving for H+pH calculations – Solving for H+
If the pH of Coke is 3.12, [HIf the pH of Coke is 3.12, [H
++
] = ???] = ???
Because pH = - log [HBecause pH = - log [H
++
] then] then
- pH = log [H- pH = log [H
++
]]
Take antilog (10Take antilog (10
xx
) of both) of both
sides and get sides and get
1010
-pH -pH
==
[H[H
++
]]
[H[H
++
] = 10] = 10
-3.12-3.12
= 7.6 x 10 = 7.6 x 10
-4-4
M M
*** to find antilog on your calculator, look for “Shift” or “2*** to find antilog on your calculator, look for “Shift” or “2
nd nd
function” and then the log buttonfunction” and then the log button
4343
pH calculations – Solving for H+pH calculations – Solving for H+
•A solution has a pH of 8.5. What is the A solution has a pH of 8.5. What is the
Molarity of hydrogen ions in the Molarity of hydrogen ions in the
solution?solution?
pH = - log [HpH = - log [H
++
]]
8.5 = - log [H8.5 = - log [H
++
]]
-8.5 = log [H-8.5 = log [H
++
]]
Antilog -8.5 = antilog (log [HAntilog -8.5 = antilog (log [H
++
])])
1010
-8.5-8.5
= [H = [H
++
]]
3.16 X 103.16 X 10
-9-9
= [H = [H
++
]]
pH calculations – Solving for H+pH calculations – Solving for H+
•A solution has a pH of 8.5. What is the A solution has a pH of 8.5. What is the
Molarity of hydrogen ions in the Molarity of hydrogen ions in the
solution?solution?
pH = - log [HpH = - log [H
++
]]
8.5 = - log [H8.5 = - log [H
++
]]
-8.5 = log [H-8.5 = log [H
++
]]
Antilog -8.5 = antilog (log [HAntilog -8.5 = antilog (log [H
++
])])
1010
-8.5-8.5
= [H = [H
++
]]
3.16 X 103.16 X 10
-9-9
= [H = [H
++
]]
4444
More About Water
HH
22O can function as both an ACID and a BASE.O can function as both an ACID and a BASE.
In pure water there can beIn pure water there can be AUTOIONIZATIONAUTOIONIZATION
Equilibrium constant for water = KEquilibrium constant for water = K
ww
KK
ww = [H = [H
33OO
++
] [OH] [OH
--
] =] = 1.00 x 101.00 x 10
-14-14
at 25 at 25
oo
CC
HONORS ONLY!
More About Water
HH
22O can function as both an ACID and a BASE.O can function as both an ACID and a BASE.
In pure water there can beIn pure water there can be AUTOIONIZATIONAUTOIONIZATION
Equilibrium constant for water = KEquilibrium constant for water = K
ww
KK
ww = [H = [H
33OO
++
] [OH] [OH
--
] =] = 1.00 x 101.00 x 10
-14-14
at 25 at 25
oo
CC
HONORS ONLY!
4545
More About Water
KK
ww = [H = [H
33OO
++
] [OH] [OH
--
] = 1.00 x 10] = 1.00 x 10
-14-14
at 25 at 25
oo
CC
In a In a neutral neutral solution [Hsolution [H
33OO
++
] = [OH] = [OH
--
]]
so Kso K
ww = [H = [H
33OO
++
]]
22
= [OH = [OH
--
]]
22
and so [Hand so [H
33OO
++
] = [OH] = [OH
--
] = 1.00 x 10] = 1.00 x 10
-7-7
M M
OH
-
H
3O
+
AutoionizationAutoionization
HONORS ONLY!
More About Water
KK
ww = [H = [H
33OO
++
] [OH] [OH
--
] = 1.00 x 10] = 1.00 x 10
-14-14
at 25 at 25
oo
CC
In a In a neutral neutral solution [Hsolution [H
33OO
++
] = [OH] = [OH
--
]]
so Kso K
ww = [H = [H
33OO
++
]]
22
= [OH = [OH
--
]]
22
and so [Hand so [H
33OO
++
] = [OH] = [OH
--
] = 1.00 x 10] = 1.00 x 10
-7-7
M M
OH
-
H
3O
+
AutoionizationAutoionization
HONORS ONLY!
4646
pOH
•Since acids and bases are Since acids and bases are
opposites, pH and pOH are opposites, pH and pOH are
opposites!opposites!
•pOH does not really exist, but it is pOH does not really exist, but it is
useful for changing bases to pH.useful for changing bases to pH.
•pOH looks at the perspective of a pOH looks at the perspective of a
basebase
pOH = - log [OHpOH = - log [OH
--
]]
Since pH and pOH are on opposite Since pH and pOH are on opposite
ends,ends,
pH + pOH = 14pH + pOH = 14
pOH
•Since acids and bases are Since acids and bases are
opposites, pH and pOH are opposites, pH and pOH are
opposites!opposites!
•pOH does not really exist, but it is pOH does not really exist, but it is
useful for changing bases to pH.useful for changing bases to pH.
•pOH looks at the perspective of a pOH looks at the perspective of a
basebase
pOH = - log [OHpOH = - log [OH
--
]]
Since pH and pOH are on opposite Since pH and pOH are on opposite
ends,ends,
pH + pOH = 14pH + pOH = 14
4747
pH [H
+
] [OH
-
] pOH
pH [H
+
] [OH
-
] pOH
4848
[H[H
33OO
++
], [OH], [OH
--
] and pH] and pH
What is the pH of the What is the pH of the 0.0010 M NaOH solution? 0.0010 M NaOH solution?
[OH-] = 0.0010 (or 1.0 X 10[OH-] = 0.0010 (or 1.0 X 10
-3-3
M) M)
pOH = - log 0.0010pOH = - log 0.0010
pOH = 3pOH = 3
pH = 14 – 3 = 11pH = 14 – 3 = 11
OR KOR K
ww = [H = [H
33OO
++
] [OH] [OH
--
]]
[H[H
3OO
++
] = 1.0 x 10] = 1.0 x 10
-11-11
M M
pH = - log (1.0 x 10pH = - log (1.0 x 10
-11-11
) = 11.00) = 11.00
[H[H
33OO
++
], [OH], [OH
--
] and pH] and pH
What is the pH of the What is the pH of the 0.0010 M NaOH solution? 0.0010 M NaOH solution?
[OH-] = 0.0010 (or 1.0 X 10[OH-] = 0.0010 (or 1.0 X 10
-3-3
M) M)
pOH = - log 0.0010pOH = - log 0.0010
pOH = 3pOH = 3
pH = 14 – 3 = 11pH = 14 – 3 = 11
OR KOR K
ww = [H = [H
33OO
++
] [OH] [OH
--
]]
[H[H
3OO
++
] = 1.0 x 10] = 1.0 x 10
-11-11
M M
pH = - log (1.0 x 10pH = - log (1.0 x 10
-11-11
) = 11.00) = 11.00
4949
The pH of rainwater collected in a certain region of the
northeastern United States on a particular day was
4.82. What is the H
+
ion concentration of the
rainwater?
The OH
-
ion concentration of a blood sample is
2.5 x 10
-7
M. What is the pH of the blood?
The pH of rainwater collected in a certain region of the
northeastern United States on a particular day was
4.82. What is the H
+
ion concentration of the
rainwater?
The OH
-
ion concentration of a blood sample is
2.5 x 10
-7
M. What is the pH of the blood?
5050
[OH[OH
--
]]
[H[H
++
]] pOHpOH
pHpH
1
0
1
0
-
p
O
H
-
p
O
H
1
0
1
0
-
p
H
-
p
H
-
L
o
g
[
H
-
L
o
g
[
H
+
+
]
]
-
L
o
g
[
O
H
L
o
g
[
O
H
-
-
]
]
1
4
- p
O
H
1
4
- p
O
H
1
4
- p
H
1
4
- p
H
1
. 0
x 1
0
1
. 0
x 1
0
- 1
4
- 1
4
[ O
H
[ O
H
-- ]]
1
. 0
x 1
0
1
. 0
x 1
0
- 1
4
- 1
4
[ H[ H
++ ]]
[OH[OH
--
]]
[H[H
++
]] pOHpOH
pHpH
1
0
1
0
-
p
O
H
-
p
O
H
1
0
1
0
-
p
H
-
p
H
-
L
o
g
[
H
-
L
o
g
[
H
+
+
]
]
-
L
o
g
[
O
H
L
o
g
[
O
H
-
-
]
]
1
4
- p
O
H
1
4
- p
O
H
1
4
- p
H
1
4
- p
H
1
. 0
x 1
0
1
. 0
x 1
0
- 1
4
- 1
4
[ O
H
[ O
H
-- ]]
1
. 0
x 1
0
1
. 0
x 1
0
- 1
4
- 1
4
[ H[ H
++ ]]
5151Calculating [H
3O
+
], pH, [OH
-
], and pOH
Problem 1: A chemist dilutes concentrated
hydrochloric acid to make two solutions: (a) 3.0
M and (b) 0.0024 M. Calculate the [H
3
O
+
], pH,
[OH
-
], and pOH of the two solutions at 25°C.
Problem 2: What is the [H
3O
+
], [OH
-
], and pOH
of a solution with pH = 3.67? Is this an acid,
base, or neutral?
Problem 3: Problem #2 with pH = 8.05?
3O
+
], pH, [OH
-
], and pOH
Problem 1: A chemist dilutes concentrated
hydrochloric acid to make two solutions: (a) 3.0
M and (b) 0.0024 M. Calculate the [H
3
O
+
], pH,
[OH
-
], and pOH of the two solutions at 25°C.
Problem 2: What is the [H
3O
+
], [OH
-
], and pOH
of a solution with pH = 3.67? Is this an acid,
base, or neutral?
Problem 3: Problem #2 with pH = 8.05?
5252
HNO
3
, HCl, H
2
SO
4
and HClO
4
are among the
only known strong acids.
Strong and Weak Acids/BasesStrong and Weak Acids/Bases
The strength of an acid (or base) is
determined by the amount of
IONIZATION.
HONORS ONLY!
HNO
3
, HCl, H
2
SO
4
and HClO
4
are among the
only known strong acids.
Strong and Weak Acids/BasesStrong and Weak Acids/Bases
The strength of an acid (or base) is
determined by the amount of
IONIZATION.
HONORS ONLY!
5353
Strong and Weak Acids/BasesStrong and Weak Acids/Bases
•Generally divide acids and bases into STRONG or Generally divide acids and bases into STRONG or
WEAK ones.WEAK ones.
STRONG ACID:STRONG ACID: HNOHNO
3 3 (aq) + H(aq) + H
22O (l) --->O (l) --->
HH
33OO
+ +
(aq) + NO(aq) + NO
33
- -
(aq)(aq)
HNOHNO
33 is about 100% dissociated in water. is about 100% dissociated in water.
HONORS ONLY!
Strong and Weak Acids/BasesStrong and Weak Acids/Bases
•Generally divide acids and bases into STRONG or Generally divide acids and bases into STRONG or
WEAK ones.WEAK ones.
STRONG ACID:STRONG ACID: HNOHNO
3 3 (aq) + H(aq) + H
22O (l) --->O (l) --->
HH
33OO
+ +
(aq) + NO(aq) + NO
33
- -
(aq)(aq)
HNOHNO
33 is about 100% dissociated in water. is about 100% dissociated in water.
HONORS ONLY!
5454
•Weak acidsWeak acids are much less than 100% ionized in are much less than 100% ionized in
water.water.
One of the best known is acetic acid = CHOne of the best known is acetic acid = CH
33COCO
22HH
Strong and Weak Acids/BasesStrong and Weak Acids/Bases
HONORS ONLY!
•Weak acidsWeak acids are much less than 100% ionized in are much less than 100% ionized in
water.water.
One of the best known is acetic acid = CHOne of the best known is acetic acid = CH
33COCO
22HH
Strong and Weak Acids/BasesStrong and Weak Acids/Bases
HONORS ONLY!
5555
•Strong Base:Strong Base: 100% dissociated in 100% dissociated in
water.water.
NaOH (aq) ---> NaNaOH (aq) ---> Na
+ +
(aq) + OH(aq) + OH
- -
(aq)(aq)
Strong and Weak Acids/BasesStrong and Weak Acids/Bases
Other common strong Other common strong
bases include KOH andbases include KOH and
Ca(OH)Ca(OH)
22..
CaO (lime) + HCaO (lime) + H
22O -->O -->
Ca(OH)Ca(OH)
22 (slaked lime) (slaked lime)
CaOCaO
HONORS ONLY!
•Strong Base:Strong Base: 100% dissociated in 100% dissociated in
water.water.
NaOH (aq) ---> NaNaOH (aq) ---> Na
+ +
(aq) + OH(aq) + OH
- -
(aq)(aq)
Strong and Weak Acids/BasesStrong and Weak Acids/Bases
Other common strong Other common strong
bases include KOH andbases include KOH and
Ca(OH)Ca(OH)
22..
CaO (lime) + HCaO (lime) + H
22O -->O -->
Ca(OH)Ca(OH)
22 (slaked lime) (slaked lime)
CaOCaO
HONORS ONLY!
5656
•Weak base:Weak base: less than 100% ionized less than 100% ionized
in waterin water
One of the best known weak bases is One of the best known weak bases is
ammoniaammonia
NHNH
3 3 (aq) + H(aq) + H
22O (l) O (l) NH NH
44
+ +
(aq) + OH(aq) + OH
- -
(aq)(aq)
Strong and Weak Acids/BasesStrong and Weak Acids/Bases
HONORS ONLY!
•Weak base:Weak base: less than 100% ionized less than 100% ionized
in waterin water
One of the best known weak bases is One of the best known weak bases is
ammoniaammonia
NHNH
3 3 (aq) + H(aq) + H
22O (l) O (l) NH NH
44
+ +
(aq) + OH(aq) + OH
- -
(aq)(aq)
Strong and Weak Acids/BasesStrong and Weak Acids/Bases
HONORS ONLY!
5757
Weak BasesWeak Bases
HONORS ONLY!
Weak BasesWeak Bases
HONORS ONLY!
5858
Equilibria Involving Equilibria Involving
Weak Acids and BasesWeak Acids and Bases
Consider acetic acid, HCConsider acetic acid, HC
22HH
33OO
22 (HOAc) (HOAc)
HCHC
22HH
33OO
22 + H + H
22O O H H
33OO
++
+ C + C
22HH
33OO
22
--
AcidAcid Conj. base Conj. base
K
a
[H
3
O
+
][OAc
-
]
[HOAc]
1.8 x 10
-5
(K is designated K(K is designated K
aa for ACID) for ACID)
K gives the ratio of ions (split up) to molecules K gives the ratio of ions (split up) to molecules
(don’t split up)(don’t split up)
HONORS ONLY!
Equilibria Involving Equilibria Involving
Weak Acids and BasesWeak Acids and Bases
Consider acetic acid, HCConsider acetic acid, HC
22HH
33OO
22 (HOAc) (HOAc)
HCHC
22HH
33OO
22 + H + H
22O O H H
33OO
++
+ C + C
22HH
33OO
22
--
AcidAcid Conj. base Conj. base
K
a
[H
3
O
+
][OAc
-
]
[HOAc]
1.8 x 10
-5
(K is designated K(K is designated K
aa for ACID) for ACID)
K gives the ratio of ions (split up) to molecules K gives the ratio of ions (split up) to molecules
(don’t split up)(don’t split up)
HONORS ONLY!
5959
Ionization Constants for Acids/Bases Ionization Constants for Acids/Bases
AcidsAcids ConjugateConjugate
BasesBases
Increase
strength
Increase
strength
HONORS ONLY!
Ionization Constants for Acids/Bases Ionization Constants for Acids/Bases
AcidsAcids ConjugateConjugate
BasesBases
Increase
strength
Increase
strength
HONORS ONLY!
6060
Equilibrium Constants Equilibrium Constants
for Weak Acidsfor Weak Acids
Weak acid has KWeak acid has K
aa < 1 < 1
Leads to small [HLeads to small [H
33OO
++
] and a pH of 2 - 7] and a pH of 2 - 7
HONORS ONLY!
Equilibrium Constants Equilibrium Constants
for Weak Acidsfor Weak Acids
Weak acid has KWeak acid has K
aa < 1 < 1
Leads to small [HLeads to small [H
33OO
++
] and a pH of 2 - 7] and a pH of 2 - 7
HONORS ONLY!
6161
Equilibrium Constants Equilibrium Constants
for Weak Basesfor Weak Bases
Weak base has KWeak base has K
bb < 1 < 1
Leads to small [OHLeads to small [OH
--
] and a pH of 12 - 7] and a pH of 12 - 7
HONORS ONLY!
Equilibrium Constants Equilibrium Constants
for Weak Basesfor Weak Bases
Weak base has KWeak base has K
bb < 1 < 1
Leads to small [OHLeads to small [OH
--
] and a pH of 12 - 7] and a pH of 12 - 7
HONORS ONLY!
6262
Relation Relation
of Kof K
aa, K, K
bb, ,
[H[H
33OO
++
] ]
and pHand pH
HONORS ONLY!
Relation Relation
of Kof K
aa, K, K
bb, ,
[H[H
33OO
++
] ]
and pHand pH
HONORS ONLY!
6363
Equilibria Involving A Weak AcidEquilibria Involving A Weak Acid
You have 1.00 M HOAc. Calc. the You have 1.00 M HOAc. Calc. the
equilibrium concs. of HOAc, Hequilibrium concs. of HOAc, H
33OO
++
, OAc, OAc
--
, ,
and the pH.and the pH.
Step 1.Step 1. Define equilibrium concs. in ICE Define equilibrium concs. in ICE
table.table.
[HOAc][HOAc] [H[H
33OO
++
]] [OAc[OAc
--
]]
initialinitial
changechange
equilibequilib
1.001.00 00 00
-x-x +x+x +x+x
1.00-x1.00-x xx xx
HONORS ONLY!
Equilibria Involving A Weak AcidEquilibria Involving A Weak Acid
You have 1.00 M HOAc. Calc. the You have 1.00 M HOAc. Calc. the
equilibrium concs. of HOAc, Hequilibrium concs. of HOAc, H
33OO
++
, OAc, OAc
--
, ,
and the pH.and the pH.
Step 1.Step 1. Define equilibrium concs. in ICE Define equilibrium concs. in ICE
table.table.
[HOAc][HOAc] [H[H
33OO
++
]] [OAc[OAc
--
]]
initialinitial
changechange
equilibequilib
1.001.00 00 00
-x-x +x+x +x+x
1.00-x1.00-x xx xx
HONORS ONLY!
6464
Equilibria Involving A Weak AcidEquilibria Involving A Weak Acid
Step 2.Step 2. Write KWrite K
aa expression expression
You have 1.00 M HOAc. Calc. the equilibrium concs. You have 1.00 M HOAc. Calc. the equilibrium concs.
of HOAc, Hof HOAc, H
33OO
++
, OAc, OAc
--
, and the pH., and the pH.
K
a1.8 x 10
-5
=
[H
3O
+
][OAc
-
]
[HOAc]
x
2
1.00 - x
This is a quadratic. Solve using quadratic This is a quadratic. Solve using quadratic
formula.formula.
or you can make an approximation if x is very or you can make an approximation if x is very
small! (Rule of thumb: 10small! (Rule of thumb: 10
-5-5
or smaller is ok) or smaller is ok)
HONORS ONLY!
Equilibria Involving A Weak AcidEquilibria Involving A Weak Acid
Step 2.Step 2. Write KWrite K
aa expression expression
You have 1.00 M HOAc. Calc. the equilibrium concs. You have 1.00 M HOAc. Calc. the equilibrium concs.
of HOAc, Hof HOAc, H
33OO
++
, OAc, OAc
--
, and the pH., and the pH.
K
a1.8 x 10
-5
=
[H
3O
+
][OAc
-
]
[HOAc]
x
2
1.00 - x
This is a quadratic. Solve using quadratic This is a quadratic. Solve using quadratic
formula.formula.
or you can make an approximation if x is very or you can make an approximation if x is very
small! (Rule of thumb: 10small! (Rule of thumb: 10
-5-5
or smaller is ok) or smaller is ok)
HONORS ONLY!
6565
Equilibria Involving A Weak AcidEquilibria Involving A Weak Acid
Step 3.Step 3. Solve KSolve K
aa expression expression
You have 1.00 M HOAc. Calc. the equilibrium concs. You have 1.00 M HOAc. Calc. the equilibrium concs.
of HOAc, Hof HOAc, H
33OO
++
, OAc, OAc
--
, and the pH., and the pH.
K
a1.8 x 10
-5
=
[H
3O
+
][OAc
-
]
[HOAc]
x
2
1.00 - x
First assume x is very small because First assume x is very small because
KK
aa is so small. is so small.
K
a
1.8 x 10
-5
=
x
2
1.00
Now we can more easily solve this Now we can more easily solve this
approximate expression.approximate expression.
HONORS ONLY!
Equilibria Involving A Weak AcidEquilibria Involving A Weak Acid
Step 3.Step 3. Solve KSolve K
aa expression expression
You have 1.00 M HOAc. Calc. the equilibrium concs. You have 1.00 M HOAc. Calc. the equilibrium concs.
of HOAc, Hof HOAc, H
33OO
++
, OAc, OAc
--
, and the pH., and the pH.
K
a1.8 x 10
-5
=
[H
3O
+
][OAc
-
]
[HOAc]
x
2
1.00 - x
First assume x is very small because First assume x is very small because
KK
aa is so small. is so small.
K
a
1.8 x 10
-5
=
x
2
1.00
Now we can more easily solve this Now we can more easily solve this
approximate expression.approximate expression.
HONORS ONLY!
6666
ApproximatingApproximating
If K is really small, the equilibrium If K is really small, the equilibrium
concentrations will be nearly the same as concentrations will be nearly the same as
the initial concentrations.the initial concentrations.
Example: 0.20 – x is just about 0.20 if Example: 0.20 – x is just about 0.20 if
x is really small.x is really small.
If the K is 10If the K is 10
-5-5
or smaller (10 or smaller (10
-6-6
, 10, 10
-7-7
, etc.), you , etc.), you
should approximate. Otherwise, you have should approximate. Otherwise, you have
to use the quadratic.to use the quadratic.
ApproximatingApproximating
If K is really small, the equilibrium If K is really small, the equilibrium
concentrations will be nearly the same as concentrations will be nearly the same as
the initial concentrations.the initial concentrations.
Example: 0.20 – x is just about 0.20 if Example: 0.20 – x is just about 0.20 if
x is really small.x is really small.
If the K is 10If the K is 10
-5-5
or smaller (10 or smaller (10
-6-6
, 10, 10
-7-7
, etc.), you , etc.), you
should approximate. Otherwise, you have should approximate. Otherwise, you have
to use the quadratic.to use the quadratic.
6767
Equilibria Involving A Weak AcidEquilibria Involving A Weak Acid
Step 3.Step 3. Solve KSolve K
aa approximateapproximate expressionexpression
You have 1.00 M HOAc. Calc. the equilibrium concs. You have 1.00 M HOAc. Calc. the equilibrium concs.
of HOAc, Hof HOAc, H
33OO
++
, OAc, OAc
--
, and the pH., and the pH.
K
a1.8 x 10
-5
=
x
2
1.00
x =x = [[HH
33OO
++
] = [] = [OAcOAc
--
] = 4.2 x 10] = 4.2 x 10
-3-3
M M
pH = - log [pH = - log [HH
33OO
++
] = -log (4.2 x 10] = -log (4.2 x 10
-3-3
) =) = 2.372.37
HONORS ONLY!
Equilibria Involving A Weak AcidEquilibria Involving A Weak Acid
Step 3.Step 3. Solve KSolve K
aa approximateapproximate expressionexpression
You have 1.00 M HOAc. Calc. the equilibrium concs. You have 1.00 M HOAc. Calc. the equilibrium concs.
of HOAc, Hof HOAc, H
33OO
++
, OAc, OAc
--
, and the pH., and the pH.
K
a1.8 x 10
-5
=
x
2
1.00
x =x = [[HH
33OO
++
] = [] = [OAcOAc
--
] = 4.2 x 10] = 4.2 x 10
-3-3
M M
pH = - log [pH = - log [HH
33OO
++
] = -log (4.2 x 10] = -log (4.2 x 10
-3-3
) =) = 2.372.37
HONORS ONLY!
6868
Equilibria Involving A Weak AcidEquilibria Involving A Weak Acid
Calculate the pH of a 0.0010 M solution of Calculate the pH of a 0.0010 M solution of
formic acid, HCOformic acid, HCO
22H.H.
HCOHCO
22H + HH + H
22O O HCO HCO
22
--
+ H + H
33OO
++
KK
aa = 1.8 x 10 = 1.8 x 10
-4-4
Approximate solutionApproximate solution
[H[H
33OO
++
] = 4.2 x 10] = 4.2 x 10
-4-4
M, M, pH = 3.37pH = 3.37
Exact SolutionExact Solution
[H[H
33OO
++
] = [HCO] = [HCO
22
--
] = 3.4 x 10] = 3.4 x 10
-4-4
M M
[HCO[HCO
22H] = 0.0010 - 3.4 x 10H] = 0.0010 - 3.4 x 10
-4-4
= 0.0007 M = 0.0007 M
pH = 3.47 pH = 3.47
HONORS ONLY!
Equilibria Involving A Weak AcidEquilibria Involving A Weak Acid
Calculate the pH of a 0.0010 M solution of Calculate the pH of a 0.0010 M solution of
formic acid, HCOformic acid, HCO
22H.H.
HCOHCO
22H + HH + H
22O O HCO HCO
22
--
+ H + H
33OO
++
KK
aa = 1.8 x 10 = 1.8 x 10
-4-4
Approximate solutionApproximate solution
[H[H
33OO
++
] = 4.2 x 10] = 4.2 x 10
-4-4
M, M, pH = 3.37pH = 3.37
Exact SolutionExact Solution
[H[H
33OO
++
] = [HCO] = [HCO
22
--
] = 3.4 x 10] = 3.4 x 10
-4-4
M M
[HCO[HCO
22H] = 0.0010 - 3.4 x 10H] = 0.0010 - 3.4 x 10
-4-4
= 0.0007 M = 0.0007 M
pH = 3.47 pH = 3.47
HONORS ONLY!
6969
Equilibria Involving A Weak BaseEquilibria Involving A Weak Base
You have 0.010 M NHYou have 0.010 M NH
33. Calc. the pH.. Calc. the pH.
NHNH
33 + H + H
22O O NH NH
44
++
+ OH + OH
--
KK
bb = 1.8 x 10 = 1.8 x 10
-5-5
Step 1.Step 1. Define equilibrium concs. in ICE tableDefine equilibrium concs. in ICE table
[NH[NH
33]] [NH[NH
44
++
]] [OH[OH
--
]]
initialinitial
changechange
equilibequilib
0.0100.010 00 00
-x-x +x+x +x+x
0.010 - x0.010 - xx x xx
HONORS ONLY!
Equilibria Involving A Weak BaseEquilibria Involving A Weak Base
You have 0.010 M NHYou have 0.010 M NH
33. Calc. the pH.. Calc. the pH.
NHNH
33 + H + H
22O O NH NH
44
++
+ OH + OH
--
KK
bb = 1.8 x 10 = 1.8 x 10
-5-5
Step 1.Step 1. Define equilibrium concs. in ICE tableDefine equilibrium concs. in ICE table
[NH[NH
33]] [NH[NH
44
++
]] [OH[OH
--
]]
initialinitial
changechange
equilibequilib
0.0100.010 00 00
-x-x +x+x +x+x
0.010 - x0.010 - xx x xx
HONORS ONLY!
7070
Equilibria Involving A Weak BaseEquilibria Involving A Weak Base
You have 0.010 M NHYou have 0.010 M NH
33. Calc. the pH.. Calc. the pH.
NHNH
33 + H + H
22O O NH NH
44
++
+ OH + OH
--
KK
bb = 1.8 x 10 = 1.8 x 10
-5-5
Step 1.Step 1. Define equilibrium concs. in ICE tableDefine equilibrium concs. in ICE table
[NH[NH
33]] [NH[NH
44
++
]] [OH[OH
--
]]
initialinitial
changechange
equilibequilib
0.0100.010 00 00
-x-x +x+x +x+x
0.010 - x0.010 - xx x xx
HONORS ONLY!
Equilibria Involving A Weak BaseEquilibria Involving A Weak Base
You have 0.010 M NHYou have 0.010 M NH
33. Calc. the pH.. Calc. the pH.
NHNH
33 + H + H
22O O NH NH
44
++
+ OH + OH
--
KK
bb = 1.8 x 10 = 1.8 x 10
-5-5
Step 1.Step 1. Define equilibrium concs. in ICE tableDefine equilibrium concs. in ICE table
[NH[NH
33]] [NH[NH
44
++
]] [OH[OH
--
]]
initialinitial
changechange
equilibequilib
0.0100.010 00 00
-x-x +x+x +x+x
0.010 - x0.010 - xx x xx
HONORS ONLY!
7171
Equilibria Involving A Weak BaseEquilibria Involving A Weak Base
You have 0.010 M NHYou have 0.010 M NH
33. Calc. the pH.. Calc. the pH.
NHNH
33 + H + H
22O O NH NH
44
++
+ OH + OH
--
KK
bb = 1.8 x 10 = 1.8 x 10
-5-5
Step 2.Step 2. Solve the equilibrium expressionSolve the equilibrium expression
K
b 1.8 x 10
-5
=
[NH
4
+
][OH
-
]
[NH
3
]
=
x
2
0.010 - x
Assume x is small, soAssume x is small, so
x = [OHx = [OH
--
] = [NH] = [NH
44
++
] = 4.2 x 10] = 4.2 x 10
-4-4
M M
and [NHand [NH
33] = 0.010 - 4.2 x 10] = 0.010 - 4.2 x 10
-4-4
≈ 0.010 M ≈ 0.010 M
The approximation is validThe approximation is valid!!
HONORS ONLY!
Equilibria Involving A Weak BaseEquilibria Involving A Weak Base
You have 0.010 M NHYou have 0.010 M NH
33. Calc. the pH.. Calc. the pH.
NHNH
33 + H + H
22O O NH NH
44
++
+ OH + OH
--
KK
bb = 1.8 x 10 = 1.8 x 10
-5-5
Step 2.Step 2. Solve the equilibrium expressionSolve the equilibrium expression
K
b 1.8 x 10
-5
=
[NH
4
+
][OH
-
]
[NH
3
]
=
x
2
0.010 - x
Assume x is small, soAssume x is small, so
x = [OHx = [OH
--
] = [NH] = [NH
44
++
] = 4.2 x 10] = 4.2 x 10
-4-4
M M
and [NHand [NH
33] = 0.010 - 4.2 x 10] = 0.010 - 4.2 x 10
-4-4
≈ 0.010 M ≈ 0.010 M
The approximation is validThe approximation is valid!!
HONORS ONLY!
7272
Equilibria Involving A Weak BaseEquilibria Involving A Weak Base
You have 0.010 M NHYou have 0.010 M NH
33. Calc. the pH.. Calc. the pH.
NHNH
33 + H + H
22O O NH NH
44
++
+ OH + OH
--
KK
bb = 1.8 x 10 = 1.8 x 10
-5-5
Step 3.Step 3. Calculate pHCalculate pH
[OH[OH
--
] = 4.2 x 10] = 4.2 x 10
-4-4
M M
so pOH = - log [OHso pOH = - log [OH
--
] = 3.37] = 3.37
Because pH + pOH = 14,Because pH + pOH = 14,
pH = 10.63pH = 10.63
HONORS ONLY!
Equilibria Involving A Weak BaseEquilibria Involving A Weak Base
You have 0.010 M NHYou have 0.010 M NH
33. Calc. the pH.. Calc. the pH.
NHNH
33 + H + H
22O O NH NH
44
++
+ OH + OH
--
KK
bb = 1.8 x 10 = 1.8 x 10
-5-5
Step 3.Step 3. Calculate pHCalculate pH
[OH[OH
--
] = 4.2 x 10] = 4.2 x 10
-4-4
M M
so pOH = - log [OHso pOH = - log [OH
--
] = 3.37] = 3.37
Because pH + pOH = 14,Because pH + pOH = 14,
pH = 10.63pH = 10.63
HONORS ONLY!
7373
Types of Acid/Base Reactions: Types of Acid/Base Reactions:
SummarySummary
HONORS ONLY!
Types of Acid/Base Reactions: Types of Acid/Base Reactions:
SummarySummary
HONORS ONLY!
7474
pH testing
•There are several ways to test pHThere are several ways to test pH
–Blue litmus paper (red = acid)Blue litmus paper (red = acid)
–Red litmus paper (blue = basic)Red litmus paper (blue = basic)
–pH paper (multi-colored)pH paper (multi-colored)
–pH meter (7 is neutral, <7 acid, >7 pH meter (7 is neutral, <7 acid, >7
base)base)
–Universal indicator (multi-colored)Universal indicator (multi-colored)
–Indicators like phenolphthaleinIndicators like phenolphthalein
–Natural indicators like red cabbage, Natural indicators like red cabbage,
radishesradishes
pH testing
•There are several ways to test pHThere are several ways to test pH
–Blue litmus paper (red = acid)Blue litmus paper (red = acid)
–Red litmus paper (blue = basic)Red litmus paper (blue = basic)
–pH paper (multi-colored)pH paper (multi-colored)
–pH meter (7 is neutral, <7 acid, >7 pH meter (7 is neutral, <7 acid, >7
base)base)
–Universal indicator (multi-colored)Universal indicator (multi-colored)
–Indicators like phenolphthaleinIndicators like phenolphthalein
–Natural indicators like red cabbage, Natural indicators like red cabbage,
radishesradishes
7575
Paper testing
•Paper tests like litmus paper and pH Paper tests like litmus paper and pH
paperpaper
–Put a stirring rod into the solution Put a stirring rod into the solution
and stir.and stir.
–Take the stirring rod out, and place Take the stirring rod out, and place
a drop of the solution from the end a drop of the solution from the end
of the stirring rod onto a piece of of the stirring rod onto a piece of
the paperthe paper
–Read and record the color change. Read and record the color change.
Note what the color indicates.Note what the color indicates.
–You should only use a small portion You should only use a small portion
of the paper. You can use one of the paper. You can use one
piece of paper for several tests.piece of paper for several tests.
Paper testing
•Paper tests like litmus paper and pH Paper tests like litmus paper and pH
paperpaper
–Put a stirring rod into the solution Put a stirring rod into the solution
and stir.and stir.
–Take the stirring rod out, and place Take the stirring rod out, and place
a drop of the solution from the end a drop of the solution from the end
of the stirring rod onto a piece of of the stirring rod onto a piece of
the paperthe paper
–Read and record the color change. Read and record the color change.
Note what the color indicates.Note what the color indicates.
–You should only use a small portion You should only use a small portion
of the paper. You can use one of the paper. You can use one
piece of paper for several tests.piece of paper for several tests.
7676
pH paper
pH paper
7777
pH meter
•Tests the voltage of the Tests the voltage of the
electrolyteelectrolyte
•Converts the voltage to Converts the voltage to
pHpH
•Very cheap, accurateVery cheap, accurate
•Must be calibrated with Must be calibrated with
a buffer solutiona buffer solution
pH meter
•Tests the voltage of the Tests the voltage of the
electrolyteelectrolyte
•Converts the voltage to Converts the voltage to
pHpH
•Very cheap, accurateVery cheap, accurate
•Must be calibrated with Must be calibrated with
a buffer solutiona buffer solution
7878
pH indicators
•Indicators are dyes that can be
added that will change color in
the presence of an acid or base.
•Some indicators only work in a
specific range of pH
•Once the drops are added, the
sample is ruined
•Some dyes are natural, like radish
skin or red cabbage
pH indicators
•Indicators are dyes that can be
added that will change color in
the presence of an acid or base.
•Some indicators only work in a
specific range of pH
•Once the drops are added, the
sample is ruined
•Some dyes are natural, like radish
skin or red cabbage
7979
ACID-BASE REACTIONSACID-BASE REACTIONS
TitrationsTitrations
HH
22CC
22OO
44(aq) + 2 NaOH(aq) --->(aq) + 2 NaOH(aq) --->
acidacid basebase
NaNa
22CC
22OO
44(aq) + 2 H(aq) + 2 H
22O(liq)O(liq)
Carry out this reaction using aCarry out this reaction using a TITRATIONTITRATION..
Oxalic acid,Oxalic acid,
HH
22CC
22OO
44
ACID-BASE REACTIONSACID-BASE REACTIONS
TitrationsTitrations
HH
22CC
22OO
44(aq) + 2 NaOH(aq) --->(aq) + 2 NaOH(aq) --->
acidacid basebase
NaNa
22CC
22OO
44(aq) + 2 H(aq) + 2 H
22O(liq)O(liq)
Carry out this reaction using aCarry out this reaction using a TITRATIONTITRATION..
Oxalic acid,Oxalic acid,
HH
22CC
22OO
44
8080
Setup for titrating an acid with a baseSetup for titrating an acid with a base
Setup for titrating an acid with a baseSetup for titrating an acid with a base
8181
TitrationTitration
1. Add solution from the buret.1. Add solution from the buret.
2. Reagent (base) reacts with 2. Reagent (base) reacts with
compound (acid) in solution compound (acid) in solution
in the flask.in the flask.
3.3.Indicator shows when exact Indicator shows when exact
stoichiometric reaction has stoichiometric reaction has
occurred. (Acid = Base)occurred. (Acid = Base)
This is called This is called
NEUTRALIZATION.NEUTRALIZATION.
TitrationTitration
1. Add solution from the buret.1. Add solution from the buret.
2. Reagent (base) reacts with 2. Reagent (base) reacts with
compound (acid) in solution compound (acid) in solution
in the flask.in the flask.
3.3.Indicator shows when exact Indicator shows when exact
stoichiometric reaction has stoichiometric reaction has
occurred. (Acid = Base)occurred. (Acid = Base)
This is called This is called
NEUTRALIZATION.NEUTRALIZATION.
8282
35.62 mL of NaOH is 35.62 mL of NaOH is
neutralized with 25.2 mL of neutralized with 25.2 mL of
0.0998 M HCl by titration to 0.0998 M HCl by titration to
an equivalence point. What an equivalence point. What
is the concentration of the is the concentration of the
NaOH?NaOH?
LAB PROBLEM #1: Standardize a LAB PROBLEM #1: Standardize a
solution of NaOH — i.e., accurately solution of NaOH — i.e., accurately
determine its concentration.determine its concentration.
35.62 mL of NaOH is 35.62 mL of NaOH is
neutralized with 25.2 mL of neutralized with 25.2 mL of
0.0998 M HCl by titration to 0.0998 M HCl by titration to
an equivalence point. What an equivalence point. What
is the concentration of the is the concentration of the
NaOH?NaOH?
LAB PROBLEM #1: Standardize a LAB PROBLEM #1: Standardize a
solution of NaOH — i.e., accurately solution of NaOH — i.e., accurately
determine its concentration.determine its concentration.
8383
PROBLEM: You have 50.0 mL of 3.0 M PROBLEM: You have 50.0 mL of 3.0 M
NaOH and you want 0.50 M NaOH. NaOH and you want 0.50 M NaOH.
What do you do?What do you do?
Add water to the 3.0 M solution to lower Add water to the 3.0 M solution to lower
its concentration to 0.50 M its concentration to 0.50 M
Dilute the solution!Dilute the solution!
PROBLEM: You have 50.0 mL of 3.0 M PROBLEM: You have 50.0 mL of 3.0 M
NaOH and you want 0.50 M NaOH. NaOH and you want 0.50 M NaOH.
What do you do?What do you do?
Add water to the 3.0 M solution to lower Add water to the 3.0 M solution to lower
its concentration to 0.50 M its concentration to 0.50 M
Dilute the solution!Dilute the solution!
8484
PROBLEM: You have 50.0 mL of 3.0 M PROBLEM: You have 50.0 mL of 3.0 M
NaOH and you want 0.50 M NaOH. What do NaOH and you want 0.50 M NaOH. What do
you do?you do?
3.0 M NaOH 0.50 M NaOH
H2O
Concentrated Dilute
But how much water But how much water
do we add?do we add?
PROBLEM: You have 50.0 mL of 3.0 M PROBLEM: You have 50.0 mL of 3.0 M
NaOH and you want 0.50 M NaOH. What do NaOH and you want 0.50 M NaOH. What do
you do?you do?
3.0 M NaOH 0.50 M NaOH
H2O
Concentrated Dilute
But how much water But how much water
do we add?do we add?
8585
PROBLEM: You have 50.0 mL of 3.0 M PROBLEM: You have 50.0 mL of 3.0 M
NaOH and you want 0.50 M NaOH. What do NaOH and you want 0.50 M NaOH. What do
you doyou do??
How much water is added?How much water is added?
The important point is that --->The important point is that --->
moles of NaOH in ORIGINAL solution = moles of NaOH in ORIGINAL solution =
moles of NaOH in FINAL solutionmoles of NaOH in FINAL solution
PROBLEM: You have 50.0 mL of 3.0 M PROBLEM: You have 50.0 mL of 3.0 M
NaOH and you want 0.50 M NaOH. What do NaOH and you want 0.50 M NaOH. What do
you doyou do??
How much water is added?How much water is added?
The important point is that --->The important point is that --->
moles of NaOH in ORIGINAL solution = moles of NaOH in ORIGINAL solution =
moles of NaOH in FINAL solutionmoles of NaOH in FINAL solution
8686
PROBLEM: You have 50.0 mL of 3.0 M NaOH and PROBLEM: You have 50.0 mL of 3.0 M NaOH and
you want 0.50 M NaOH. What do you do?you want 0.50 M NaOH. What do you do?
Amount of NaOH in original solution = Amount of NaOH in original solution =
M • VM • V = =
(3.0 mol/L)(0.050 L) = 0.5 M NaOH X V(3.0 mol/L)(0.050 L) = 0.5 M NaOH X V
Amount of NaOH in final solution must also = Amount of NaOH in final solution must also =
0.15 mol NaOH0.15 mol NaOH
Volume of final solution =Volume of final solution =
(0.15 mol NaOH) / (0.50 M) = 0.30 L(0.15 mol NaOH) / (0.50 M) = 0.30 L
or or 300 mL300 mL
PROBLEM: You have 50.0 mL of 3.0 M NaOH and PROBLEM: You have 50.0 mL of 3.0 M NaOH and
you want 0.50 M NaOH. What do you do?you want 0.50 M NaOH. What do you do?
Amount of NaOH in original solution = Amount of NaOH in original solution =
M • VM • V = =
(3.0 mol/L)(0.050 L) = 0.5 M NaOH X V(3.0 mol/L)(0.050 L) = 0.5 M NaOH X V
Amount of NaOH in final solution must also = Amount of NaOH in final solution must also =
0.15 mol NaOH0.15 mol NaOH
Volume of final solution =Volume of final solution =
(0.15 mol NaOH) / (0.50 M) = 0.30 L(0.15 mol NaOH) / (0.50 M) = 0.30 L
or or 300 mL300 mL
8787
PROBLEM: You have 50.0 mL of 3.0 M PROBLEM: You have 50.0 mL of 3.0 M
NaOH and you want 0.50 M NaOH. What do NaOH and you want 0.50 M NaOH. What do
you do?you do?
Conclusion:Conclusion:
add 250 mL add 250 mL
of waterof water to to
50.0 mL of 3.0 50.0 mL of 3.0
M NaOH to M NaOH to
make 300 mL make 300 mL
of 0.50 M of 0.50 M
NaOH.NaOH.
3.0 M NaOH 0.50 M NaOH
H2O
Concentrated Dilute
PROBLEM: You have 50.0 mL of 3.0 M PROBLEM: You have 50.0 mL of 3.0 M
NaOH and you want 0.50 M NaOH. What do NaOH and you want 0.50 M NaOH. What do
you do?you do?
Conclusion:Conclusion:
add 250 mL add 250 mL
of waterof water to to
50.0 mL of 3.0 50.0 mL of 3.0
M NaOH to M NaOH to
make 300 mL make 300 mL
of 0.50 M of 0.50 M
NaOH.NaOH.
3.0 M NaOH 0.50 M NaOH
H2O
Concentrated Dilute
8888
A shortcutA shortcut
MM
11 • V • V
11 = M = M
22 • V • V
22
Preparing Solutions by Preparing Solutions by
DilutionDilution
A shortcutA shortcut
MM
11 • V • V
11 = M = M
22 • V • V
22
Preparing Solutions by Preparing Solutions by
DilutionDilution
8989
You try this dilution problem
•You have a stock bottle of hydrochloric acid,
which is 12.1 M. You need 400. mL of 0.10 M
HCl. How much of the acid and how much
water will you need?
You try this dilution problem
•You have a stock bottle of hydrochloric acid,
which is 12.1 M. You need 400. mL of 0.10 M
HCl. How much of the acid and how much
water will you need?
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