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11The Chemistry The Chemistry of Acids and of Acids and BasesBases
Chapter 17Chapter 17
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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
55Strong and Weak Strong and Weak
Acids/BasesAcids/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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HNO3, HCl, H2SO4 and HClO4 are among the only known strong acids.
Strong and Weak Strong and Weak Acids/BasesAcids/Bases
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• Weak acids are much less than 100% ionized in Weak acids 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
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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 Strong and Weak Acids/BasesAcids/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) ee NH NH44++(aq) + OH(aq) + OH--(aq)(aq)
Strong and Weak Strong and Weak Acids/BasesAcids/Bases
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ACID-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 + H2O
ACID-BASE THEORIESACID-BASE THEORIES
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ACID-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 by the gain or loss of Hrelated by the gain or loss of H++
Every acid has a conjugate Every acid has a conjugate base - and vice-versa.base - and vice-versa.
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Conjugate PairsConjugate Pairs
1414More 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
Equilibrium constant for autoion = KEquilibrium constant for autoion = Kww
KKww = [H = [H33OO++] [OH] [OH--] = ] = 1.00 x 101.00 x 10-14-14 at 25 at 25 ooCC
1515More 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+
AutoionizationAutoionization
1616Calculating [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--].].
SolutionSolution2 H2 H22O(liq) O(liq) e e 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 ICE table.Set up a ICE table.
1717Calculating [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) ee H H33OO++(aq) + OH(aq) + OH--(aq)(aq) initialinitial 00 0.00100.0010 changechange +x+x +x+x equilibequilib xx 0.0010 + x0.0010 + xKKww = (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[H33OO++] = K] = Kww / 0.0010 = 1.0 x 10 / 0.0010 = 1.0 x 10-11-11 M M
1818Calculating [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--].].
SolutionSolution2 H2 H22O(liq) O(liq) e e 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 _________
because
[H3O+] < [OH-]
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[H[H33OO++], [OH], [OH--] and pH] and pHA 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 [Hlog [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 pHWhat 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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Figure 17.1Figure 17.1
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[H[H33OO++], [OH], [OH--] and pH] and pHIf 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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pH of Common pH of Common SubstancesSubstances
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Other pX ScalesOther pX ScalesIn generalIn general pX = -log XpX = -log Xand 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--])])
pKpKww = 14 = pH + pOH = 14 = pH + pOH
2525Equilibria Involving Equilibria Involving Weak Acids and Weak Acids and
BasesBases
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
AcidAcid Conjugate BaseConjugate Baseacetic, CHacetic, CH33COCO22HH CHCH33COCO22
--, acetate, acetateammonium, NHammonium, NH44
++ NHNH33, ammonia, ammoniabicarbonate, HCObicarbonate, HCO33
-- COCO332-2-, 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
Consider acetic acid, CHConsider acetic acid, CH33COCO22H (HOAc)H (HOAc)HOAc + HHOAc + H22O O ee H H33OO++ + OAc + OAc--
AcidAcid Conj. base Conj. base
Ka [H3O+][OAc- ][HOAc]
1.8 x 10-5
(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.
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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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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
3030Ionization Constants for Ionization Constants for Acids/Bases Acids/Bases
AcidsAcids ConjugateConjugateBasesBases
Increase strength
Increase strength
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Relation Relation of Kof Kaa, K, Kbb, ,
[H[H33OO++] ] and pHand pH
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K and Acid-Base Reactions K and Acid-Base Reactions
Reactions always go from the Reactions always go from the stronger A-B pair (larger K) to the stronger A-B pair (larger K) to the
weaker A-B pair (smaller K).weaker A-B pair (smaller K).
ACIDSACIDS CONJUGATE BASESCONJUGATE BASESSTRONGSTRONG
weakweak
weakweak
STRONGSTRONG
3333K and Acid-Base ReactionsK and Acid-Base ReactionsA 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— donor—
must have a must have a WEAK CONJUGATE BASEWEAK CONJUGATE BASE—a —a poor Hpoor H++ acceptor. acceptor.
HNOHNO33(aq) + H(aq) + H22O(liq) O(liq) ee H H33OO++(aq) + NO(aq) + NO33--(aq)(aq)
STRONG ASTRONG A basebase acid acid weak Bweak B
•Every A-B reaction has two acids and two Every A-B reaction has two acids and two bases.bases.•Equilibrium always lies toward the weaker pair.Equilibrium always lies toward the weaker pair.•Here K is very large.Here K is very large.
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K and Acid-Base ReactionsK and Acid-Base Reactions
We know from experiment that HNOWe know from experiment that HNO33 is a strong is a strong acid.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--
3.3. K for this reaction is largeK for this reaction is large
WEAK BASE
ACID
STRONG ACID
BASEH3O+ + NO3
-HNO3 + H2O
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K and Acid-Base ReactionsK and Acid-Base ReactionsAcetic acid is only 0.42% ionized when [HOAc] = 1.0 M. It Acetic acid is only 0.42% ionized when [HOAc] = 1.0 M. It
is a is a WEAK ACIDWEAK ACID
HOAc + HHOAc + H22O O ee H H33OO++ + OAc + OAc--
WEAK AWEAK A basebase acid acid STRONG BSTRONG BBecause [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
3.3. K for this reaction is smallK for this reaction is small
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Types of Acid/Base ReactionsTypes of Acid/Base ReactionsStrong acid + Strong baseStrong acid + Strong baseHH++ + Cl + Cl-- + Na + Na++ + OH + OH-- ee H H22O + NaO + Na++ + Cl + Cl--
Net ionic equationNet ionic equation
HH++(aq) + OH(aq) + OH--(aq) (aq) ee H H22O(liq)O(liq)
K = 1/KK = 1/Kww = 1 x 10 = 1 x 101414
Mixing equal molar quantities of a strong acid and strong
base produces a neutral solution.
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Types of Acid/Base ReactionsTypes of Acid/Base ReactionsWeak acid + Strong baseWeak acid + Strong baseCHCH33COCO22H + OHH + OH- - ee H H22O + CHO + CH33COCO22
--
This is the reverse of the reaction of CHThis is the reverse of the reaction of CH33COCO22--
(conjugate base) with H(conjugate base) with H22O.O.
OHOH-- stronger base than CH stronger base than CH33COCO22--
K = 1/KK = 1/Kbb = 5.6 x 10 = 5.6 x 1044
Mixing equal molar quantities of a weak acid and strong base produces the acid’s conjugate
base. The solution is basic.
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Types of Acid/Base ReactionsTypes of Acid/Base ReactionsStrong acid + Weak baseStrong acid + Weak baseHH33OO++ + NH + NH33 ee H H22O + NHO + NH44
++
This is the reverse of the reaction of NHThis is the reverse of the reaction of NH44++
(conjugate acid of NH(conjugate acid of NH33) with H) with H22O.O.
HH33OO++ stronger acid than NH stronger acid than NH44++
K = 1/KK = 1/Kaa = 5.6 x 10 = 5.6 x 1044
Mixing equal molar quantities of a strong acid and weak base produces the bases’s conjugate
acid. The solution is acid.
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Types of Acid/Base ReactionsTypes of Acid/Base Reactions
Weak acid + Weak baseWeak acid + Weak base
•Product cation = conjugate acid of weak base.Product cation = conjugate acid of weak base.•Product anion = conjugate base of weak acid.Product anion = conjugate base of weak acid.•pH of solution depends on relative strengths pH of solution depends on relative strengths of cation and anion.of cation and anion.
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Types of Acid/Base Reactions: Types of Acid/Base Reactions: SummarySummary
4141
Calculations Calculations with with
Equilibrium Equilibrium ConstantsConstants
pH of an acetic pH of an acetic acid solution.acid solution.
What are your What are your observations?observations?
0.0001 M0.0001 M
0.003 M0.003 M
0.06 M0.06 M
2.0 M2.0 M
a pH meter, Screen 17.9a pH meter, Screen 17.9
4242Equilibria 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
4343Equilibria Involving A Weak Equilibria Involving A Weak AcidAcid
You have 1.00 M HOAc. Calc. the equilibrium You have 1.00 M HOAc. Calc. the equilibrium concs. of HOAc, Hconcs. 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 1.001.00 00 00changechange -x-x +x+x +x+xequilibequilib 1.00-x1.00-x xx xxNote that we neglect [HNote that we neglect [H33OO++] from H] from H22O.O.
4444Equilibria Involving A Weak Equilibria Involving A Weak AcidAcid
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 - x
This is a quadratic. Solve using quadratic This is a quadratic. Solve using quadratic formula or method of approximations (see formula or method of approximations (see
Appendix A).Appendix A).
4545Equilibria Involving A Weak Equilibria Involving A Weak AcidAcid
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 - 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
1.00
Now we can more easily solve this Now we can more easily solve this approximate expression.approximate expression.
4646Equilibria Involving A Weak AcidEquilibria Involving A Weak Acid
Step 3.Step 3. Solve K Solve Kaa approximateapproximate 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 = x2
1.00
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
4747Equilibria Involving A Weak Equilibria Involving A Weak AcidAcid
Consider the approximate expression Consider the approximate expression
Ka 1.8 x 10-5 = x2
1.00 x [H3O+ ] = [Ka • 1.00]1/2
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•Ka a < C< Coo, then [H, then [H33OO++] = [K] = [Kaa•C•Coo]]1/21/2
4848Equilibria 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 ee 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
4949
Weak BasesWeak Bases
5050Equilibria 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 ee NH NH44++ + OH + OH--
KKbb = 1.8 x 10 = 1.8 x 10-5-5
Step 1.Step 1. Define equilibrium concs. in ICE table Define equilibrium concs. in ICE table
[NH[NH33]] [NH[NH44++]] [OH[OH--]]
initialinitial
changechange
equilibequilib
5151Equilibria 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 ee NH NH44++ + OH + OH--
KKbb = 1.8 x 10 = 1.8 x 10-5-5
Step 1.Step 1. Define equilibrium concs. in ICE table Define equilibrium concs. in ICE table
[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
5252Equilibria 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 e e 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 - x
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 Mand [NHand [NH33] = 0.010 - 4.2 x 10] = 0.010 - 4.2 x 10-4-4 ≈ 0.010 M ≈ 0.010 MThe approximation is validThe approximation is valid !!
5353Equilibria 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 ee 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 Mso pOH = - log [OHso pOH = - log [OH--] = 3.37] = 3.37Because pH + pOH = 14,Because pH + pOH = 14,
pH = 10.63pH = 10.63
5454
MX + HMX + H22O ----> acidic or basic solution?O ----> acidic or basic solution?Consider NHConsider NH44ClClNHNH44Cl(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 basebaseClCl-- 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 Salts
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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 acidacidNHNH44
++ 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 17.4 for a summary of acid-See TABLE 17.4 for a summary of acid-base properties of ions.base properties of ions.
Acid-Base Properties of SaltsAcid-Base Properties of Salts
5656Acid-Base Properties of Acid-Base Properties of
SaltsSalts
5757
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 ---> neutralCOCO33
2-2- + + HH22O O ee HCOHCO33-- + OH + OH--
basebase acidacid acidacidbasebase KKbb = 2.1 x 10 = 2.1 x 10-4-4
Step 1.Step 1. Set up concentration tableSet up concentration table [CO[CO33
2-2-]] [HCO[HCO33--]] [OH[OH--] ] initialinitial
changechange equilibequilib
Acid-Base Properties of SaltsAcid-Base Properties of Salts
5858
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- + + HH22O O ee HCOHCO33
-- + OH + OH--
basebase acidacid acidacid basebase KKbb = 2.1 x 10 = 2.1 x 10-4-4
Step 1.Step 1. Set up ICE tableSet up ICE 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 Salts
5959
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 ---> neutralCOCO33
2-2- ++ HH22O O ee HCOHCO33-- ++ OHOH--
basebase acidacid acidacidbasebase KKbb = 2.1 x 10 = 2.1 x 10-4-4
Acid-Base Properties of SaltsAcid-Base Properties of Salts
Kb = 2.1 x 10-4 = [HCO3- ][OH- ]
[CO32 ] x2
0.10 - x
Assume 0.10 - x ≈ 0.10, because 100•Kb < Co
x = [HCO3-] = [OH-] = 0.0046 M
Step 2.Step 2. Solve the equilibrium expressionSolve the equilibrium expression
6060
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 ---> neutralCOCO33
2-2- + H+ H22O O ee HCOHCO33-- + OH + OH--
basebase acidacid acidacid basebase KKbb = 2.1 x 10 = 2.1 x 10-4-4
Acid-Base Properties of SaltsAcid-Base Properties of Salts
Step 3.Step 3. Calculate the pHCalculate the pH[OH[OH--] = 0.0046 M] = 0.0046 MpOH = - log [OHpOH = - log [OH--] = 2.34] = 2.34pH + pOH = 14, pH + pOH = 14,
so so pH = 11.66pH = 11.66, and the solution is ________., and the solution is ________.