1 The Chemistry of Acids and Bases. 2 Acid and Bases.
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Transcript of 1 The Chemistry of Acids and Bases. 2 Acid and Bases.
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11The Chemistry The Chemistry of Acids and of Acids and BasesBases
The Chemistry The Chemistry of Acids and of Acids and BasesBases
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Acid and BasesAcid and Bases
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Acid and BasesAcid and Bases
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Acid and BasesAcid and Bases
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55Acids
Have a sour taste. Vinegar is a solution of acetic acid. CitrusHave a sour taste. Vinegar is a solution of acetic acid. Citrusfruits 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
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66
Some Properties of Acids
Produce H+ (as H3O+) ions in water (the hydronium ion is a
hydrogen ion attached to a water molecule)
Taste sour
Corrode metals
Are electrolytes
React with bases to form a salt and water
Have a pH is less than 7
Turns blue litmus paper to red “Blue to Red A-CID”
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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
Have a pH greater than 7Have a pH greater than 7
Turns red litmus paper to blue “Basic Blue”Turns red litmus paper to blue “Basic Blue”
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88
Some Common Bases
NaOHNaOH sodium hydroxidesodium hydroxide lyelye
KOHKOH potassium hydroxidepotassium hydroxide liquid 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)33 aluminum hydroxidealuminum hydroxide Maalox (antacid)Maalox (antacid)
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99The pH scale is a way of The pH scale 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
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1010
pH of Common pH of Common SubstancesSubstances
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1111Calculating the pH
pH = - log [H+](Remember that the [ ] mean Molarity)
Example: If [H+] = 1.4 X 10-3
pH = - log (1.4 X 10-3)
Using your calculator:
- log 1.4 x 2nd , -3 (enter)
pH = 2.85
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1212pH calculations – Solving for pH calculations – Solving for H+H+
• A solution has a pH of 7.41. What is the A solution has a pH of 7.41. What is the Molarity of hydrogen ions in the solution?Molarity of hydrogen ions in the solution?
pH = - log [HpH = - log [H++]]
7.41 = - log [H7.41 = - log [H++]]
-7.41 = log [H-7.41 = log [H++]]
Antilog -7.41 = antilog (log [HAntilog -7.41 = antilog (log [H++])])
1010-7.41-7.41 = [H = [H++]]
3.9 X 103.9 X 10-8-8 M = [H M = [H++]]
pH = - log [HpH = - log [H++]]
7.41 = - log [H7.41 = - log [H++]]
-7.41 = log [H-7.41 = log [H++]]
Antilog -7.41 = antilog (log [HAntilog -7.41 = antilog (log [H++])])
1010-7.41-7.41 = [H = [H++]]
3.9 X 103.9 X 10-8-8 M = [H M = [H++]]
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1313pH calculations – Solving for pH calculations – Solving for H+H+
• A solution has a pH of 7.41. What is the Molarity of A solution has a pH of 7.41. What is the Molarity of hydrogen ions in the solution?hydrogen ions in the solution?
[H[H++] = antilog (- pH)] = antilog (- pH)
Use your calculator:
2nd log - 7.41
(enter)
[H[H++] = antilog (- pH)] = antilog (- pH)
Use your calculator:
2nd log - 7.41
(enter)
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1414pOH
• 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
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[H[H++], [OH], [OH--] and pH] and pHWhat is the [HWhat is the [H++] and [OH] and [OH--] when pH = 7.41? ] when pH = 7.41?
pH = pH + pOH pH = pH + pOH
14 = 7.41 + pOH 14 = 7.41 + pOH
pOH = 14 – 7.41 = 6.59pOH = 14 – 7.41 = 6.59
[H[H++] = antilog (-pH)] = antilog (-pH)
2nd log -7.412nd log -7.41
3.89 x 103.89 x 10-8-8
[OH[OH--] = antilog (-pOH)] = antilog (-pOH)
2nd log - 6.592nd log - 6.59
2.57 x 102.57 x 10-7-7
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1616pH 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
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1717Paper 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 Take the stirring rod out, and
place a drop of the solution from place a drop of the solution from the end of the stirring rod onto a the end of the stirring rod onto a piece of the paperpiece of the paper
– Read and record the color Read and record the color change. Note what the color change. Note what the color indicates.indicates.
– You should only use a small You should only use a small portion of the paper. You can use portion of the paper. You can use one piece of paper for several one piece of paper for several tests.tests.
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1818pH paperpH paper
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1919
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
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2020pH 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
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2121
ACID-BASE REACTIONSACID-BASE REACTIONSTitrationsTitrations
ACID-BASE REACTIONSACID-BASE REACTIONSTitrationsTitrations
HH22CC22OO44(aq) + 2 NaOH(aq) --->(aq) + 2 NaOH(aq) --->
acidacid basebase
NaNa22CC22OO44(aq) + 2 H(aq) + 2 H22O(liq)O(liq)
Carry out this reaction using aCarry out this reaction using a TITRATIONTITRATION..
Oxalic acid,Oxalic acid,
HH22CC22OO44
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2222Setup for titrating an acid with a baseSetup for titrating an acid with a base
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2323
TitrationTitrationTitrationTitration
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.
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2424
Anion Ending Acid Name
-ide hydro-(stem)-ic acid
-ate (stem)-ic acid
-ite (stem)-ous acid
Acid Nomenclature Review
Binary Binary
TernaryTernary
An easy way to remember which goes with which…An easy way to remember which goes with which…
““In the cafeteria, youIn the cafeteria, you ATEATE somethingsomething ICICky”ky”
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2525Acid Nomenclature Flowchart
h yd ro - p re fix-ic en d in g
2 e lem en ts
-a te en d in gb ecom es-ic en d in g
-ite en d in gb ecom es
-o u s en d in g
n o h yd ro - p re fix
3 e lem en ts
AC ID Ss ta rt w ith 'H '
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2626
• HBr HBr (aq)(aq)
• HH22COCO33
• HH22SOSO33
hydrohydrobromicbromic acidacid
carboncarbonicic acidacid
sulfursulfurousous acidacid
Acid Nomenclature Review
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2727
Name ‘Em!
• HI HI (aq)(aq)
• HCl HCl (aq)(aq)
• HH22SOSO33
• HNOHNO33
• HIOHIO44
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2828
Name ‘Em!
• HI HI (aq) (aq) HydoiodicHydoiodic AcidAcid
• HCl HCl (aq) (aq) Hydrochloric AcidHydrochloric Acid
• HH22SOSO3 3 Sulfurous AcidSulfurous Acid
• HH22SOSO4 4 Sulfuric AcidSulfuric Acid
• HNOHNO3 3 Nitric AcidNitric Acid
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2929
Acid/Base Definitions
• Definition #1: Arrhenius (traditional)
Acids – produce H+ ions (or hydronium ions H3O+)
Bases – produce OH- ions
(problem: some bases don’t have hydroxide ions!)
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3030Arrhenius acid is a substance that produces H+ (H3O+) in water
Arrhenius base is a substance that produces OH- in water
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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!
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A Brønsted-Lowry acid is a proton donorA Brønsted-Lowry base is a proton acceptor
acidconjugate
basebase conjugate
acid
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Acid-Base TheoriesAcid-Base TheoriesAcid-Base TheoriesAcid-Base Theories
The Brønsted definition means NHThe Brønsted definition means NH33 is is a a BaseBase in water — and water is itself in water — and water is itself an an AcidAcid
BaseAcidAcidBaseNH4
+ + OH-NH3 + H2OBaseAcidAcidBase
NH4+ + OH-NH3 + H2O
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3434
Conjugate PairsConjugate Pairs
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3535
Learning Check!
Label the acid, base, conjugate acid, and Label the acid, base, conjugate acid, and conjugate base in each reaction:conjugate base in each reaction:
HONORS ONLY!HONORS ONLY!
HCl + OHHCl + OH-- Cl Cl-- + H + H22OO HCl + OHHCl + OH-- Cl Cl-- + H + H22OO
HH22O + HO + H22SOSO44 HSO HSO44-- + H + H33OO
++ HH22O + HO + H22SOSO44 HSO HSO44-- + H + H33OO
++
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3636Acids & Base Acids & Base DefinitionsDefinitions
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 Definition #3 – Lewis
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3737
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.
HH
H
BASE
••••••
O—HO—H
H+
ACID
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Lewis Acid/Base ReactionLewis Acid/Base Reaction
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Lewis Acid-Base Lewis Acid-Base Interactions in BiologyInteractions in Biology
• The heme group The heme group in hemoglobin in hemoglobin can interact with can interact with OO22 and CO. and CO.
• The Fe ion in The Fe ion in hemoglobin is a hemoglobin is a Lewis acidLewis acid
• OO22 and CO can act and CO can act as Lewis basesas Lewis bases
Heme group
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4040
Try These!Try These!
Find the pH of these:Find the pH of these:
1) A 0.15 M solution of Hydrochloric acid1) A 0.15 M solution of Hydrochloric acid
2) A 3.00 X 102) A 3.00 X 10-7-7 M solution of Nitric acid M solution of Nitric acid
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4141
Try These!Try These!
Find the pH of these:Find the pH of these:
1) A 0.15 M solution of Hydrochloric acid1) A 0.15 M solution of Hydrochloric acid
2) A 3.00 X 102) A 3.00 X 10-7-7 M solution of Nitric acid M solution of Nitric acid
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4242pH calculations – Solving for pH calculations – Solving for H+H+pH calculations – Solving for pH calculations – Solving for H+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 (10xx) 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 “2nd nd
function” and then the log buttonfunction” and then the log button
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4343
More About WaterMore About Water
HH22O 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 = Kww
KKww = [H = [H33OO++] [OH] [OH--] =] = 1.00 x 101.00 x 10-14-14 at 25 at 25 ooCC
HONORS ONLY!HONORS ONLY!
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4444
More About WaterMore About Water
KKww = [H = [H33OO++] [OH] [OH--] = 1.00 x 10] = 1.00 x 10-14-14 at 25 at 25 ooCC
In a In a neutral neutral solution [Hsolution [H33OO++] = [OH] = [OH--]]
so Kso Kww = [H = [H33OO++]]22 = [OH = [OH--]]22
and so [Hand so [H33OO++] = [OH] = [OH--] = 1.00 x 10] = 1.00 x 10-7-7 M M
OH-
H3O+
OH-
H3O+
AutoionizationAutoionization
HONORS ONLY!HONORS ONLY!
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4545
pHpH [H+][H+] [OH-][OH-] pOHpOH
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4646The 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?
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4747
[OH[OH--]]
[H[H++]] pOHpOH
pHpH
1010 -pOH
-pOH
1010 -pH-pH-Log[H
-Log[H++]]
-Log[OH
Log[OH
--]]
14 -
pOH
14 -
pOH
14 -
pH
14 -
pH
1.0
x 10
1.0
x 10-1
4-14
[OH[O
H-- ]]
1.0
x 10
1.0
x 10-1
4-14
[H[H
++ ]]
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4848Calculating [H3O+], 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 [H3O+], pH, [OH-], and pOH of the two solutions at 25°C.
Problem 2: What is the [H3O+], [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?
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4949
HNO3, HCl, H2SO4 and HClO4 are among the only known strong acids.
Strong and Weak Strong and Weak Acids/BasesAcids/Bases
The strength of an acid (or base) is determined by the amount of IONIZATION.
The strength of an acid (or base) is determined by the amount of IONIZATION.
HONORS ONLY!HONORS ONLY!
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5050
Strong and Weak Strong and Weak Acids/BasesAcids/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: HNOHNO3 3 (aq) + H(aq) + H22O (l) --->O (l) --->
HH33OO+ + (aq) + NO(aq) + NO33- - (aq)(aq)
HNOHNO33 is about 100% dissociated in water. is about 100% dissociated in water.
HONORS ONLY!HONORS ONLY!
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5151
• 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 = CH33COCO22HH
Strong and Weak Strong and Weak Acids/BasesAcids/Bases
Strong and Weak Strong and Weak Acids/BasesAcids/Bases
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5252
• 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 Strong and Weak Acids/BasesAcids/Bases
Other common strong Other common strong bases include KOH andbases include KOH and Ca(OH)Ca(OH)22..
CaO (lime) + HCaO (lime) + H22O -->O -->
Ca(OH)Ca(OH)22 (slaked lime) (slaked lime)CaOCaO
HONORS ONLY!HONORS ONLY!
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5353
• 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
NHNH3 3 (aq) + H(aq) + H22O (l) O (l) NH NH44+ + (aq) + OH(aq) + OH- - (aq)(aq)
Strong and Weak Strong and Weak Acids/BasesAcids/Bases
Strong and Weak Strong and Weak Acids/BasesAcids/Bases
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5454
Weak BasesWeak Bases
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5555
Equilibria Involving Equilibria Involving Weak Acids and BasesWeak Acids and Bases
Consider acetic acid, HCConsider acetic acid, HC22HH33OO22 (HOAc) (HOAc)
HCHC22HH33OO22 + H + H22O O H H33OO++ + C + C22HH33OO22 --
AcidAcid Conj. base Conj. base
Ka [H3O+][OAc- ]
[HOAc] 1.8 x 10-5Ka
[H3O+][OAc- ][HOAc]
1.8 x 10-5
(K is designated K(K is designated Kaa 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!HONORS ONLY!
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5656Ionization Constants for Ionization Constants for Acids/Bases Acids/Bases
AcidsAcids ConjugateConjugateBasesBases
Increase strength
Increase strength
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5757
Equilibrium Constants Equilibrium Constants for Weak Acidsfor Weak Acids
Equilibrium Constants Equilibrium Constants for Weak Acidsfor Weak Acids
Weak acid has KWeak acid has Kaa < 1 < 1
Leads to small [HLeads to small [H33OO++] and a pH of 2 - 7] and a pH of 2 - 7
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5858
Equilibrium Constants Equilibrium Constants for Weak Basesfor Weak Bases
Equilibrium Constants Equilibrium Constants for Weak Basesfor Weak Bases
Weak base has KWeak base has Kbb < 1 < 1
Leads to small [OHLeads to small [OH--] and a pH of 12 - 7] and a pH of 12 - 7
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5959
Relation Relation
of Kof Kaa, K, Kbb, ,
[H[H33OO++] ]
and pHand pH
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6060Equilibria Involving A Weak Equilibria Involving A Weak AcidAcid
Equilibria Involving A Weak Equilibria Involving A Weak AcidAcid
You have 1.00 M HOAc. Calc. the You have 1.00 M HOAc. Calc. the equilibrium concs. of HOAc, Hequilibrium concs. of HOAc, H33OO++, 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[H33OO++]] [OAc[OAc--]]
initialinitial
changechange
equilibequilib
1.001.00 00 001.001.00 00 00
-x-x +x+x +x+x-x-x +x+x +x+x
1.00-x1.00-x xx xx1.00-x1.00-x xx xx
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6161Equilibria Involving A Weak Equilibria Involving A Weak AcidAcid
Equilibria Involving A Weak Equilibria Involving A Weak AcidAcid
Step 2.Step 2. Write KWrite Kaa 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, H33OO++, OAc, OAc--, and the pH., and the pH.
Ka 1.8 x 10-5 = [H3O+][OAc- ]
[HOAc]
x2
1.00 - xKa 1.8 x 10-5 =
[H3O+][OAc- ][HOAc]
x2
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)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!HONORS ONLY!
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6262Equilibria Involving A Weak Equilibria Involving A Weak AcidAcid
Equilibria Involving A Weak Equilibria Involving A Weak AcidAcid
Step 3.Step 3. Solve KSolve Kaa 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, H33OO++, OAc, OAc--, and the pH., and the pH.
Ka 1.8 x 10-5 = [H3O+][OAc- ]
[HOAc]
x2
1.00 - xKa 1.8 x 10-5 =
[H3O+][OAc- ][HOAc]
x2
1.00 - x
First assume x is very small because First assume x is very small because KKaa is so small. is so small.
Ka 1.8 x 10-5 = x2
1.00Ka 1.8 x 10-5 =
x2
1.00
Now we can more easily solve this Now we can more easily solve this approximate expression.approximate expression.
HONORS ONLY!HONORS ONLY!
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6363
Equilibria Involving A Weak AcidEquilibria Involving A Weak AcidEquilibria Involving A Weak AcidEquilibria Involving A Weak Acid
Step 3.Step 3. Solve KSolve Kaa 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, H33OO++, OAc, OAc--, and the pH., and the pH.
Ka 1.8 x 10-5 = x2
1.00Ka 1.8 x 10-5 =
x2
1.00
x =x = [[HH33OO++] = [] = [OAcOAc--] = 4.2 x 10] = 4.2 x 10-3-3 M M
pH = - log [pH = - log [HH33OO++] = -log (4.2 x 10] = -log (4.2 x 10-3-3) =) = 2.372.37
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6464Equilibria Involving A Weak Equilibria Involving A Weak AcidAcid
Equilibria Involving A Weak Equilibria Involving A Weak AcidAcid
Calculate the pH of a 0.0010 M solution of Calculate the pH of a 0.0010 M solution of formic acid, HCOformic acid, HCO22H.H.
HCOHCO22H + HH + H22O O HCO HCO22-- + H + H33OO++
KKaa = 1.8 x 10 = 1.8 x 10-4-4
Approximate solutionApproximate solution
[H[H33OO++] = 4.2 x 10] = 4.2 x 10-4-4 M, M, pH = 3.37pH = 3.37
Exact SolutionExact Solution
[H[H33OO++] = [HCO] = [HCO22--] = 3.4 x 10] = 3.4 x 10-4-4 M M
[HCO[HCO22H] = 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!HONORS ONLY!
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6565Equilibria Involving A Weak Equilibria Involving A Weak BaseBase
You have 0.010 M NHYou have 0.010 M NH33. Calc. the pH.. Calc. the pH.
NHNH33 + H + H22O O NH NH44++ + OH + OH--
KKbb = 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[NH33]] [NH[NH44++]] [OH[OH--]]
initialinitial
changechange
equilibequilib
0.0100.010 00 000.0100.010 00 00
-x-x +x+x +x+x-x-x +x+x +x+x
0.010 - x0.010 - x x x xx0.010 - x0.010 - x x x xx
HONORS ONLY!HONORS ONLY!
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6666Equilibria Involving A Weak Equilibria Involving A Weak BaseBase
You have 0.010 M NHYou have 0.010 M NH33. Calc. the pH.. Calc. the pH.
NHNH33 + H + H22O O NH NH44++ + OH + OH--
KKbb = 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[NH33]] [NH[NH44++]] [OH[OH--]]
initialinitial
changechange
equilibequilib
0.0100.010 00 000.0100.010 00 00
-x-x +x+x +x+x-x-x +x+x +x+x
0.010 - x0.010 - x x x xx0.010 - x0.010 - x x x xx
HONORS ONLY!HONORS ONLY!
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6767Equilibria Involving A Weak Equilibria Involving A Weak BaseBase
You have 0.010 M NHYou have 0.010 M NH33. Calc. the pH.. Calc. the pH.
NHNH33 + H + H22O O NH NH44++ + OH + OH--
KKbb = 1.8 x 10 = 1.8 x 10-5-5
Step 2.Step 2. Solve the equilibrium expressionSolve the equilibrium expression
Kb 1.8 x 10-5 = [NH4
+][OH- ][NH3 ]
= x2
0.010 - xKb 1.8 x 10-5 =
[NH4+][OH- ]
[NH3 ] =
x2
0.010 - x
Assume x is small, soAssume x is small, so x = [OHx = [OH--] = [NH] = [NH44
++] = 4.2 x 10] = 4.2 x 10-4-4 M M
and [NHand [NH33] = 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!HONORS ONLY!
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6868Equilibria Involving A Weak Equilibria Involving A Weak BaseBase
You have 0.010 M NHYou have 0.010 M NH33. Calc. the pH.. Calc. the pH.
NHNH33 + H + H22O O NH NH44++ + OH + OH--
KKbb = 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!HONORS ONLY!
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6969
Types of Acid/Base Reactions: Types of Acid/Base Reactions: SummarySummary
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7070
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.
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.
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7171
PROBLEM: You have 50.0 mL of PROBLEM: You have 50.0 mL of 3.0 M NaOH and you want 0.50 M 3.0 M NaOH and you want 0.50 M NaOH. What do you do?NaOH. What do you do?
PROBLEM: You have 50.0 mL of PROBLEM: You have 50.0 mL of 3.0 M NaOH and you want 0.50 M 3.0 M NaOH and you want 0.50 M NaOH. What do you do?NaOH. 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!
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7272
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?
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?
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?
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7373
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 doWhat do you do??
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 doWhat do you 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
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7474
PROBLEM: You have 50.0 mL of 3.0 M NaOH PROBLEM: You have 50.0 mL of 3.0 M NaOH and you want 0.50 M NaOH. What do you do?and you want 0.50 M NaOH. What do you do?PROBLEM: You have 50.0 mL of 3.0 M NaOH PROBLEM: You have 50.0 mL of 3.0 M NaOH and you want 0.50 M NaOH. What do you do?and 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) = 0.15 mol NaOH(3.0 mol/L)(0.050 L) = 0.15 mol NaOH
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)(1 L/0.50 mol) = 0.30 L(0.15 mol NaOH)(1 L/0.50 mol) = 0.30 L
or or 300 mL300 mL
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7575
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?
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?
Conclusion:Conclusion:
add 250 mL add 250 mL of waterof water to to 50.0 mL of 50.0 mL of 3.0 M NaOH 3.0 M NaOH to make 300 to make 300 mL of 0.50 M mL of 0.50 M NaOH.NaOH.
3.0 M NaOH 0.50 M NaOH
H2O
Concentrated Dilute
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7676
A shortcutA shortcut
MM11 • V • V11 = M = M22 • V • V22
Preparing Solutions Preparing Solutions by Dilutionby Dilution
Preparing Solutions Preparing Solutions by Dilutionby Dilution
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7777You 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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7878
Sources
• www.mccsc.edu/~nrapp/chemistrypowerpoint/Student%20Ch%2017%20AcidsBases.ppt