chapter 14 the chemistry of acids and bases. "acid" latin word acidus, meaning sour....
TRANSCRIPT
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CHAPTER 14CHAPTER 14
THE CHEMISTRY OF THE CHEMISTRY OF ACIDS AND BASESACIDS AND BASES
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"ACID""ACID"
Latin word acidus, meaning Latin word acidus, meaning sour. sour.
(lemon)(lemon)
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"ALKALI""ALKALI"
Arabic word for the ashes that Arabic word for the ashes that come come
from burning certain plants. from burning certain plants.
Water solutions feel slippery and Water solutions feel slippery and
taste bitter (soap).taste bitter (soap).
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Acids and bases are extremely Acids and bases are extremely
important in many everyday important in many everyday
applications: our own bloodstream, applications: our own bloodstream,
our environment, cleaning materials, our environment, cleaning materials,
industry. industry.
(sulfuric acid is an economic (sulfuric acid is an economic indicator!)indicator!)
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ACID-BASEACID-BASE
THEORIESTHEORIES
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Arrhenius Definition
acidacid----donates a hydrogen ion donates a hydrogen ion (H(H++) in water) in water
basebase--donates a hydroxide ion --donates a hydroxide ion in water (OHin water (OH--))
This theory was limited to This theory was limited to
substances with those "parts"; substances with those "parts";
ammonia is a MAJOR exception!ammonia is a MAJOR exception!
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Bronsted-Lowry Bronsted-Lowry DefinitionDefinitionacidacid----donates a proton in waterdonates a proton in water
basebase----accepts a proton in water accepts a proton in water
This theory is better; it explains This theory is better; it explains ammonia as a base! This is the main ammonia as a base! This is the main theory that we will use for our theory that we will use for our acid/base discussion.acid/base discussion.
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Lewis DefinitionLewis Definition
acidacid----accepts an electron pairaccepts an electron pair
basebase--donates an electron pair--donates an electron pair
This theory explains This theory explains allall traditional traditional acids and bases + a host of acids and bases + a host of coordination compounds and is coordination compounds and is used used
widely in organic chemistry. Uses widely in organic chemistry. Uses coordinate covalent bondscoordinate covalent bonds
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The Bronsted-Lowry The Bronsted-Lowry Concept of Acids and Bases Concept of Acids and Bases
Using this theory, you should be Using this theory, you should be
able to write weak acid/base able to write weak acid/base
dissociation equations and dissociation equations and identify identify
acid, base, conjugate acid and acid, base, conjugate acid and
conjugate base.conjugate base.
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Conjugate Acid-Base PairConjugate Acid-Base Pair
A pair of compounds that differ A pair of compounds that differ by the presence of one Hby the presence of one H++ unit. unit.
This idea is critical when it comes This idea is critical when it comes to understanding buffer systems. to understanding buffer systems.
Pay close attention here! Pay close attention here!
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AcidsAcids
donate a proton (Hdonate a proton (H++))
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Neutral CompoundNeutral Compound
HNOHNO33 + H + H22O O H H33OO++ + + NONO33
--
acid base CA CBacid base CA CB
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CationCation
NHNH44++ + H + H22O O H H33OO++ + NH + NH33
acid base CA CBacid base CA CB
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AnionAnion
HH22POPO44-- + H + H220 0 H H33OO+ + + HPO + HPO44
2-2-
acid base CA CBacid base CA CB
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In each of the acid examples---notice In each of the acid examples---notice
the formation of the formation of HH33OO++
This species is named the This species is named the hydronium hydronium
ionion. .
It lets you know that the solution is It lets you know that the solution is
acidicacidic!!
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Hydronium, HHydronium, H33OO++
--H--H++ riding piggy-back on a water riding piggy-back on a water
molecule. molecule.
Water is polar and the + charge of Water is polar and the + charge of the the
naked proton is greatly attracted to naked proton is greatly attracted to
Mickey's chin!)Mickey's chin!)
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BasesBases
accept a proton (Haccept a proton (H++))
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Neutral CompoundNeutral Compound
NHNH33 + H + H22O O NH NH44++ + OH + OH--
base acid CA CBbase acid CA CB
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AnionAnion
COCO332-2- + H + H22O O HCO HCO33
-- + + OHOH--
base acid CA base acid CA CBCB
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AnionAnion
POPO443- 3- + H + H22O O HPO HPO44
2-2- + OH + OH--
base acid CA base acid CA CBCB
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In each of the basic examples--In each of the basic examples--
notice the formation of notice the formation of OH-OH- -- this -- this
species is named the species is named the hydroxide hydroxide
ionion. It lets you know that the . It lets you know that the
solution is solution is basicbasic! !
You try!!You try!!
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Exercise 1Exercise 1
In the following reaction, identify In the following reaction, identify
the acid on the left and its CB on the acid on the left and its CB on
the right. Similarly identify the the right. Similarly identify the base base
on the left and its CA on the right.on the left and its CA on the right.
HBr + NHHBr + NH33 NH NH44++ + Br + Br--
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What is the conjugate base of What is the conjugate base of HH22S?S?
What is the conjugate acid of What is the conjugate acid of NONO33
--??
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ACIDS ONLY DONATE ACIDS ONLY DONATE
ONE PROTON AT A ONE PROTON AT A
TIME!!!TIME!!!
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– monoproticmonoprotic--acids donating one --acids donating one HH++ (ex. HC (ex. HC22HH33OO22))
– diproticdiprotic--acids donating two --acids donating two HH++'s (ex. H's (ex. H22CC22OO44))
– polyproticpolyprotic--acids donating --acids donating many Hmany H++'s (ex. H's (ex. H33POPO44))
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Polyprotic BasesPolyprotic Bases
accept more than one Haccept more than one H++
anions with -2 and -3 charges anions with -2 and -3 charges
(example: PO(example: PO443-3- ; HPO ; HPO44
2-2-))
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Amphiprotic or Amphiprotic or AmphotericAmphoteric
molecules or ions that can behave as molecules or ions that can behave as
EITHER acids or bases: EITHER acids or bases:
water, anions of weak acids water, anions of weak acids
(look at the examples above—sometimes (look at the examples above—sometimes
water was an acid, sometimes it acted as water was an acid, sometimes it acted as
a base)a base)
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Exercise 2Exercise 2 Acid Acid Dissociation (Ionization) Dissociation (Ionization) ReactionsReactions
Write the simple dissociation Write the simple dissociation
(ionization) reaction (omitting (ionization) reaction (omitting
water) for each of the following water) for each of the following
acids.acids.
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a. Hydrochloric acid (HCl)a. Hydrochloric acid (HCl)
b. Acetic acid (HCb. Acetic acid (HC22HH33OO22))
c. The ammonium ion (NHc. The ammonium ion (NH44++))
d. The anilinium ion (Cd. The anilinium ion (C66HH55NHNH33++))
e. The hydrated aluminum(III) ion e. The hydrated aluminum(III) ion
[Al(H[Al(H22O)O)66]]3+3+
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SolutionSolution
A: HCl(aq) A: HCl(aq) H H++(aq) + Cl(aq) + Cl--(aq)(aq)
B: HCB: HC22HH33OO22(aq) (aq)
HH++(aq) + C(aq) + C22HH33OO22--(aq)(aq)
C: NHC: NH44++(aq) (aq) H H++(aq) + NH(aq) + NH33(aq)(aq)
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Solution, cont.Solution, cont.
D: CD: C66HH55NHNH33++(aq) (aq) H H++(aq) + (aq) +
CC66HH55NHNH22(aq)(aq)
E: Al(HE: Al(H22O)O)663+3+(aq) (aq)
HH++(aq) + Al(H(aq) + Al(H22O)O)55OHOH22++(aq) (aq)
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Relative Strengths of Relative Strengths of Acids and BasesAcids and Bases
Strength is determined by the Strength is determined by the
position of the "dissociation" position of the "dissociation"
equilibrium.equilibrium.
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Strong acids/Strong basesStrong acids/Strong bases
dissociate completely in waterdissociate completely in water
have very large K valueshave very large K values
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Weak acids/Weak basesWeak acids/Weak bases
dissociate only to a slight extent dissociate only to a slight extent in in
waterwater
dissociation constant is very dissociation constant is very small small
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Strong
Weak
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Do NotDo Not
confuse concentration confuse concentration
with strength! with strength!
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Strong AcidsStrong Acids
Hydrohalic acids: Hydrohalic acids: HCl, HBr, HIHCl, HBr, HI
Nitric: HNONitric: HNO33
Sulfuric: HSulfuric: H22SOSO44
Perchloric: HClOPerchloric: HClO44
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The more The more oxygen present oxygen present in the in the polyatomic ion, polyatomic ion,
the stronger its the stronger its acid WITHIN that acid WITHIN that group. group.
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Strong BasesStrong Bases
Hydroxides OR oxides of IA Hydroxides OR oxides of IA and and
IIA metalsIIA metals
Solubility plays a role (those Solubility plays a role (those that that
are very soluble are strong!)are very soluble are strong!)
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The stronger The stronger the acid, the the acid, the weaker its CB. weaker its CB.
The converse The converse is also true.is also true.
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Weak Acids and Bases - Weak Acids and Bases - Equilibrium Equilibrium expressionsexpressions
The vast majority of acid/bases are The vast majority of acid/bases are
weak.weak.
Remember, this means they do Remember, this means they do not not
ionize much.ionize much.
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The equilibrium expression for The equilibrium expression for acids acids
is known as the is known as the KKaa (the acid (the acid
dissociation constantdissociation constant). ).
It is set up the same way as in It is set up the same way as in
general equilibrium.general equilibrium.
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Many common weak acids are Many common weak acids are
oxyacids, oxyacids,
like phosphoric acid and like phosphoric acid and
nitrous acid.nitrous acid.
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Other common Other common weak acids are weak acids are organic acids,organic acids,those that contain a those that contain a carboxyl groupcarboxyl group
COOH group COOH group
like acetic acid and like acetic acid and benzoic acid. benzoic acid.
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For Weak Acid Reactions:For Weak Acid Reactions:
HA + HHA + H22O O H H33OO+ + + A + A--
KKaa = = [H [H33OO++][A][A--]] < 1 < 1
[HA][HA]
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Write the KWrite the Kaa expression for acetic expression for acetic
acid using Bronsted-Lowry. acid using Bronsted-Lowry.
(Note: Water is a pure liquid and (Note: Water is a pure liquid and
is thus, left out of the equilibrium is thus, left out of the equilibrium expression.)expression.)
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Weak bases (bases without Weak bases (bases without OHOH--) )
react with water to produce a react with water to produce a
hydroxide ion.hydroxide ion.
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Common examples of weak bases are Common examples of weak bases are
ammonia (NHammonia (NH33), methylamine ), methylamine
(CH(CH33NHNH22), and ethylamine (C), and ethylamine (C22HH55NHNH22). ).
The lone pair on N forms a bond with The lone pair on N forms a bond with
an Han H++. Most weak bases involve N.. Most weak bases involve N.
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The equilibrium expression The equilibrium expression for for
basesbases is known as the is known as the KKbb..
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For Weak Base For Weak Base Reactions:Reactions:
B + HB + H22O O HB HB+ + + OH + OH--
KKbb = = [H [H33OO++][OH][OH--] ] <1 <1
[B][B]
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Set up the KSet up the Kbb expression for expression for
ammonia using Bronsted-ammonia using Bronsted-Lowry.Lowry.
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Notice that KNotice that Kaa and K and Kbb expressions expressions
look very similar. look very similar.
The difference is that a base The difference is that a base
produces the hydroxide ion in produces the hydroxide ion in
solution, while the acid produces the solution, while the acid produces the
hydronium ion in solution. hydronium ion in solution.
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Another note on this Another note on this point:point:
HH++ and H and H33OO++ are both equivalent are both equivalent terms here. Often water is left terms here. Often water is left completely out of the equation completely out of the equation since since
it does not appear in the it does not appear in the equilibrium. equilibrium.
This has become an accepted This has become an accepted practice. practice.
(*However, water is very important (*However, water is very important in causing the acid to dissociate.)in causing the acid to dissociate.)
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Exercise 3Exercise 3 Relative Base StrengthRelative Base Strength
Using table 14.2, arrange the Using table 14.2, arrange the
following species according to following species according to their their
strength as bases: strength as bases:
HH22O, FO, F--, Cl, Cl--, NO, NO22--, and CN, and CN--
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SolutionSolution
ClCl-- < H < H22O < FO < F-- < NO < NO22-- < CN < CN--
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WATER WATER
THE HYDRONIUM ION THE HYDRONIUM ION
AUTO-IONIZATION AUTO-IONIZATION
THE pH SCALETHE pH SCALE
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Fredrich Kohlrausch, around Fredrich Kohlrausch, around
1900, found that no matter how 1900, found that no matter how
pure water is, it still conducts a pure water is, it still conducts a
minute amount of electric minute amount of electric
current. This proves that water current. This proves that water
self-ionizes. self-ionizes.
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Since the water molecule is Since the water molecule is
amphoteric, it may dissociate amphoteric, it may dissociate
with itself to a slight extent. with itself to a slight extent.
Only about 2 out of a billion Only about 2 out of a billion
water molecules are ionized at water molecules are ionized at
any instant!any instant!
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HH22O(l) + HO(l) + H22O(l) <=> HO(l) <=> H33OO++(aq) + OH(aq) + OH--
(aq)(aq)
The equilibrium expression used The equilibrium expression used
here is referred to as the here is referred to as the KKww
(ionization constant for (ionization constant for water)water). .
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In pure water or dilute aqueous In pure water or dilute aqueous
solutions, the concentration of water solutions, the concentration of water
can be considered to be a constant can be considered to be a constant
(55.4 M), so we include that with the (55.4 M), so we include that with the
equilibrium constant and write the equilibrium constant and write the
expression as:expression as:
KKeqeq[H[H22O]O]22 = K = Kww = [H = [H33OO++][OH][OH--]]
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KKww = 1.0 x 10 = 1.0 x 10-14-14
(K(Kww = 1.008 x 10 = 1.008 x 10-14-14 @ 25° Celsius) @ 25° Celsius)
Knowing this value allows us to Knowing this value allows us to
calculate the OHcalculate the OH-- and H and H++
concentration for concentration for various various situationsituations.s.
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[OH[OH--] = [H] = [H++] : solution is neutral (in ] : solution is neutral (in
pure water, each of these is 1.0 x 10pure water, each of these is 1.0 x 10--
77))
[OH[OH--] > [H] > [H++] : solution is basic ] : solution is basic
[OH[OH--] < [H] < [H++] : solution is acidic ] : solution is acidic
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KKww = K = Kaa x K x Kbb
another very beneficial another very beneficial equationequation
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Exercise 5Exercise 5 Autoionization of WaterAutoionization of Water
At 60°C, the value of KAt 60°C, the value of Kww is 1 X 10 is 1 X 10-13-13..
a. Using Le Chatelier’s principle, a. Using Le Chatelier’s principle,
predict whether the reaction predict whether the reaction
2H2H22O(l) O(l) H H33OO++(aq) + OH(aq) + OH--(aq)(aq)
is exothermic or endothermic.is exothermic or endothermic.
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Exercise 5, cont.Exercise 5, cont.
b. Calculate [Hb. Calculate [H++] and [OH] and [OH--] in a ] in a
neutral solution at 60°C.neutral solution at 60°C.
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SolutionSolution
A: endothermicA: endothermic
B: [HB: [H++] = [OH] = [OH--] = 3 X 10] = 3 X 10-7-7 MM
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The pH ScaleThe pH Scale
Used to Used to designate the designate the [H[H++] in most ] in most aqueous aqueous solutions solutions where Hwhere H++ is is small.small.
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pH = - log [HpH = - log [H++] ]
pOH = - log [OHpOH = - log [OH--] ]
pH + pOH = 14 pH + pOH = 14
pH = 6.9 and lower (acidic) pH = 6.9 and lower (acidic)
= 7.0 (neutral) = 7.0 (neutral)
= 7.1 and greater = 7.1 and greater (basic)(basic)
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Use as many decimal places as Use as many decimal places as
there are sig.figs. in the problem!there are sig.figs. in the problem!
The negative base 10 logarithm of The negative base 10 logarithm of
the hydronium ion concentration the hydronium ion concentration
becomes the whole number; becomes the whole number;
therefore, only the decimals to the therefore, only the decimals to the
right are significant. right are significant.
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Exercise 6Exercise 6 Calculating [HCalculating [H++] and ] and [OH[OH--]]Calculate [HCalculate [H++] or [OH] or [OH--] as required ] as required
for for
each of the following solutions at each of the following solutions at
25°C, and state whether the solution 25°C, and state whether the solution
is neutral, acidic, or basic.is neutral, acidic, or basic.
a. 1.0 X 10a. 1.0 X 10-5-5 MM OH OH--
b. 1.0 X 10b. 1.0 X 10-7-7 MM OH OH--
c. 10.0c. 10.0 M M H H++
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SolutionSolution
A: [HA: [H++] = 1.0 X 10] = 1.0 X 10-9-9 M, M, basic basic
B: [HB: [H++] = 1.0 X 10] = 1.0 X 10-7-7 M,M, neutral neutral
C: [OHC: [OH--] = 1.0 X 10] = 1.0 X 10-15-15 M, M, acidicacidic
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Exercise 7Exercise 7 Calculating pH and Calculating pH and pOHpOHCalculate pH and pOH for each of Calculate pH and pOH for each of
the following solutions at 25°C.the following solutions at 25°C.
a. 1.0 X 10a. 1.0 X 10-3-3 MM OH OH--
b. 1.0b. 1.0 M M H H++
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SolutionSolution
A: pH = 11.00A: pH = 11.00
pOH = 3.00pOH = 3.00
B: pH = 0.00B: pH = 0.00
pOH = 14.00pOH = 14.00
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Exercise 8Exercise 8 Calculating Calculating pHpH
The pH of a sample of human The pH of a sample of human blood blood
was measured to be 7.41 at 25°C. was measured to be 7.41 at 25°C.
Calculate pOH, [HCalculate pOH, [H++], and [OH], and [OH--] for ] for
the sample.the sample.
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SolutionSolution
pOH = 6.59pOH = 6.59
[H[H++] = 3.9 X 10] = 3.9 X 10-8-8
[OH[OH--] = 2.6 X 10] = 2.6 X 10-7-7 MM
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Exercise 9Exercise 9 pH of Strong Acids pH of Strong Acids
Calculate the pH of:Calculate the pH of:
a. 0.10 a. 0.10 MM HNO HNO33
b. 1.0 X 10b. 1.0 X 10-10-10 MM HCl HCl
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SolutionSolution
A: pH = 1.00A: pH = 1.00
B: pH = 7.00B: pH = 7.00
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Exercise 10Exercise 10 The pH of Strong The pH of Strong BasesBases
Calculate the pH of a 5.0 X 10Calculate the pH of a 5.0 X 10-2-2 MM
NaOH solution.NaOH solution.
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SolutionSolution
pH = 12.70pH = 12.70
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Calculating pH of Weak Calculating pH of Weak Acid SolutionsAcid Solutions
Calculating pH of weak acids Calculating pH of weak acids
involves setting up an involves setting up an equilibrium. equilibrium.
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Always start by…Always start by…
1) writing the equation… 1) writing the equation…
2) setting up the acid equilibrium 2) setting up the acid equilibrium
expression (Kexpression (Kaa)…)…
3) defining initial concentrations, 3) defining initial concentrations, changes, and final changes, and final concentrations in terms of X …concentrations in terms of X …
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4) substituting values and 4) substituting values and variables variables
into the Kinto the Kaa expression… expression…
5) solving for X 5) solving for X
(use the (use the RICERICE diagram learned in diagram learned in
general equilibrium!)general equilibrium!)
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Example:Example:
Calculate the pH of a 1.00 x 10Calculate the pH of a 1.00 x 10--
44 M M
solution of acetic acid.solution of acetic acid.
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The KThe Kaa of acetic acid is 1.8 x 10 of acetic acid is 1.8 x 10-5-5
HCHC22HH33OO22 H H++ + C + C22HH33OO22--
KKaa = = [H[H++][C][C22HH33OO22--]] = 1.8 x 10 = 1.8 x 10-5-5
[HC[HC22HH33OO22]]
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RReaction HCeaction HC22HH33OO22 H+ + C H+ + C22HH33OO22--
IInitial 1.00 x 10nitial 1.00 x 10-4-4 0 0 0 0
CChange -x +x +xhange -x +x +x
EEquilibrium 1.00 x 10quilibrium 1.00 x 10-4 -4 - x x x- x x x
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1.8 x 101.8 x 10-5-5 = = (x)(x) _ (x)(x) _
1.00x101.00x10-4-4 - - xx
1.8 x 10-5 1.8 x 10-5 (x)(x) _ (x)(x) _
1.00 x 101.00 x 10-4-4
x = 4.2 x 10x = 4.2 x 10-5-5
Often, the -x in a KOften, the -x in a Kaa expression can be treated as expression can be treated as negligible.negligible.
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When you assume that x is When you assume that x is
negligible, you must check the negligible, you must check the
validity of this assumption. validity of this assumption.
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To be valid, x must be less than To be valid, x must be less than
5% of the number that it was to be 5% of the number that it was to be
subtracted from.subtracted from.
% dissociation = % dissociation = "x" "x" x 100 x 100
[original][original]
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In this example, 4.2 x 10In this example, 4.2 x 10-5-5 is greater is greater
than 5% of 1.00 x 10than 5% of 1.00 x 10-4-4. .
This means that the assumption that This means that the assumption that
x was negligible was invalid and x x was negligible was invalid and x
must be solved for using the must be solved for using the
quadratic equation or the method of quadratic equation or the method of
successive approximation.successive approximation.
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Use of the Use of the Quadratic Equation Quadratic Equation
a
acbbx
2
42
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ax2 + bx + c = 0
0108.1108.1 952 xx
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5105.3 x 5102.5 x
)1(2
)108.1)(1(4)108.1(108.1 9255 x
and
Using the values: a = 1, b = 1.8x10-5, c= -1.8x10-9
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Since a concentration Since a concentration can not be negative…can not be negative…
x = 3.5 x 10x = 3.5 x 10-5-5 M M
x = [Hx = [H++] = 3.5 x 10] = 3.5 x 10-5-5
pH = -log 3.5 x 10pH = -log 3.5 x 10-5-5 = 4.46 = 4.46
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Another method which some people Another method which some people
prefer is the method of successive prefer is the method of successive
approximations. In this method, you approximations. In this method, you
start out assuming that x is start out assuming that x is
negligible, solve for x, and repeatedly negligible, solve for x, and repeatedly
plug your value of x into the plug your value of x into the
equation again until you get the equation again until you get the
same value of x two successive times.same value of x two successive times.
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Exercise 11Exercise 11 The pH of Weak Acids The pH of Weak Acids
The hypochlorite ion (OClThe hypochlorite ion (OCl--) is a ) is a
strong oxidizing agent often found strong oxidizing agent often found
in household bleaches and in household bleaches and
disinfectants. It is also the active disinfectants. It is also the active
ingredient that forms when ingredient that forms when
swimming pool water is treated swimming pool water is treated with with
chlorine. chlorine.
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In addition to its oxidizing abilities, In addition to its oxidizing abilities,
the hypochlorite ion has a relatively the hypochlorite ion has a relatively
high affinity for protons (it is a high affinity for protons (it is a
much stronger base than Clmuch stronger base than Cl--, for , for
example) and forms the weakly example) and forms the weakly
acidic hypochlorous acid (HOCl, acidic hypochlorous acid (HOCl,
KKaa = 3.5 X 10 = 3.5 X 10-8-8).).
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Calculate the pH of a 0.100Calculate the pH of a 0.100 M M
aqueous solution of aqueous solution of hypochlorous hypochlorous
acid.acid.
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SolutionSolution
pH = 4.23pH = 4.23
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Determination of the pH Determination of the pH of a Mixture of Weak of a Mixture of Weak AcidsAcidsOnly the acid with the largest KOnly the acid with the largest Kaa
value will contribute an value will contribute an appreciable appreciable
[H[H++]. ].
Determine the pH based on Determine the pH based on
this acid and ignore any others.this acid and ignore any others.
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Exercise 12Exercise 12 The pH The pH of Weak Acid Mixturesof Weak Acid Mixtures
Calculate the pH of a solution that Calculate the pH of a solution that
contains: contains:
1.00 1.00 MM HCN (K HCN (Kaa = 6.2 X 10 = 6.2 X 10-10-10) ) and and
5.00 5.00 MM HNO HNO2 2 (K(Kaa = 4.0 X 10 = 4.0 X 10-4-4). ).
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Exercise 12, cont.Exercise 12, cont.
Also, calculate the Also, calculate the concentration of concentration of
cyanide ion (CNcyanide ion (CN--) in this ) in this solution at solution at
equilibrium.equilibrium.
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SolutionSolution
pH = 1.35pH = 1.35
[CN[CN--] = 1.4 X 10] = 1.4 X 10-8-8 MM
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Exercise 13Exercise 13 Calculating Calculating Percent DissociationPercent Dissociation
Calculate the percent dissociation Calculate the percent dissociation of of
acetic acid (Kacetic acid (Kaa = 1.8 X 10 = 1.8 X 10-5-5) in ) in each of the following solutions.each of the following solutions.
a. 1.00 a. 1.00 MM HC HC22HH33OO22
b. 0.100 b. 0.100 MM HC HC22HH33OO22
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SolutionSolution
A: = 0.42 %A: = 0.42 %
B: = 1.3 %B: = 1.3 %
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Exercise 14Exercise 14 Calculating Calculating KKaa from Percent from Percent DissociationDissociation
Lactic acid (HCLactic acid (HC33HH55OO33) is a waste ) is a waste
product that accumulates in product that accumulates in muscle muscle
tissue during exertion, leading tissue during exertion, leading to to
pain and a feeling of fatigue. pain and a feeling of fatigue.
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Exercise 14, cont.Exercise 14, cont.
In a 0.100 In a 0.100 M M aqueous solution, aqueous solution,
lactic acid is 3.7% dissociated. lactic acid is 3.7% dissociated.
Calculate the value of KCalculate the value of Kaa for for this this
acid.acid.
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SolutionSolution
KKaa= 1.4 X 10= 1.4 X 10-4-4
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Determination of the pH of a Determination of the pH of a weak weak
basebase is very similar to the is very similar to the
determination of the pH of a weak determination of the pH of a weak
acid. acid.
Follow the same steps. Follow the same steps.
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Remember, however, that Remember, however, that xx is is the the
[OH[OH--]] and taking the negative and taking the negative log log
of of xx will give you the will give you the pOH pOH and and not not
the pH!the pH!
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Exercise 15Exercise 15 The pH of Weak Bases I The pH of Weak Bases I
Calculate the pH for a 15.0 Calculate the pH for a 15.0 MM
solution of NHsolution of NH33 (K (Kbb = 1.8 X 10 = 1.8 X 10-5-5).).
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SolutionSolution
pH = 12.20pH = 12.20
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Exercise 16Exercise 16 The pH of Weak Bases The pH of Weak Bases IIIICalculate the pH of a 1.0 Calculate the pH of a 1.0 MM
solution solution
of methylamine (Kof methylamine (Kbb = 4.38 X = 4.38 X 1010-4-4).).
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SolutionSolution
pH = 12.32pH = 12.32
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Calculating pH of Calculating pH of polyprotic acidspolyprotic acids
Acids with more than one Acids with more than one ionizable ionizable
hydrogen will ionize in steps. hydrogen will ionize in steps.
Each dissociation has its own KEach dissociation has its own Kaa
value. value.
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The The firstfirst dissociation will be dissociation will be the the
greatestgreatest and subsequent and subsequent
dissociations will have much dissociations will have much smaller equilibrium smaller equilibrium constants.constants.
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As each H is removed, the As each H is removed, the
remaining acid gets weaker remaining acid gets weaker and and
therefore has a smaller Ktherefore has a smaller Kaa..
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As the negative charge on the As the negative charge on the acid acid
increases, it becomes more increases, it becomes more difficult difficult
to remove the positively to remove the positively charged charged
proton.proton.
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Example:Example:
Consider the dissociation of Consider the dissociation of phosphoric acid. phosphoric acid.
HH33POPO4(aq)4(aq) + H + H22OO(l)(l) <=> <=>
HH33OO++(aq)(aq) + H + H22POPO44
-- (aq)(aq)
KKa1a1 = 7.5 x 10 = 7.5 x 10-3-3
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HH22POPO44--(aq)(aq) + H + H22OO(l) (l) <=> <=>
HH33OO++(aq)(aq) + HPO + HPO44
2-2-
(aq)(aq)
KKa2a2 = 6.2 x 10 = 6.2 x 10-8-8
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HPOHPO442-2-
(aq)(aq) + H + H22OO(l) (l) <=> <=>
HH33OO++(aq)(aq) + PO + PO44
3-3-(aq)(aq)
KKaa33 = 4.8 x 10 = 4.8 x 10-13-13
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Looking at the KLooking at the Kaa values, it is values, it is
obvious that only the first obvious that only the first
dissociation will be important in dissociation will be important in
determining the pH of the solution.determining the pH of the solution.
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Except for HExcept for H22SOSO44, polyprotic acids , polyprotic acids
have Khave Ka2a2 and K and Ka3a3 values so much values so much
weaker than their Kweaker than their Ka1a1 value that value that
the 2nd and 3rd (if applicable) the 2nd and 3rd (if applicable)
dissociation can be ignored. dissociation can be ignored.
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The [HThe [H++] obtained from this 2nd ] obtained from this 2nd
and 3rd dissociation is and 3rd dissociation is negligible negligible
compared to the [Hcompared to the [H++] from the ] from the 1st 1st
dissociation.dissociation.
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Because HBecause H22SOSO44 is a strong acid in its is a strong acid in its
first dissociation and a weak acid in first dissociation and a weak acid in
its second, we need to consider both its second, we need to consider both
if the concentration is more dilute if the concentration is more dilute
than 1.0 M. than 1.0 M.
The quadratic equation is needed to The quadratic equation is needed to
work this type of problem.work this type of problem.
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Exercise 17Exercise 17 The pH of a Polyprotic The pH of a Polyprotic AcidAcidCalculate the pH of a 5.0 Calculate the pH of a 5.0 MM H H33POPO44
solution and the equilibrium solution and the equilibrium
concentrations of the species: concentrations of the species:
HH33POPO44, H, H22POPO44--, HPO, HPO44
2-2-, and PO, and PO443-3-
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SolutionSolution
pH = 0.72pH = 0.72
[H[H33POPO44] = 4.8 ] = 4.8 MM
[H[H22POPO44--] = 0.19] = 0.19 M M
[HPO[HPO442-2-] = 6.2 X 10] = 6.2 X 10-8 -8 MM
[PO[PO443-3-] = 1.6 X 10] = 1.6 X 10-19-19 MM
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Exercise 18Exercise 18 The pH of a Sulfuric The pH of a Sulfuric AcidAcidCalculate the pH of a 1.0 Calculate the pH of a 1.0 MM
HH22SOSO44
solution.solution.
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SolutionSolution
pH = 0.00pH = 0.00
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Exercise 19Exercise 19 The pH of a Sulfuric The pH of a Sulfuric AcidAcidCalculate the pH of a 1.0 X 10Calculate the pH of a 1.0 X 10-2-2
MM
HH22SOSO44 solution. solution.
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SolutionSolution
pH = 1.84pH = 1.84
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ACID-BASE ACID-BASE PROPERTIES OF PROPERTIES OF
SALTS:SALTS:
HYDROLYSISHYDROLYSIS
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Salts are produced from the reaction Salts are produced from the reaction
of an acid and a base. of an acid and a base. (neutralization) (neutralization)
Salts are Salts are not alwaysnot always neutral. Some neutral. Some
hydrolyze with water to produce hydrolyze with water to produce
acidic and basic solutions.acidic and basic solutions.
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Neutral SaltsNeutral Salts
Salts that are formed from the Salts that are formed from the cation of a strong base and the cation of a strong base and the anion of a strong acid form anion of a strong acid form neutral solutions when dissolved neutral solutions when dissolved in water. in water.
A salt such as NaNOA salt such as NaNO33 gives a gives a neutral solution.neutral solution.
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Basic SaltsBasic Salts
Salts that are formed from the Salts that are formed from the
cation of a strong base and the cation of a strong base and the
anion of a weak acid form basic anion of a weak acid form basic
solutions when dissolved in solutions when dissolved in water.water.
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The anion hydrolyzes the The anion hydrolyzes the water water
molecule to produce molecule to produce hydroxide hydroxide
ions and thus a basic ions and thus a basic solution. solution.
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KK22S should be basic since SS should be basic since S-2-2 is is
the CB of the very weak acid the CB of the very weak acid HSHS--, ,
while Kwhile K++ does not hydrolyze does not hydrolyze
appreciably.appreciably.
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SS2-2- + H + H22O O OH OH-- + HS + HS--
strong base weak strong base weak acidacid
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Acid SaltsAcid Salts
Salts that are formed from the Salts that are formed from the
cation of a weak base and the cation of a weak base and the
anion of a strong acid form anion of a strong acid form
acidic solutions when acidic solutions when dissolved in dissolved in
water.water.
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The cation hydrolyzes the The cation hydrolyzes the water water
molecule to produce molecule to produce hydronium hydronium
ions and thus an acidic ions and thus an acidic solution. solution.
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NHNH44Cl should be weakly acidic, Cl should be weakly acidic,
since NHsince NH44++ hydrolyzes to give an hydrolyzes to give an
acidic solution, while Clacidic solution, while Cl-- does not does not
hydrolyze.hydrolyze.
NHNH44++ + H + H22O O H H33OO++ + NH + NH33
strong acid weak strong acid weak basebase
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If both the cation If both the cation
and the anion and the anion
contribute to the contribute to the
pH situation, pH situation,
compare Kcompare Kaa to K to Kbb..
If KIf Kbb is larger, basic! is larger, basic!
The converse is also true.The converse is also true.
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The following will help predict The following will help predict
acidic, basic, or neutral. acidic, basic, or neutral.
However, you must explain However, you must explain using using
appropriate equations as appropriate equations as proof!!!proof!!!
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1. Strong acid + Strong base 1. Strong acid + Strong base
= =
Neutral saltNeutral salt
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2. Strong acid + Weak base 2. Strong acid + Weak base
= =
Acidic salt Acidic salt
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3. Weak acid + Strong base 3. Weak acid + Strong base
= =
Basic salt Basic salt
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4. Weak acid + Weak base 4. Weak acid + Weak base
= =
??? ???
(must look at K values to decide)(must look at K values to decide)
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Exercise 20Exercise 20 The Acid-The Acid-Base Properties of SaltsBase Properties of Salts
Predict whether an aqueous Predict whether an aqueous solution solution
of each of the following salts will be of each of the following salts will be
acidic, basic, or neutral. Prove with acidic, basic, or neutral. Prove with
appropriate equations. appropriate equations.
a. NHa. NH44CC22HH33OO22
b. NHb. NH44CNCN
c. Alc. Al22(SO(SO44))33
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SolutionSolution
A: neutralA: neutral
B: basicB: basic
C: acidicC: acidic
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Exercise 21Exercise 21 Salts as Weak Bases Salts as Weak Bases
Calculate the pH of a 0.30 Calculate the pH of a 0.30 MM NaF NaF
solution. solution.
The KThe Kaa value for HF is 7.2 X 10 value for HF is 7.2 X 10-4-4..
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SolutionSolution
pH = 8.31pH = 8.31
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Exercise 22Exercise 22 Salts as Weak Acids I Salts as Weak Acids I
Calculate the pH of a 0.10 Calculate the pH of a 0.10 MM NHNH44Cl Cl
solution. solution.
The KThe Kbb value for NH value for NH33 is 1.8 X 10 is 1.8 X 10--
55..
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SolutionSolution
pH = 5.13 pH = 5.13
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Exercise 23Exercise 23 Salts as Weak Acids II Salts as Weak Acids II
Calculate the pH of a 0.010 Calculate the pH of a 0.010 MM AlClAlCl33
solution. solution.
The KThe Kaa value for Al(H value for Al(H22O)O)663+3+ is is
1.4 X 101.4 X 10-5-5..
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SolutionSolution
pH = 3.43pH = 3.43
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The Lewis Concept of The Lewis Concept of Acids and BasesAcids and Bases
acidacid--can --can acceptaccept a pair of a pair of
electrons to form a coordinate electrons to form a coordinate
covalent bondcovalent bond
basebase--can --can donatedonate a pair of a pair of
electrons to form a coordinate electrons to form a coordinate
covalent bondcovalent bond
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Yes, this is the dot guy and the Yes, this is the dot guy and the
structures guy.structures guy.
He was extremely busy He was extremely busy making making
your life difficult!your life difficult!
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BFBF3 3 — the most famous of — the most famous of all!!all!!
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Exercise 24Exercise 24
Tell whether each of the following is Tell whether each of the following is
a Lewis acid or base. a Lewis acid or base.
Draw structures as proof. Draw structures as proof.
a) PHa) PH33 c) H c) H22SS
b) BClb) BCl33 d) SF d) SF44
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Exercise 25Exercise 25Lewis Acids and BasisLewis Acids and Basis
For each reaction, identify the For each reaction, identify the Lewis Lewis
acid and base.acid and base.
a. Nia. Ni22++(aq) + 6NH(aq) + 6NH33(aq) (aq) Ni(NHNi(NH33))66
2+2+(aq)(aq)
b. Hb. H++(aq) + H(aq) + H22O(aq) O(aq) H H33OO++(aq) (aq)
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SolutionSolution
A: Lewis acid = nickel(II) ionA: Lewis acid = nickel(II) ion
Lewis base = ammoniaLewis base = ammonia
B: Lewis acid = protonB: Lewis acid = proton
Lewis base = water Lewis base = water moleculemolecule
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