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Chapter 13
Acids and Bases
Some Properties of Acids
Produce H+ (as H3O+) ions in water (the hydronium ion is a
hydrogen ion attached to a water molecule)
Taste sour
Corrode metals
Electrolytes
React with bases to form a salt and water
pH is less than 7
Turns blue litmus paper to red “Blue to Red A-CID”
Some Properties of Bases
Produce OHProduce OH-- ions in water ions in water
Taste bitter, chalkyTaste bitter, chalky
Are electrolytesAre electrolytes
Feel soapy, slipperyFeel soapy, slippery
React with acids to form salts and waterReact with acids to form salts and water
pH greater than 7pH greater than 7
Turns red litmus paper to blue “Turns red litmus paper to blue “BBasic asic BBlue”lue”
Anion Ending Acid Name
-ide hydro-(stem)-ic acid
-ate (stem)-ic acid
-ite (stem)-ous acid
Acid Nomenclature Review
No OxygenNo Oxygen
w/Oxygen w/Oxygen
An easy way to remember which goes with which…An easy way to remember which goes with which…
““In the cafeteria, youIn the cafeteria, you ATEATE somethingsomething ICICky”ky”
Arrhenius acid is a substance that produces H+ (H3O+) in water
Arrhenius base is a substance that produces OH- in water
Acid/Base definitionsDefinition 1: Arrhenius
Acid/Base Definitions
• Definition #2: Brønsted – LowryDefinition #2: Brønsted – Lowry
Acids – proton donorAcids – proton donor
Bases – proton acceptorBases – proton acceptor
A “proton” is really just a hydrogen A “proton” is really just a hydrogen atom that has lost it’s electron!atom that has lost it’s electron!
A Brønsted-Lowry acid is a proton donorA Brønsted-Lowry base is a proton acceptor
acidconjugate
basebase conjugate
acid
ACID-BASE THEORIESACID-BASE THEORIESACID-BASE THEORIESACID-BASE THEORIES
The Brønsted definition means NHThe Brønsted definition means NH33 is a is a BASE BASE in in
water — and water is itself anwater — and water is itself an ACIDACID
BaseAcidAcidBaseNH4
+ + OH-NH3 + H2OBaseAcidAcidBase
NH4+ + OH-NH3 + H2O
Conjugate PairsConjugate Pairs
Learning Check!
Label the acid, base, conjugate acid, and Label the acid, base, conjugate acid, and conjugate base in each reaction:conjugate base in each reaction:
HCl + OHHCl + OH-- Cl Cl-- + H + H22OO HCl + OHHCl + OH-- Cl Cl-- + H + H22OO
HH22O + HO + H22SOSO44 HSO HSO44-- + H + H33OO
++ HH22O + HO + H22SOSO44 HSO HSO44-- + H + H33OO
++
AcidAcidAcidAcid
AcidAcidAcidAcid
BaseBaseBaseBase
BaseBaseBaseBase
Conj.Conj.BaseBaseConj.Conj.BaseBase
Conj.Conj.BaseBaseConj.Conj.BaseBase
Conj.Conj.AcidAcidConj.Conj.AcidAcid
Conj.Conj.AcidAcidConj.Conj.AcidAcid
Acids & Base DefinitionsAcids & Base Definitions
Lewis acid - a substance that Lewis acid - a substance that accepts an electron pairaccepts an electron pair
Lewis base - a substance Lewis base - a substance that donates an electron that donates an electron pairpair
Definition #3 – Lewis Definition #3 – Lewis
Formation ofFormation of hydronium ion hydronium ion is also an is also an excellent example.excellent example.
Lewis Acids & BasesLewis Acids & Bases
•Electron pair of the new O-H bond Electron pair of the new O-H bond originates on the Lewis base.originates on the Lewis base.
HH
H
BASE
••••••
O—HO—H
H+
ACID
Lewis Acid/Base ReactionLewis Acid/Base Reaction
The The pH scalepH scale is a way of is a way of expressing the strength of expressing the strength of acids and bases. Instead of acids and bases. Instead of using very small numbers, using very small numbers, we just use the NEGATIVE we just use the NEGATIVE power of 10 on the Molarity power of 10 on the Molarity of the Hof the H++ (or OH (or OH--) ion.) ion.
Under 7 = acidUnder 7 = acid7 = neutral7 = neutral
Over 7 = base Over 7 = base
Calculating the pH
pH = - log [H+](Remember that the [ ] mean Molarity)
Example: If [H+] = 1 X 10-10
pH = - log 1 X 10-10
pH = - (- 10)
pH = 10
Example: If [H+] = 1.8 X 10-5
pH = - log 1.8 X 10-5
pH = - (- 4.74)
pH = 4.74
Try These!Try These!
Find the pH of these:Find the pH of these:
1)1) A 0.15 M solution of A 0.15 M solution of Hydrochloric acidHydrochloric acid
2) A 3.00 X 102) A 3.00 X 10-7-7 M M solution of Nitric solution of Nitric acidacid
pH = - log [HpH = - log [H++]]
pH = - log 0.15pH = - log 0.15
pH = - (- 0.82)pH = - (- 0.82)
pH = 0.82pH = 0.82
pH = - log 3 X 10pH = - log 3 X 10-7-7
pH = - (- 6.52)pH = - (- 6.52)
pH = 6.52pH = 6.52
pH calculations – Solving for H+pH calculations – Solving for H+pH calculations – Solving for H+pH calculations – Solving for H+
If the pH of Coke is 3.12, [HIf the pH of Coke is 3.12, [H++] = ???] = ???
Because pH = - log [HBecause pH = - log [H++] then] then
- pH = log [H- pH = log [H++]]
Take antilog (10Take antilog (10xx) of both) of both sides and get sides and get
1010-pH -pH == [H[H++]]
[H[H++] = 10] = 10-3.12-3.12 = 7.6 x 10 = 7.6 x 10-4-4 M M
*** to find antilog on your calculator, look for “Shift” *** to find antilog on your calculator, look for “Shift” or “2or “2nd nd function” and then the log buttonfunction” and then the log button
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 beIn pure water there can be AUTOIONIZATIONAUTOIONIZATION
Equilibrium constant for water = KEquilibrium constant for water = Kww
KKww = [H = [H33OO++] [OH] [OH--] =] = 1.00 x 101.00 x 10-14-14 at 25 at 25 ooCC
More About WaterMore About Water
KKww = [H = [H33OO++] [OH] [OH--] = 1.00 x 10] = 1.00 x 10-14-14 at 25 at 25 ooCC
In a neutral solution [HIn a neutral solution [H33OO++] = [OH] = [OH--]]
and so [Hand so [H33OO++] = [OH] = [OH--] = 1.00 x 10] = 1.00 x 10-7-7 M M
OH-
H3O+
OH-
H3O+
AutoionizationAutoionization
pOH• Since acids and bases are Since acids and bases are
opposites, pH and pOH are opposites, pH and pOH are opposites!opposites!
• pOH does not really exist, but it is pOH does not really exist, but it is useful for changing bases to pH.useful for changing bases to pH.
• pOH looks at the perspective of a pOH looks at the perspective of a basebase
pOH = - log [OHpOH = - log [OH--]]Since pH and pOH are on opposite Since pH and pOH are on opposite
ends,ends,pH + pOH = 14pH + pOH = 14
pHpH [H+][H+] [OH-][OH-] pOHpOH
[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?
[OH[OH--] = 0.0010 (or 1.0 X 10] = 0.0010 (or 1.0 X 10-3-3 M) M)
pOH = - log 0.0010pOH = - log 0.0010
pOH = 3pOH = 3
pH = 14 – 3 = 11pH = 14 – 3 = 11
OR KOR Kww = [H = [H33OO++] [OH] [OH--]]
[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
What is the pH of a 2 x 10-3 M HNO3 solution?
HNO3 is a strong acid – 100% dissociation.
HNO3 (aq) + H2O (l) H3O+ (aq) + NO3- (aq)
pH = -log [H+] = -log [H3O+] = -log(0.002) = 2.7
Start
End
0.002 M
0.002 M 0.002 M0.0 M
0.0 M 0.0 M
What is the pH of a 1.8 x 10-2 M Ba(OH)2 solution?
Ba(OH)2 is a strong base – 100% dissociation.
Ba(OH)2 (s) Ba2+ (aq) + 2OH- (aq)
Start
End
0.018 M
0.018 M 0.036 M0.0 M
0.0 M 0.0 M
pH = 14.00 – pOH = 14.00 + log(0.036) = 12.56
HNO3, HCl, HBr, HI, H2SO4 and HClO4 are the strong acids.
Strong and Weak Acids/BasesStrong and Weak Acids/Bases
The strength of an acid (or base) is determined by the amount of IONIZATION.
The strength of an acid (or base) is determined by the amount of IONIZATION.
• Generally divide acids and bases into STRONG or Generally divide acids and bases into STRONG or WEAK ones.WEAK ones.
STRONG ACID:STRONG ACID: HNOHNO3 3 (aq) + H(aq) + H22O (l) O (l) HH33OO+ + (aq) + NO(aq) + NO33
- - (aq)(aq)
HNOHNO33 is about 100% dissociated in water. is about 100% dissociated in water.
Strong and Weak Acids/BasesStrong and Weak Acids/Bases
• Weak acidsWeak acids are much less than 100% ionized in are much less than 100% ionized in
water.water.
*One of the best known is acetic acid = CH*One of the best known is acetic acid = CH33COCO22HH
Strong and Weak Acids/BasesStrong and Weak Acids/Bases
• Strong Base:Strong Base: 100% dissociated in water.100% dissociated in water.
NaOH (aq) NaOH (aq) Na Na+ + (aq) + OH(aq) + OH- - (aq)(aq)
Other common strong Other common strong bases include KOH andbases include KOH and Ca(OH)Ca(OH)22..
CaO (lime) + HCaO (lime) + H22O -->O -->
Ca(OH)Ca(OH)22 (slaked lime) (slaked lime)
CaOCaO
Strong and Weak Acids/BasesStrong and Weak Acids/Bases
Strong bases are the group I hydroxidesStrong bases are the group I hydroxides
Calcium, strontium, and barium hydroxides are Calcium, strontium, and barium hydroxides are strong, but only soluble in water to 0.01 Mstrong, but only soluble in water to 0.01 M
• Weak base:Weak base: less than 100% ionized in waterless than 100% ionized in water
One of the best known weak bases is ammoniaOne of the best known weak bases is ammonia
NHNH3 3 (aq) + H(aq) + H22O (l) O (l) ↔↔ 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
Weak BasesWeak Bases
Equilibria Involving Equilibria Involving Weak Acids and BasesWeak Acids and Bases
Consider acetic acid, HCConsider acetic acid, HC22HH33OO22 (HOAc) (HOAc)
HCHC22HH33OO22 + H + H22O O ↔↔ H H33OO++ + C + C22HH33OO22 --
AcidAcid Conj. base Conj. base
Ka [H3O+][OAc- ]
[HOAc] 1.8 x 10-5Ka
[H3O+][OAc- ][HOAc]
1.8 x 10-5
(K is designated K(K is designated Kaa for ACID) for ACID)
K gives the ratio of ions (split up) to molecules (don’t split up)K gives the ratio of ions (split up) to molecules (don’t split up)
Ionization Constants for Acids/BasesIonization Constants for Acids/Bases
AcidsAcids ConjugateConjugateBasesBases
Increase strength
Increase strength
Equilibrium Constants Equilibrium Constants for Weak Acidsfor Weak Acids
Equilibrium Constants Equilibrium Constants for Weak Acidsfor Weak Acids
Weak acid has KWeak acid has Kaa < 1 < 1
Leads to small [HLeads to small [H33OO++] and a pH of 2 - 7] and a pH of 2 - 7
Equilibria Involving A Weak AcidEquilibria Involving A Weak AcidEquilibria Involving A Weak AcidEquilibria Involving A Weak AcidYou have 1.00 M HOAc. Calc. the You have 1.00 M HOAc. Calc. the
equilibrium concs. of HOAc, Hequilibrium concs. of HOAc, H33OO++, OAc, OAc--, ,
and the pH.and the pH.
Step 1.Step 1. Define equilibrium concs. in ICE Define equilibrium concs. in ICE
table.table.
[HOAc][HOAc] [H[H33OO++]] [OAc[OAc--]]
initialinitial
changechange
equilibequilib
1.001.00 00 001.001.00 00 00
-x-x +x+x +x+x-x-x +x+x +x+x
1.00-x1.00-x xx xx1.00-x1.00-x xx xx
Step 2.Step 2. Write KWrite Kaa expression expression
You have 1.00 M HOAc. Calc. the equilibrium concs. You have 1.00 M HOAc. Calc. the equilibrium concs. of HOAc, Hof HOAc, H33OO++, OAc, OAc--, and the pH., and the pH.
Ka 1.8 x 10-5 = [H3O+][OAc- ]
[HOAc]
x2
1.00 - xKa 1.8 x 10-5 =
[H3O+][OAc- ][HOAc]
x2
1.00 - x
This is a quadratic. Solve using quadratic This is a quadratic. Solve using quadratic formula.formula.
or you can make an approximation if x is very or you can make an approximation if x is very small! (Rule of thumb: 10small! (Rule of thumb: 10-5-5 or smaller is ok) or smaller is ok)or you can make an approximation if x is very or you can make an approximation if x is very small! (Rule of thumb: 10small! (Rule of thumb: 10-5-5 or smaller is ok) or smaller is ok)
Equilibria Involving A Weak AcidEquilibria Involving A Weak AcidEquilibria Involving A Weak AcidEquilibria Involving A Weak Acid
Equilibria Involving A Weak AcidEquilibria Involving A Weak AcidEquilibria Involving A Weak AcidEquilibria Involving A Weak Acid
Step 3.Step 3. Solve KSolve Kaa expression expression
You have 1.00 M HOAc. Calc. the equilibrium concs. You have 1.00 M HOAc. Calc. the equilibrium concs. of HOAc, Hof HOAc, H33OO++, OAc, OAc--, and the pH., and the pH.
Ka 1.8 x 10-5 = [H3O+][OAc- ]
[HOAc]
x2
1.00 - xKa 1.8 x 10-5 =
[H3O+][OAc- ][HOAc]
x2
1.00 - x
First assume x is very small because First assume x is very small because KKaa is so small. is so small.
Ka 1.8 x 10-5 = x2
1.00Ka 1.8 x 10-5 =
x2
1.00
Now we can more easily solve this Now we can more easily solve this approximate expression.approximate expression.
Step 3.Step 3. Solve KSolve Kaa approximateapproximate expressionexpression
You have 1.00 M HOAc. Calc. the equilibrium concs. You have 1.00 M HOAc. Calc. the equilibrium concs. of HOAc, Hof HOAc, H33OO++, OAc, OAc--, and the pH., and the pH.
Ka 1.8 x 10-5 = x2
1.00Ka 1.8 x 10-5 =
x2
1.00
x =x = [[HH33OO++] = [] = [OAcOAc--] = 4.2 x 10] = 4.2 x 10-3-3 M M
pH = - log [pH = - log [HH33OO++] = -log (4.2 x 10] = -log (4.2 x 10-3-3) =) = 2.372.37
Equilibria Involving A Weak AcidEquilibria Involving A Weak AcidEquilibria Involving A Weak AcidEquilibria Involving A Weak Acid
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 formic Calculate the pH of a 0.0010 M solution of formic
acid, HCOacid, HCO22H.H.
HCOHCO22H + HH + H22O O ↔↔ HCO HCO22-- + H + H33OO++
KKaa = 1.8 x 10 = 1.8 x 10-4-4
Approximate solutionApproximate solution
[H[H33OO++] = 4.2 x 10] = 4.2 x 10-4-4 M, M, pH = 3.37pH = 3.37
Exact SolutionExact Solution
[H[H33OO++] = [HCO] = [HCO22--] = 3.4 x 10] = 3.4 x 10-4-4 M M
[HCO[HCO22H] = 0.0010 - 3.4 x 10H] = 0.0010 - 3.4 x 10-4-4 = 0.0007 M = 0.0007 M
pH = 3.47 pH = 3.47
Equilibrium Constants Equilibrium Constants for Weak Basesfor Weak Bases
Equilibrium Constants Equilibrium Constants for Weak Basesfor Weak Bases
Weak base has KWeak base has Kbb < 1 < 1
Leads to small [OHLeads to small [OH--] and a pH of 12 - 7] and a pH of 12 - 7
RelatioRelatio
n of Kn of Kaa, ,
KKbb, ,
[H[H33OO++] ]
and pHand pH
Equilibria 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 ↔ ↔ NHNH44++ + OH + OH--
KKbb = 1.8 x 10 = 1.8 x 10-5-5
Step 1.Step 1. Define equilibrium concs. in ICE tableDefine equilibrium concs. in ICE table
[NH[NH33]] [NH[NH44++]] [OH[OH--]]
initialinitial
changechange
equilibequilib
0.0100.010 00 000.0100.010 00 00
-x-x +x+x +x+x-x-x +x+x +x+x
0.010 - x0.010 - x x x xx0.010 - x0.010 - x x x xx
Equilibria 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 expressionSolve the equilibrium expression
Kb 1.8 x 10-5 = [NH4
+][OH- ][NH3 ]
= x2
0.010 - xKb 1.8 x 10-5 =
[NH4+][OH- ]
[NH3 ] =
x2
0.010 - x
Assume x is small, soAssume x is small, so x = [OHx = [OH--] = [NH] = [NH44
++] = 4.2 x 10] = 4.2 x 10-4-4 M M
and [NHand [NH33] = 0.010 - 4.2 x 10] = 0.010 - 4.2 x 10-4-4 ≈ 0.010 M ≈ 0.010 M
The approximation is validThe approximation is valid !!
Equilibria Involving A Weak Equilibria Involving A Weak BaseBase
You have 0.010 M NHYou have 0.010 M NH33. Calc. the pH.. Calc. the pH.
NHNH33 + H + H22O O NH NH44++ + OH + OH--
KKbb = 1.8 x 10 = 1.8 x 10-5-5
Step 3.Step 3. Calculate pHCalculate pH
[OH[OH--] = 4.2 x 10] = 4.2 x 10-4-4 M M
so pOH = - log [OHso pOH = - log [OH--] = 3.37] = 3.37
Because pH + pOH = 14,Because pH + pOH = 14,
pH = 10.63pH = 10.63
Types of Acid/Base Reactions: Types of Acid/Base Reactions: SummarySummary
F- (aq) + H2O (l) OH- (aq) + HF (aq)
Weak Bases are weak electrolytes
NO2- (aq) + H2O (l) OH- (aq) + HNO2 (aq)
Conjugate acid-base pairs:
• The conjugate base of a strong acid has no measurable strength.
• H3O+ is the strongest acidstrongest acid that can exist in aqueous solution.
• The OH- ion is the strongest basestrongest base that can exist in aqueous solution.
Strong Acid Weak Acid
percent ionization = Ionized acid concentration at equilibrium
Initial concentration of acidx 100%
For a monoprotic acid HA
Percent ionization = [H+]
[HA]0
x 100% [HA]0 = initial concentration
Ionization Constants of Conjugate Acid-Base Pairs
HA (aq) H+ (aq) + A- (aq)
A- (aq) + H2O (l) OH- (aq) + HA (aq)
Ka
Kb
H2O (l) H+ (aq) + OH- (aq) Kw
KaKb = Kw
Weak Acid and Its Conjugate Base
Ka = Kw
Kb
Kb = Kw
Ka
Molecular Structure and Acid Strength
H X H+ + X-
The stronger the bond
The weaker the acid
HF << HCl < HBr < HI
• Bond strength
• Polarity
Molecular Structure and Acid Strength
Z O H Z O- + H+- +
The O-H bond will be more polar and easier to break if:
• Z is very electronegative or
• Z is in a high oxidation state
Molecular Structure and Acid Strength
1. Oxoacids having different central atoms (Z) that are from the same group and that have the same oxidation number.
Acid strength increases with increasing electronegativity of Z
H O Cl O
O••
••••••
••
••••
••••
H O Br O
O••
••••••
••
••••
••••Cl is more electronegative than Br
HClO3 > HBrO3
15.9
Molecular Structure and Acid Strength
2. Oxoacids having the same central atom (Z) but different numbers of attached groups.
Acid strength increases as the oxidation number of Z increases.
HClO4 > HClO3 > HClO2 > HClO
Acid-Base Properties of SaltsNeutral Solutions:
Salts containing an alkali metal or alkaline earth metal ion (except Be2+) and the conjugate base of a strong acid (e.g. Cl-, Br-, and NO3
-).
NaCl (s) Na+ (aq) + Cl- (aq)H2O
Basic Solutions:
Salts derived from a strong base and a weak acid.
NaCH3COO (s) Na+ (aq) + CH3COO- (aq)H2O
CH3COO- (aq) + H2O (l) CH3COOH (aq) + OH- (aq)
Acid-Base Properties of Salts
Acid Solutions:
Salts derived from a strong acid and a weak base.
NH4Cl (s) NH4+ (aq) + Cl- (aq)
H2O
NH4+ (aq) NH3 (aq) + H+ (aq)
Salts with small, highly charged metal cations (e.g. Al3+, Cr3+, and Be2+) and the conjugate base of a strong acid.
Al(H2O)6 (aq) Al(OH)(H2O)5 (aq) + H+ (aq)3+ 2+
Acid-Base Properties of Salts
Solutions in which both the cation and the anion hydrolyze:
• Kb for the anion > Ka for the cation, solution will be basic
• Kb for the anion < Ka for the cation, solution will be acidic
• Kb for the anion Ka for the cation, solution will be neutral