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The Chemistry of The Chemistry of Acids and BasesAcids and BasesThe Chemistry of The Chemistry of Acids and BasesAcids and Bases
Chapter 17Chapter 17Copyright © 1999 by Harcourt Brace & CompanyAll rights reserved.Requests for permission to make copies of any part of the work should be mailed to: Permissions Department, Harcourt Brace & Company, 6277 Sea Harbor Drive, Orlando, Florida
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Strong and Weak Acids/BasesStrong and Weak Acids/BasesStrong and Weak Acids/BasesStrong and Weak Acids/Bases
• Generally divide acids and bases into Generally divide acids and bases into STRONG or WEAK ones.STRONG or WEAK ones.
STRONG ACID:STRONG ACID: HNO HNO33(aq) + H(aq) + H22O(liq) --->O(liq) --->
HH33OO++(aq) + NO(aq) + NO33--(aq)(aq)
HNOHNO33 is about 100% dissociated in water. is about 100% dissociated in water.
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Strong and Weak Acids/BasesStrong and Weak Acids/BasesStrong and Weak Acids/BasesStrong and Weak Acids/Bases
• Generally divide acids and bases into Generally divide acids and bases into STRONG or WEAK ones.STRONG or WEAK ones.
STRONG ACID:STRONG ACID: HNO HNO33(aq) + H(aq) + H22O(liq) --->O(liq) --->
HH33OO++(aq) + NO(aq) + NO33--(aq)(aq)
HNOHNO33 is about 100% dissociated in water. is about 100% dissociated in water.
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H2O
H—Cl
Cl-
H3O+
H2O
H—Cl
Cl-
H3O+
HNO3, HCl, H2SO4 and HClO4 are among the only known strong acids.
Strong and Weak Acids/BasesStrong and Weak Acids/BasesStrong and Weak Acids/BasesStrong and Weak Acids/Bases
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• Weak acids are much less than 100% ionized Weak acids are much less than 100% ionized
in water.in water.
One of the best known is acetic acid = One of the best known is acetic acid = CHCH33COCO22H = HOAcH = HOAc
HOAc(aq) + HHOAc(aq) + H22O(liq) O(liq) OAcOAc--(aq) + H(aq) + H33OO++(aq)(aq)
OAcOAc-- = CH = CH33COCO22-- = acetate ion = acetate ion
Strong and Weak Acids/BasesStrong and Weak Acids/BasesStrong and Weak Acids/BasesStrong and Weak Acids/Bases
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• 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/BasesStrong and Weak Acids/BasesStrong and Weak Acids/Bases
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• 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/BasesStrong and Weak Acids/BasesStrong and Weak Acids/Bases
Other common strong Other common strong bases include KOH and bases include KOH and Ca(OH)Ca(OH)22..
CaO (lime) + HCaO (lime) + H22O -->O -->
Ca(OH)Ca(OH)22 (slaked lime) (slaked lime)CaOCaO
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• 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
NHNH33(aq) + H(aq) + H22O(liq) O(liq) NH NH44++(aq) + OH(aq) + OH--
(aq)(aq)
Strong and Weak Acids/BasesStrong and Weak Acids/BasesStrong and Weak Acids/BasesStrong and Weak Acids/Bases
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• Weak base:Weak base: less than 100% less than 100% ionized in waterionized in water
One of the best known weak bases is One of the best known weak bases is ammoniaammonia
NHNH33(aq) + H(aq) + H22O(liq) O(liq) NH NH44++(aq) + OH(aq) + OH--
(aq)(aq)
Strong and Weak Acids/BasesStrong and Weak Acids/BasesStrong and Weak Acids/BasesStrong and Weak Acids/Bases
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ACID-BASE THEORIESACID-BASE THEORIESACID-BASE THEORIESACID-BASE THEORIES• The most general theory for common The most general theory for common
aqueous acids and bases is the aqueous acids and bases is the BRØNSTED - LOWRY BRØNSTED - LOWRY theorytheory
• ACIDS DONATE HACIDS DONATE H++ IONS IONS
• BASES ACCEPT HBASES ACCEPT H++ IONS IONS
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The Brønsted definition means NHThe Brønsted definition means NH33 is a is a BASEBASE in water — and water is itself an in water — and water is itself an ACIDACID
BaseAcidAcidBaseNH4
+ + OH-NH3 + H2OBaseAcidAcidBase
NH4+ + OH-NH3 + H2O
ACID-BASE THEORIESACID-BASE THEORIESACID-BASE THEORIESACID-BASE THEORIES
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ACID-BASE THEORIESACID-BASE THEORIESACID-BASE THEORIESACID-BASE THEORIESNHNH33 is a is 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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ACID-BASE THEORIESACID-BASE THEORIESACID-BASE THEORIESACID-BASE THEORIESNHNH33 is a is a BASEBASE in water — and water is itself in water — and water is itself
an an ACIDACID
NHNH3 3 / NH/ NH44++ is a is a conjugate pairconjugate pair — related — related
by the gain or loss of Hby the gain or loss of H++
BaseAcidAcidBaseNH4
+ + OH-NH3 + H2OBaseAcidAcidBase
NH4+ + OH-NH3 + H2O
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ACID-BASE THEORIESACID-BASE THEORIESACID-BASE THEORIESACID-BASE THEORIESNHNH33 is a is a BASEBASE in water — and water is itself an in water — and water is itself an ACIDACID
NHNH3 3 / NH/ NH44++ is a is a conjugate pairconjugate pair — related by the — related by the
gain or loss of Hgain or loss of H++
Every acid has a conjugate base - and Every acid has a conjugate base - and vice-versa.vice-versa.
BaseAcidAcidBaseNH4
+ + OH-NH3 + H2OBaseAcidAcidBase
NH4+ + OH-NH3 + H2O
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ACID-BASE THEORIESACID-BASE THEORIESACID-BASE THEORIESACID-BASE THEORIESA strong acid is 100% dissociated.A strong acid is 100% dissociated.
Therefore, a Therefore, a STRONG ACIDSTRONG ACID—a good H—a good H++ donor—must donor—must have a have a WEAK CONJUGATE BASEWEAK CONJUGATE BASE—a poor H—a poor H++ acceptor. acceptor.
HNOHNO33(aq) + H(aq) + H22O(liq) O(liq) H H33OO++(aq) + NO(aq) + NO33--(aq)(aq)
STRONG ASTRONG A basebase acid acid weak Bweak B
Notice that every A-B reaction has two acids Notice that every A-B reaction has two acids and two bases!and two bases!
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ACID-BASE THEORIESACID-BASE THEORIESACID-BASE THEORIESACID-BASE THEORIES
We know from experiment that HNOWe know from experiment that HNO33 is a is a strong acid.strong acid.
1.1. It is a stronger acid than HIt is a stronger acid than H33OO++
2.2. HH22O is a stronger base than NOO is a stronger base than NO33--
WEAK BASE
ACID
STRONG ACID
BASEH3O+ + NO3
-HNO3 + H2OWEAK BASE
ACID
STRONG ACID
BASEH3O+ + NO3
-HNO3 + H2O
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ACID-BASE THEORIESACID-BASE THEORIESACID-BASE THEORIESACID-BASE THEORIESAcetic acid is only 0.42% ionized when [HOAc] = Acetic acid is only 0.42% ionized when [HOAc] =
1.0 M. It is a 1.0 M. It is a WEAK ACIDWEAK ACID
HOAc + HHOAc + H22O O H H33OO++ + OAc + OAc--
WEAK AWEAK A basebase acid acid STRONG BSTRONG B
Because [HBecause [H33OO++] is small, this must mean] is small, this must mean
1.1. HH33OO++ is a stronger acid than HOAc is a stronger acid than HOAc
2.2. OAcOAc-- is a stronger base than H is a stronger base than H22OO
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Acid-Base ReactionsAcid-Base ReactionsAcid-Base ReactionsAcid-Base Reactions• Now we can describe
reactions of acids with bases and the direction of such reaction.
• Consider the acid HF reacting with the base NH3.
• HF + NH3 --> NH4+ + F-
• Now we can describe reactions of acids with bases and the direction of such reaction.
• Consider the acid HF reacting with the base NH3.
• HF + NH3 --> NH4+ + F-
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Acid-Base ReactionsAcid-Base ReactionsAcid-Base ReactionsAcid-Base Reactions• Now we can describe
reactions of acids with bases and the direction of such reaction.
• Consider the acid HF reacting with the base NH3.
• HF + NH3 --> NH4+ + F-
• Now we can describe reactions of acids with bases and the direction of such reaction.
• Consider the acid HF reacting with the base NH3.
• HF + NH3 --> NH4+ + F-
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Predicting the Direction of Predicting the Direction of Acid-Base Reactions Acid-Base Reactions
Predicting the Direction of Predicting the Direction of Acid-Base Reactions Acid-Base Reactions
Based on experiment, we can put acids and bases on a Based on experiment, we can put acids and bases on a chart.chart.
See Table 17.3 (page 794)See Table 17.3 (page 794)
ACIDSACIDS CONJUGATE BASESCONJUGATE BASES
STRONGSTRONG weakweak
weakweak STRONGSTRONG
This chart can be used to predict the direction of This chart can be used to predict the direction of reactions between any A-B pair.reactions between any A-B pair.
Reactions always go from the stronger A-B pair to the Reactions always go from the stronger A-B pair to the weaker A-B pair.weaker A-B pair.
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ACID-BASE THEORIESACID-BASE THEORIESACID-BASE THEORIESACID-BASE THEORIESPredicting the direction of an acid-base reaction.Predicting the direction of an acid-base reaction.
ACID1 BASE 2 ACID2 BASE1+ +
STRONG weak
Use Table 17.3Use Table 17.3Reactions always go from Reactions always go from the stronger A-B pair to the stronger A-B pair to the weaker A-B pair.the weaker A-B pair.
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MORE ABOUT WATERMORE ABOUT WATERHH22O can function as both an ACID and a BASE.O can function as both an ACID and a BASE.
In pure water there can be In pure water there can be AUTOIONIZATIONAUTOIONIZATION
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MORE ABOUT WATERMORE ABOUT WATERMORE 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 neutralneutral 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
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Calculating [HCalculating [H33OO++] & [OH] & [OH--]]Calculating [HCalculating [H33OO++] & [OH] & [OH--]]You add 0.0010 mol of NaOH to 1.0 L of You add 0.0010 mol of NaOH to 1.0 L of
pure water. Calculate [Hpure water. Calculate [H3OO++] and [OH] and [OH--].].
SolutionSolution
2 H2 H22O(liq) O(liq) H H3OO++(aq) + OH(aq) + OH--(aq)(aq)
Le Chatelier predicts equilibrium shifts to Le Chatelier predicts equilibrium shifts to the ____________. the ____________.
[H[H3OO++] < 10] < 10-7-7 at equilibrium. at equilibrium.
Set up a concentration table.Set up a concentration table.
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Calculating [HCalculating [H33OO++] & [OH] & [OH--]]Calculating [HCalculating [H33OO++] & [OH] & [OH--]]You add 0.0010 mol of NaOH to 1.0 L of pure You add 0.0010 mol of NaOH to 1.0 L of pure
water. Calculate [Hwater. Calculate [H3OO++] and [OH] and [OH--].].
SolutionSolution
2 H2 H22O(liq) O(liq) H H3OO++(aq) + OH(aq) + OH--(aq)(aq)
initialinitial 00 0.00100.0010
changechange +x+x +x+x
equilibequilib xx 0.0010 + x0.0010 + x
KKww = (x) (0.0010 + x) = (x) (0.0010 + x)
Because x << 0.0010 M, assume Because x << 0.0010 M, assume [OH[OH--] = 0.0010 M] = 0.0010 M
[H[H3OO++] = K] = Kww / 0.0010 = 1.0 x 10 / 0.0010 = 1.0 x 10-11-11 M M
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Calculating [HCalculating [H33OO++] & [OH] & [OH--]]Calculating [HCalculating [H33OO++] & [OH] & [OH--]]You add 0.0010 mol of NaOH to 1.0 L of You add 0.0010 mol of NaOH to 1.0 L of
pure water. Calculate [Hpure water. Calculate [H3OO++] and [OH] and [OH--].].
SolutionSolution
2 H2 H22O(liq) O(liq) H H3OO++(aq) + OH(aq) + OH--(aq)(aq)
[H[H3OO++] = K] = Kww / 0.0010 = 1.0 x 10 / 0.0010 = 1.0 x 10-11-11 M M
This solution is This solution is BASICBASIC because because [H[H3OO++] < [OH] < [OH--]]
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[H[H33OO++], [OH], [OH--] and pH] and pH[H[H33OO++], [OH], [OH--] and pH] and pH
A common way to express acidity and basicity A common way to express acidity and basicity is with pHis with pH
pH = log (1/ [HpH = log (1/ [H33OO++]) ]) = =
- log [H - log [H33OO++]]
In a neutral solution, In a neutral solution, [H[H3OO++] = [OH] = [OH--] = ] =
1.00 x 101.00 x 10-7-7 at 25 at 25 ooCC
pH = -log (1.00 x 10pH = -log (1.00 x 10-7-7) ) = = - (-7) = 7- (-7) = 7
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[H[H33OO++], [OH], [OH--] and pH] and pH[H[H33OO++], [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?
[H[H3OO++] = 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
General conclusion —General conclusion —
Basic solution Basic solution pH > 7pH > 7
Neutral Neutral pH = 7pH = 7
Acidic solutionAcidic solution pH < 7pH < 7
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[H[H33OO++], [OH], [OH--] and pH] and pH[H[H33OO++], [OH], [OH--] and pH] and pH
If the pH of Coke is 3.12, it is ____________.If the pH of Coke is 3.12, it is ____________.
Because pH = - log [HBecause pH = - log [H3OO++] then] then
log [Hlog [H3OO++] = - pH] = - pH
Take antilog and getTake antilog and get
[H[H3OO++] = 10] = 10-pH-pH
[H[H3OO++] = 10] = 10-3.12-3.12 = =
7.6 x 107.6 x 10-4-4 M M
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Other pX ScalesOther pX ScalesOther pX ScalesOther pX Scales
In generalIn general pX = -log XpX = -log X
and so and so pOH = - log [OHpOH = - log [OH--]]
KKww = [H = [H3OO++] [OH] [OH--] = 1.00 x 10] = 1.00 x 10-14-14 at 25 at 25 ooCC
Take the log of both sidesTake the log of both sides
-log (10-log (10-14-14) = - log [H) = - log [H3OO++] + (-log [OH] + (-log [OH--])])
14 = pH + pOH14 = pH + pOH
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Equilibria Involving Equilibria Involving Weak Acids and BasesWeak Acids and Bases
Equilibria Involving Equilibria Involving Weak Acids and BasesWeak Acids and Bases
Aspirin is a good example Aspirin is a good example of a weak acid, of a weak acid, KKaa = 3.2 x 10 = 3.2 x 10-4-4
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Equilibria Involving Equilibria Involving Weak Acids and BasesWeak Acids and Bases
Equilibria Involving Equilibria Involving Weak Acids and BasesWeak Acids and Bases
AcidAcid Conjugate BaseConjugate Base
acetic, CHacetic, CH33COCO22HH CHCH33COCO22--, acetate, acetate
ammonium, NHammonium, NH44++ NHNH33, ammonia, ammonia
bicarbonate, HCObicarbonate, HCO33-- COCO33
2-2-, carbonate, carbonate
A weak acid (or base) is one that ionizes A weak acid (or base) is one that ionizes to a VERY small extent (< 5%).to a VERY small extent (< 5%).
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Equilibria Involving Equilibria Involving Weak Acids and BasesWeak Acids and Bases
Equilibria Involving Equilibria Involving Weak Acids and BasesWeak Acids and Bases
Consider acetic acidConsider acetic acid
HOAc + HHOAc + H22O O H H33OO++ + OAc + OAc--
AcidAcid Conj. Conj. basebase
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Equilibria Involving Equilibria Involving Weak Acids and BasesWeak Acids and Bases
Equilibria Involving Equilibria Involving Weak Acids and BasesWeak Acids and Bases
Consider acetic acidConsider acetic acid
HOAc + HHOAc + H22O HO H33OO++ + OAc + OAc--
AcidAcid Conj. base Conj. base
(K is designated K(K is designated Kaa for for ACIDACID))
Because [HBecause [H33OO++] and [OAc] and [OAc--] are SMALL, K] are SMALL, Kaa << 1. << 1.
Ka [H3O+][OAc- ]
[HOAc] 1.8 x 10-5Ka
[H3O+][OAc- ][HOAc]
1.8 x 10-5
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Equilibria Involving Equilibria Involving Weak Acids and BasesWeak Acids and Bases
Equilibria Involving Equilibria Involving Weak Acids and BasesWeak Acids and Bases
Values of KValues of Kaa for acid and K for acid and Kbb for bases are for bases are
found in found in TABLE 17.4TABLE 17.4 — page 799 — page 799
Notice the relation of TABLE 17.4 to the Notice the relation of TABLE 17.4 to the
table of relative acid/base strengths table of relative acid/base strengths
(Table 17.3).(Table 17.3).
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Equilibria Involving A Weak Acid
Equilibria Involving A Weak Acid
Determining the pH of Determining the pH of
an acetic acid an acetic acid
solution.solution.
See Screen 17.8.See Screen 17.8.
Determining the pH of Determining the pH of
an acetic acid an acetic acid
solution.solution.
See Screen 17.8.See Screen 17.8.
0.0001 M0.0001 M
0.003 M0.003 M
0.06 M0.06 M
2.0 M2.0 M
a pH metera pH meter
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Equilibria Involving A Weak AcidEquilibria Involving A Weak AcidEquilibria 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, H33OO++, OAc, OAc--, ,
and the pH.and the pH.
Step 1.Step 1. Define equilibrium concs. Define equilibrium concs.
[HOAc][HOAc] [H[H33OO++]] [OAc[OAc--]]
initialinitial
changechange
equilibequilib
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Equilibria Involving A Weak AcidEquilibria Involving A Weak AcidEquilibria 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, H33OO++, OAc, OAc--, ,
and the pH.and the pH.
Step 1.Step 1. Define equilibrium concs. Define equilibrium concs.
[HOAc][HOAc] [H[H33OO++]] [OAc[OAc--]]
initialinitial 1.001.00 00 00
changechange -x-x +x+x +x+x
equilibequilib 1.00-x1.00-x xx xx
Note that we neglect [HNote that we neglect [H33OO++] from H] from H22O.O.
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Equilibria Involving A Weak AcidEquilibria Involving A Weak AcidEquilibria Involving A Weak AcidEquilibria Involving A Weak Acid
Step 2.Step 2. Write K Write 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.
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Equilibria Involving A Weak AcidEquilibria Involving A Weak AcidEquilibria Involving A Weak AcidEquilibria Involving A Weak Acid
Step 2.Step 2. Write K Write 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
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Equilibria Involving A Weak AcidEquilibria Involving A Weak AcidEquilibria Involving A Weak AcidEquilibria Involving A Weak Acid
Step 2.Step 2. Write K Write Kaa expression expression
This is a quadratic. Solve using quadratic formula or method of This is a quadratic. Solve using quadratic formula or method of approximations (see Appendix A).approximations (see Appendix A).
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
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Equilibria Involving A Weak AcidEquilibria Involving A Weak AcidEquilibria Involving A Weak AcidEquilibria Involving A Weak Acid
Step 3.Step 3. Solve K Solve 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
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Equilibria Involving A Weak AcidEquilibria Involving A Weak AcidEquilibria Involving A Weak AcidEquilibria Involving A Weak Acid
Step 3.Step 3. Solve K Solve Kaa expression expression
First assume x is very small because KFirst assume x is very small because Kaa is so small. is so small.
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
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Equilibria Involving A Weak AcidEquilibria Involving A Weak AcidEquilibria Involving A Weak AcidEquilibria Involving A Weak Acid
Step 3.Step 3. Solve K Solve Kaa expression expression
First assume x is very small because KFirst assume x is very small because Kaa is so small. is so small.
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
Ka 1.8 x 10-5 = x2
1.00Ka 1.8 x 10-5 =
x2
1.00
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Copyright (c) 1999 by Harcourt Brace & CompanyAll rights reserved
Equilibria Involving A Weak AcidEquilibria Involving A Weak AcidEquilibria Involving A Weak AcidEquilibria Involving A Weak Acid
Step 3.Step 3. Solve K Solve Kaa expression expression
First assume x is very small because KFirst assume x is very small because Kaa is so small. is so small.
And so x = [And so x = [HH33OO++] = [] = [OAcOAc--] = [K] = [Kaa • 1.00] • 1.00]1/21/2
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
Ka 1.8 x 10-5 = x2
1.00Ka 1.8 x 10-5 =
x2
1.00
4646
Copyright (c) 1999 by Harcourt Brace & CompanyAll rights reserved
Equilibria Involving A Weak AcidEquilibria Involving A Weak AcidEquilibria Involving A Weak AcidEquilibria Involving A Weak Acid
Step 3.Step 3. Solve K Solve Kaa approximateapproximate expression expression
x = [x = [HH33OO++] = [] = [OAcOAc--] = [K] = [Kaa • 1.00] • 1.00]1/21/2
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
4747
Copyright (c) 1999 by Harcourt Brace & CompanyAll rights reserved
Equilibria Involving A Weak AcidEquilibria Involving A Weak AcidEquilibria Involving A Weak AcidEquilibria Involving A Weak Acid
Step 3.Step 3. Solve K Solve Kaa approximateapproximate expression expression
x = [x = [HH33OO++] = [] = [OAcOAc--] = [K] = [Kaa • 1.00] • 1.00]1/21/2
x = x = [[HH33OO++] = [] = [OAcOAc--] = 4.2 x 10] = 4.2 x 10-3-3 M M
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
4848
Copyright (c) 1999 by Harcourt Brace & CompanyAll rights reserved
Equilibria Involving A Weak AcidEquilibria Involving A Weak AcidEquilibria Involving A Weak AcidEquilibria Involving A Weak Acid
Step 3.Step 3. Solve K Solve Kaa approximateapproximate expression expression
x = [x = [HH33OO++] = [] = [OAcOAc--] = [K] = [Kaa • 1.00] • 1.00]1/21/2
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
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
4949
Copyright (c) 1999 by Harcourt Brace & CompanyAll rights reserved
Equilibria Involving A Weak AcidEquilibria Involving A Weak AcidEquilibria Involving A Weak AcidEquilibria Involving A Weak Acid
Consider the approximate expression Consider the approximate expression
x [H3O+ ] = [Ka • 1.00]1/2x [H3O+ ] = [Ka • 1.00]1/2Ka 1.8 x 10-5 =
x2
1.00Ka 1.8 x 10-5 =
x2
1.00
5050
Copyright (c) 1999 by Harcourt Brace & CompanyAll rights reserved
Equilibria Involving A Weak AcidEquilibria Involving A Weak AcidEquilibria Involving A Weak AcidEquilibria Involving A Weak Acid
Consider the approximate expression Consider the approximate expression
For many weak acidsFor many weak acids
[H[H33OO++] = [conj. base] = [K] = [conj. base] = [Kaa • C • Coo]]1/21/2
where Cwhere C00 = initial conc. of acid = initial conc. of acid
Ka 1.8 x 10-5 = x2
1.00Ka 1.8 x 10-5 =
x2
1.00 x [H3O+ ] = [Ka • 1.00]1/2x [H3O+ ] = [Ka • 1.00]1/2
5151
Copyright (c) 1999 by Harcourt Brace & CompanyAll rights reserved
Equilibria Involving A Weak AcidEquilibria Involving A Weak AcidEquilibria Involving A Weak AcidEquilibria Involving A Weak Acid
Consider the approximate expression Consider the approximate expression
For many weak acidsFor many weak acids
[H[H33OO++] = [conj. base] = [K] = [conj. base] = [Kaa • C • Coo]]1/21/2
where Cwhere C00 = initial conc. of acid = initial conc. of acid
Useful Rule of Thumb:Useful Rule of Thumb:
x [H3O+ ] = [Ka • 1.00]1/2x [H3O+ ] = [Ka • 1.00]1/2Ka 1.8 x 10-5 =
x2
1.00Ka 1.8 x 10-5 =
x2
1.00
5252
Copyright (c) 1999 by Harcourt Brace & CompanyAll rights reserved
Equilibria Involving A Weak AcidEquilibria Involving A Weak AcidEquilibria Involving A Weak AcidEquilibria Involving A Weak Acid
Consider the approximate expression Consider the approximate expression
For many weak acidsFor many weak acids
[H[H33OO++] = [conj. base] = [K] = [conj. base] = [Kaa • C • Coo]]1/21/2
where Cwhere C00 = initial conc. of acid = initial conc. of acid
Useful Rule of Thumb:Useful Rule of Thumb:
If 100 • KIf 100 • Kaa < C < Coo, then [H, then [H33OO++] = [K] = [Kaa • C • Coo]]1/21/2
Ka 1.8 x 10-5 = x2
1.00Ka 1.8 x 10-5 =
x2
1.00 x [H3O+ ] = [Ka • 1.00]1/2x [H3O+ ] = [Ka • 1.00]1/2
5353
Copyright (c) 1999 by Harcourt Brace & CompanyAll rights reserved
Equilibria Involving A Weak AcidEquilibria Involving A Weak AcidEquilibria Involving A Weak AcidEquilibria Involving A Weak AcidCalculate 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++] ] = = [K[Kaa • C • Coo]]1/21/2 = 4.2 x 10 = 4.2 x 10-4-4 M, M, pH = 3.37pH = 3.37
Exact SolutionExact Solution
[H[H33OO++] = [] = [HCOHCO22--] = 3.4 x 10] = 3.4 x 10-4-4 M M
[[HCOHCO22HH] = 0.0010 - 3.4 x 10] = 0.0010 - 3.4 x 10-4-4 = 0.0007 M = 0.0007 M
pH = 3.47 pH = 3.47
5454
Copyright (c) 1999 by Harcourt Brace & CompanyAll rights reserved
Equilibria Involving A Weak BaseEquilibria Involving A Weak BaseEquilibria Involving A Weak BaseEquilibria Involving A Weak Base
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. Define equilibrium concs.
[NH[NH33]] [NH[NH44++]] [OH[OH--]]
initialinitial
changechange
equilibequilib
5555
Copyright (c) 1999 by Harcourt Brace & CompanyAll rights reserved
Equilibria Involving A Weak BaseEquilibria Involving A Weak BaseEquilibria Involving A Weak BaseEquilibria Involving A Weak Base
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. Define equilibrium concs.
[NH[NH33]] [NH[NH44++]] [OH[OH--]]
initialinitial 0.0100.010 00 00
changechange -x-x +x+x +x+x
equilibequilib 0.010 - x0.010 - x x x xx
5656
Copyright (c) 1999 by Harcourt Brace & CompanyAll rights reserved
Equilibria Involving A Weak BaseEquilibria Involving A Weak BaseEquilibria Involving A Weak BaseEquilibria Involving A Weak Base
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 expression Solve 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
5757
Copyright (c) 1999 by Harcourt Brace & CompanyAll rights reserved
Equilibria Involving A Weak BaseEquilibria Involving A Weak BaseEquilibria Involving A Weak BaseEquilibria Involving A Weak Base
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 expression Solve the equilibrium expression
Assume x is small (100•KAssume x is small (100•Kbb < C < Coo), so ), so x x
= [OH= [OH--] = [NH] = [NH44++] = 4.2 x 10] = 4.2 x 10-4-4 M M
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
5858
Copyright (c) 1999 by Harcourt Brace & CompanyAll rights reserved
Equilibria Involving A Weak BaseEquilibria Involving A Weak BaseEquilibria Involving A Weak BaseEquilibria Involving A Weak Base
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 expression Solve the equilibrium expression
Assume x is small (100•KAssume x is small (100•Kbb < C < Coo), so ), so x = x =
[OH[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
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
5959
Copyright (c) 1999 by Harcourt Brace & CompanyAll rights reserved
Equilibria Involving A Weak BaseEquilibria Involving A Weak BaseEquilibria Involving A Weak BaseEquilibria Involving A Weak BaseYou 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 expression Solve the equilibrium expression
Assume x is small (100•KAssume x is small (100•Kbb < C < Coo), so ), 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 !!
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
6060
Copyright (c) 1999 by Harcourt Brace & CompanyAll rights reserved
Equilibria Involving A Weak BaseEquilibria Involving A Weak BaseEquilibria Involving A Weak BaseEquilibria Involving A Weak Base
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 pH Calculate 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
6161
Copyright (c) 1999 by Harcourt Brace & CompanyAll rights reserved
MX + HMX + H22O ----> acidic or basic solution?O ----> acidic or basic solution?
Consider NHConsider NH44ClCl
NHNH44Cl(aq) ----> NHCl(aq) ----> NH44++(aq) + Cl(aq) + Cl--(aq)(aq)
(a)(a) Reaction of ClReaction of Cl-- with H with H22OO
ClCl-- + + HH22O ---->O ----> HCl + HCl + OHOH--
basebase acidacid acidacid basebase
ClCl-- ion is a VERY weak base because its ion is a VERY weak base because its conjugate acid is strong. conjugate acid is strong.
Therefore, ClTherefore, Cl-- ----> neutral solution ----> neutral solution
Acid-Base Properties of SaltsAcid-Base Properties of SaltsAcid-Base Properties of SaltsAcid-Base Properties of Salts
6262
Copyright (c) 1999 by Harcourt Brace & CompanyAll rights reserved
NHNH44Cl(aq) ----> NHCl(aq) ----> NH44++(aq) + Cl(aq) + Cl--(aq)(aq)
(b)(b) Reaction of NHReaction of NH44++ with H with H22OO
NHNH44++ + H + H22O ---->O ----> NHNH33 + + HH33OO++
acidacid basebase basebase acidacid
NHNH44++ ion is a moderate acid because its ion is a moderate acid because its
conjugate base is weak. conjugate base is weak.
Therefore, NHTherefore, NH44++ ----> acidic solution ----> acidic solution
See TABLE 17.5 for a summary of See TABLE 17.5 for a summary of acid-base properties of ions.acid-base properties of ions.
Acid-Base Properties of SaltsAcid-Base Properties of SaltsAcid-Base Properties of SaltsAcid-Base Properties of Salts
6363
Copyright (c) 1999 by Harcourt Brace & CompanyAll rights reserved
Calculate the pH of a 0.10 M solution of NaCalculate the pH of a 0.10 M solution of Na22COCO33. .
NaNa++ + H + H22O ---> neutralO ---> neutral
COCO332-2- ++ HH22OO HCOHCO33
-- ++ OHOH--
basebase acidacid acidacid basebase
KKbb = 2.1 x 10 = 2.1 x 10-4-4
Step 1.Step 1. Set up concentration tableSet up concentration table
[CO[CO332-2-]] [HCO[HCO33
--]] [OH[OH--]]
initialinitial
changechange
equilibequilib
Acid-Base Properties of SaltsAcid-Base Properties of SaltsAcid-Base Properties of SaltsAcid-Base Properties of Salts
6464
Copyright (c) 1999 by Harcourt Brace & CompanyAll rights reserved
Calculate the pH of a 0.10 M solution of NaCalculate the pH of a 0.10 M solution of Na22COCO33. .
NaNa++ + H + H22O ---> neutralO ---> neutral
COCO332-2- ++ HH22OO HCOHCO33
-- ++ OHOH--
basebase acidacid acidacid basebase
KKbb = 2.1 x 10 = 2.1 x 10-4-4
Step 1.Step 1. Set up concentration tableSet up concentration table
[CO[CO332-2-]] [HCO[HCO33
--]] [OH[OH--]]
initialinitial 0.100.10 00 00
changechange -x-x +x+x +x+x
equilibequilib 0.10 - x0.10 - x xx xx
Acid-Base Properties of SaltsAcid-Base Properties of SaltsAcid-Base Properties of SaltsAcid-Base Properties of Salts
6565
Copyright (c) 1999 by Harcourt Brace & CompanyAll rights reserved
Calculate the pH of a 0.10 M solution of NaCalculate the pH of a 0.10 M solution of Na22COCO33. .
NaNa++ + H + H22O ---> neutralO ---> neutral
COCO332-2- ++ HH22OO HCOHCO33
-- ++ OHOH--
basebase acidacid acidacid basebase
KKbb = 2.1 x 10 = 2.1 x 10-4-4
Step 2.Step 2. Solve the equilibrium expressionSolve the equilibrium expression
Acid-Base Properties of SaltsAcid-Base Properties of SaltsAcid-Base Properties of SaltsAcid-Base Properties of Salts
6666
Copyright (c) 1999 by Harcourt Brace & CompanyAll rights reserved
Calculate the pH of a 0.10 M solution of NaCalculate the pH of a 0.10 M solution of Na22COCO33. .
NaNa++ + H + H22O ---> neutralO ---> neutral
COCO332-2- ++ HH22OO HCOHCO33
-- ++ OHOH--
basebase acidacid acidacid basebase
KKbb = 2.1 x 10 = 2.1 x 10-4-4
Step 2.Step 2. Solve the equilibrium expressionSolve the equilibrium expression
Acid-Base Properties of SaltsAcid-Base Properties of SaltsAcid-Base Properties of SaltsAcid-Base Properties of Salts
Kb = 2.1 x 10-4 = [HCO3
- ][OH- ]
[CO32 ]
x2
0.10 - xKb = 2.1 x 10-4 =
[HCO3- ][OH- ]
[CO32 ]
x2
0.10 - x
6767
Copyright (c) 1999 by Harcourt Brace & CompanyAll rights reserved
Calculate the pH of a 0.10 M solution of NaCalculate the pH of a 0.10 M solution of Na22COCO33. .
NaNa++ + H + H22O ---> neutralO ---> neutral
COCO332-2- ++ HH22OO HCOHCO33
-- ++ OHOH--
basebase acidacid acidacid basebase
KKbb = 2.1 x 10 = 2.1 x 10-4-4
Step 2.Step 2. Solve the equilibrium expressionSolve the equilibrium expression
Assume 0.10 - x 0.10, because 100•KAssume 0.10 - x 0.10, because 100•Kbb < C < Coo
Acid-Base Properties of SaltsAcid-Base Properties of SaltsAcid-Base Properties of SaltsAcid-Base Properties of Salts
Kb = 2.1 x 10-4 = [HCO3
- ][OH- ]
[CO32 ]
x2
0.10 - xKb = 2.1 x 10-4 =
[HCO3- ][OH- ]
[CO32 ]
x2
0.10 - x
6868
Copyright (c) 1999 by Harcourt Brace & CompanyAll rights reserved
Calculate the pH of a 0.10 M solution of NaCalculate the pH of a 0.10 M solution of Na22COCO33. .
NaNa++ + H + H22O ---> neutralO ---> neutral
COCO332-2- ++ HH22OO HCOHCO33
-- ++ OHOH--
basebase acidacid acidacid basebase
KKbb = 2.1 x 10 = 2.1 x 10-4-4
Step 2.Step 2. Solve the equilibrium expressionSolve the equilibrium expression
Assume 0.10 - x 0.10, because 100•KAssume 0.10 - x 0.10, because 100•Kbb < C < Coo
x = [HCOx = [HCO33--] = [OH] = [OH--] = 0.0046 M ] = 0.0046 M
Acid-Base Properties of SaltsAcid-Base Properties of SaltsAcid-Base Properties of SaltsAcid-Base Properties of Salts
Kb = 2.1 x 10-4 = [HCO3
- ][OH- ]
[CO32 ]
x2
0.10 - xKb = 2.1 x 10-4 =
[HCO3- ][OH- ]
[CO32 ]
x2
0.10 - x
6969
Copyright (c) 1999 by Harcourt Brace & CompanyAll rights reserved
Calculate the pH of a 0.10 M solution of NaCalculate the pH of a 0.10 M solution of Na22COCO33. .
NaNa++ + H + H22O ---> neutralO ---> neutral
COCO332-2- ++ HH22OO HCOHCO33
-- ++ OHOH--
basebase acidacid acidacid basebase
KKbb = 2.1 x 10 = 2.1 x 10-4-4
Step 3.Step 3. Calculate the pHCalculate the pH
[OH[OH--] = 0.0046 M] = 0.0046 M
Acid-Base Properties of SaltsAcid-Base Properties of SaltsAcid-Base Properties of SaltsAcid-Base Properties of Salts
7070
Copyright (c) 1999 by Harcourt Brace & CompanyAll rights reserved
Calculate the pH of a 0.10 M solution of NaCalculate the pH of a 0.10 M solution of Na22COCO33. .
NaNa++ + H + H22O ---> neutralO ---> neutral
COCO332-2- ++ HH22OO HCOHCO33
-- ++ OHOH--
basebase acidacid acidacid basebase
KKbb = 2.1 x 10 = 2.1 x 10-4-4
Step 3.Step 3. Calculate the pHCalculate the pH
[OH[OH--] = 0.0046 M] = 0.0046 M
pOH = - log [OHpOH = - log [OH--] = 2.34] = 2.34
Acid-Base Properties of SaltsAcid-Base Properties of SaltsAcid-Base Properties of SaltsAcid-Base Properties of Salts
7171
Copyright (c) 1999 by Harcourt Brace & CompanyAll rights reserved
Calculate the pH of a 0.10 M solution of NaCalculate the pH of a 0.10 M solution of Na22COCO33. .
NaNa++ + H + H22O ---> neutralO ---> neutral
COCO332-2- ++ HH22OO HCOHCO33
-- ++ OHOH--
basebase acidacid acidacid basebase
KKbb = 2.1 x 10 = 2.1 x 10-4-4
Step 3.Step 3. Calculate the pHCalculate the pH
[OH[OH--] = 0.0046 M] = 0.0046 M
pOH = - log [OHpOH = - log [OH--] = 2.34] = 2.34
pH + pOH = 14, pH + pOH = 14,
Acid-Base Properties of SaltsAcid-Base Properties of SaltsAcid-Base Properties of SaltsAcid-Base Properties of Salts
7272
Copyright (c) 1999 by Harcourt Brace & CompanyAll rights reserved
Calculate the pH of a 0.10 M solution of NaCalculate the pH of a 0.10 M solution of Na22COCO33. .
NaNa++ + H + H22O ---> neutralO ---> neutral
COCO332-2- ++ HH22OO HCOHCO33
-- ++ OHOH--
basebase acidacid acidacid basebase
KKbb = 2.1 x 10 = 2.1 x 10-4-4
Step 3.Step 3. Calculate the pHCalculate the pH
[OH[OH--] = 0.0046 M] = 0.0046 M
pOH = - log [OHpOH = - log [OH--] = 2.34] = 2.34
pH + pOH = 14, pH + pOH = 14,
so so pH = 11.66pH = 11.66, and the solution is ________., and the solution is ________.
Acid-Base Properties of SaltsAcid-Base Properties of SaltsAcid-Base Properties of SaltsAcid-Base Properties of Salts