Download - Unit 17: Acid-Base Equilibria
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AcidsandBases
Unit 17:Acid-Base Equilibria
Dr. Jorge L. AlonsoMiami-Dade College –
Kendall CampusMiami, FL
CHM 1046: General Chemistry and Qualitative Analysis
Textbook Reference:
•Chapter 18
•Module # 6
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AcidsandBases
Acid-Base Theories
Arrhenius (1883)
Brønsted–Lowry (1923)
Lewis (1923-38)
HCl NaOH
NH3
BF3
H2O
H2SO4
AlI3
CO32-
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AcidsandBases
Acid-Base Definitions• Svante Arrhenius (1883)
Acid: Substance that, when dissolved in water, increases the concentration of hydrogen ions.
HX H+ + X-
{acid}
Base: Substance that, when dissolved in water, increases the concentration of hydroxide ions.
MOH M+ + OH-
Neutralization:
HX + MOH MX + HOH
acid base salt water
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AcidsandBases
• Brønsted–Lowry (1923)Acid: Proton (H+) donor.
… have a removable (acidic) proton which are donated to a bases.
{base}
Acid-Base Definitions
HX + H2O X- + H3O+
acid base conjugate b conjugate a.
Neutralization:
HX + B X- + BH+
acid base conj. b conj. a
B + H2O BH+ + OH-
base acid conj. a conj. b
Base: Proton (H+) acceptor.
…must have a pair of nonbonding electrons which accepts protons.
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AcidsandBases
Strong AcidsHCl, HBr, HI, HNO3, H2SO4, HClO3, & HClO4
SOLUBILITY RULES: for Ionic Compounds (Salts)• All salts containing the anions: NO3
-, ClO3-, ClO4
-, (C2H3O2-) are soluble.
• All Cl-, Br-, and I- are soluble except for Hg22+, Ag+, and Pb2+ salts.
• All SO42- are soluble except for Pb2+, Ba2+, and Sr2+.
Are strong electrolytes and exist totally as ions in aqueous solution.
For the monoprotic strong acids, [acid]i = [H3O+] or [H+]
HX H+(aq) + X-
(aq)
HX + H2O X- + H3O+
acid base conj. b conj. a
Strong acids have very weak conjugate bases
Arrhenius:
Brønsted–Lowry:
100%
100%
For strong acid: [1M HX]i = [1M H3O+] or [1M H+] For weak acid: [1M HA]i > [0.1M H3O+] or [0.1M H+]
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AcidsandBases
these substances dissociate completely (100%) in aqueous solution.
SOLUBILITY RULES: for Ionic Compounds (Salts)All OH- are insoluble except for IA metals, (NH4
+), Ca2+, Ba2+ , & Sr2+ (heavy IIA).
MOH M+ + OH-
Na+OH- + H2O Na+ + HOH + OH-
base acid conjugate acid conjugate base
Arrhenius:
Brønsted–Lowry:
these substances accept protons from water to form very weak conjugate acids.
For the monobasic strong bases, [base]i = [OH-] [1M NaOH] = [1M OH-]
Strong Bases
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AcidsandBases
Review of Strong Acid & Bases
Strong acids have very weak conjugate bases
HX + H2O H3O+ + X-
acid base conj. a conj. b
Strong bases have very weak conjugate acids
MOH M+ + OH-
base conj. a conj. b
Salts of Strong acids & bases have very weak conjugate acids & bases
MX + H2O M+(aq)
+ X- (aq)
salt conj. a conj. b
Has no affinity for H+
Has no affinity for OH-
Have no affinity for H+ or OH-
Example: HCl
Example: NaOH
Example: NaCl
Later on this course: Salts that are neutral
(according Brønsted–Lowry)
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AcidsandBases
Weak Acid Dissociation
HX
HA
Strong Acid Dissociation
HXH3O
+ X-
Concentrated 10M Strong Acid
Concentrated 10M Very Weak Acid
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AcidsandBases
Weak Acids
Reactions between acids and bases always yield their corresponding conjugate bases and conjugate acids.
HA + H2O ↔ H3O+ + A-
acid base conj. a conj. b
H+
H+
Weak acids (HA) have very strong conjugate bases (A-)
•The weaker the acid, the stronger their conjugate base.
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AcidsandBases
H+
H+
Weak Bases
Bases react with water to produce hydroxide ion.
B + HOH BH+ + OH-
• Weak bases have strong conjugate acids
base acid conj. a conj. b
•The weaker the base, the stronger their conjugate acid.
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AcidsandBases
Acid / Base Strength
The stronger an acid, the weaker its conjugate base.
The stronger an base, the weaker its conjugate acid.
conjugate
HA + H2O ↔ H3O+ + A-
acid base conj. a conj. b
B + HOH ↔ BH+ + OH-
base acid conj. a conj. b
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AcidsandBases
The Leveling Effect of Water
Strong acids react with water to produce H3O+, the strongest acid that can exist in an aqueous solution.
Strong bases react with water to produce OH-, the strongest base that can exist in an aqueous solution.
+
+
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AcidsandBases
Salts: (1) of strong base & weak acid:
MA M+(aq) + A-
(aq)
A- + HOH HA + OH- (hydrolysis)
(2) of a weak base & strong acid:
BHX BH+(aq) + X-
(aq)
BH+ + HOH B + H3O+ (hydrolysis)
(3) of a weak base & weak acid:
BHA ↔ BH+(aq) + A-
(aq)
Weak acids (HA) have very strong conjugate bases (A-)
HA + H2O ↔ H3O+ + A-
acid base conj. a conj. b
Review of Weak Acid & Bases
Weak bases (B) have very strong conjugate acids (BH+)
B + HOH ↔ BH+ + OH-
base acids conj. a conj. b
Has affinity for H+
Has affinity for OH-
Example: HF
Example: NH3
Example: NaFExample: NH4ClExample: NH4F
Later on this course: Salts that are acidic or basic
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AcidsandBases
• Lewis acids are electron-pair acceptors (have empty valance orbitals).
• Lewis bases are electron-pair donors.
{Lewis A-B}
Lewis Acids and Bases (1923-38)
+
H+ + O H- H O H..
: : : : ......
+
+
+
• Lewis acid-base neutralization reactions
+
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AcidsandBases
Is Water an Acid or a Base? The Autoionization of Water
• In pure water, a few water molecules act as acids and a few act as bases.
• This is referred to as autoionization.
H2O(l) + H2O(l) H3O+(aq) + OH−(aq)
H2O(l) H+(aq) + OH−(aq)
{Movie}
1 H3O+ and 1 OH- for every 10 million (107) H2O molecules
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AcidsandBases
Autoionization of Water
H2O(l) + H2O(l) H3O+(aq) + OH−(aq)
• The equilibrium expression for this process is
[H3O+] [OH−]
• Kw is referred to as the ion-product constant for water (@ 25°C).
[H2O]2Kw = = 1.0 10−14Kw = = 1.0 10−14
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AcidsandBases
What proportion of H2O molecules
dissociate into H3O+ & OH- ?
• In pure water,
Kw = [H3O+] [OH−] = 1.0 10−14
• Because in pure water [H3O+] = [OH−],
Kw = [1.0 10−7 ] [1.0 10−7 ] = 1.0 10−14
1 H3O+ and 1 OH- for every 10 million (107) H2O molecules
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AcidsandBases
pH…..defined as the negative base-10 logarithm of the hydronium ion concentration.
pH = −log [H3O+]
Problem: Calculate pH when [H3O+] = 2.3 x 10-3 M
Problem: Calculate [H3O+] when pH = 2.3 ?
10x
log
antilog10^ -2.3
10x
logbase 10 log log - 2.3 * 10^- 3
For pure H2O: [1.0 10−7 ] = 7.0
= 2.64
= 5.0 * 10-3
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AcidsandBases
Other “p” Scales
• The “p” in pH tells us to take the negative log of the quantity (in this case, hydrogen ions).
• Some examples of other “p” scales are: pH = −log [H3O+]
pOH = −log [OH−]
pKw = −log Kw
pKa = −log Ka
pKb = −log Kb
[H3O+] = 1 x 10-7
[OH-] = 1 x 10-7
Kw = 1 x 10-14
Ka = 6.8 x 10-4
Kb = 1.8 x 10-5
pH = 7
pOH = 7
pKw = 14
pKa = 3.2
pKb = 4.7
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AcidsandBases
pH and pOH equilibrium in pure Water
−log [H3O+] + −log [OH−] = −log Kw
• In pure water,
[H3O+] [OH−] = Kw
• Because in pure water [H3O+] = [OH−],
Kw = [1.0 10−7 ] [1.0 10−7 ] = 1.0 10−14
pH + pOH = pKw
7 + 7 = 14
[1.0 10−7 ] [1.0 10−7 ] = 1.0 10−14
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AcidsandBases
pH and pOH equilibrium in Water to which
Acids & Bases are Added
Kw = [1.0 10−6 ] [1.0 10−8 ] = 1.0 10−14
pH + pOH = pKw
6 + 8 = 14
H2O(l) + H2O(l) H3O+(aq) + OH−(aq)
H3O+
Add acid
[H3O+] [OH−]H2O
Kw = [1.0 10−7 ] [1.0 10−7 ] = 1.0 10−14
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AcidsandBases
pH and pOH equilibrium in Water to which
Acids & Bases are Added
Kw = [1.0 10−8 ] [1.0 10−6 ] = 1.0 10−14
pH + pOH = pKw
8 + 6 = 14
H2O(l) + H2O(l) H3O+(aq) + OH−(aq)
OH-Add base
[H3O+] [OH−]H2O
Kw = [1.0 10−7 ] [1.0 10−7 ] = 1.0 10−14
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AcidsandBases
pH
0 1.0 (100) 0.00000000000001
1 1.0 (10-1) 0.00000000000012 0.01 0.000000000001
3 0.001 0.00000000001 4 0.0001 0.0000000001
5 0.00001 0.000000001
6 0.000001 0.00000001
7 0.0000001 (10-7) 0.0000001 (10-7)8 0.00000001 0.0000001
9 0.000000001 0.00001
10 0.0000000001 0.0001
11 0.00000000001 0.001 12 0.000000000001 0.01
13 0.0000000000001 1.0 (10-1)
14 0.00000000000001 1.0 (100)
[H3O+] [OH−]
1
[OH−][H3O+]∝
Notice the relationship
between
[H3O+] and [OH−]
It is an inverse relationship!
pH + pOH = 14
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AcidsandBases
pH
These are the pH values for several common substances.
pH + pOH = 14
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AcidsandBases
Dissociation Constants for Weak Acids
• For a generalized acid dissociation,
the equilibrium expression would be
• This equilibrium constant is called the acid-dissociation (ionization) constant, Ka.
[H3O+] [A−][HA]
Ka =
HA(aq) + H2O(l) A−(aq) + H3O+(aq)
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AcidsandBases
Dissociation Constants
The greater the value of Ka, the stronger the acid.
pKa
3.2
3.3
4.2
4.7
7.5
9.3
9.9
pKa
Salts of weak acids are strong bases
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AcidsandBases
Calculating Ka from the pH
A 0.10 M solution of formic acid, HCOOH, at 25°C has a pH = 2.38. Calculate Ka for formic acid at this temperature. Also calculate % ionization.
Problem: (to determine Ka, simply dissolve a known [HA]i then measure pH)
HCOOH (aq) + H2O(l) HCOO−(aq) + H3O+
(aq)
pH = −log [H3O+]
2.38 = −log [H3O+]
−2.38 = log [H3O+]
10−2.38 = 10log [H3O+] = [H3O+]
4.2 10−3 = [H3O+] = [HCOO−]
[H3O+] [A−][HA]
Ka =
What is the Ka expression?
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AcidsandBases
Calculating Ka from pH
[HCOOH], M [H3O+], M [HCOO−], M
Initially 0.10 0 0
Change −4.2 10-3 +4.2 10-
3
+4.2 10−3
Equilibrium 0.0958 = 0.10 4.2 10−3 4.2 10−3
[H3O+] [A−][HA]
Ka = [4.2 10−3] [4.2 10−3][0.10]
= = 1.8 10−4
[H3O+] = [HCOO−] = 4.2 10−3
• The pH of a 0.10 M solution of formic acid, HCOOH, at 25°C is 2.38. Calculate Ka for formic acid at this temperature. Also calculate % ionization.
Problem:
HCOOH (aq) + H2O(l) HCOO−(aq) + H3O+
(aq)
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AcidsandBases
Calculating Percent Ionization
• Percent Ionization = 100
• In this example
[H3O+]eq
[HA]initial
Percent Ionization = 1004.2 10−3
0.10
= 4.2%
HCOOH(aq) + H2O(l) HCOO−(aq) + H3O
+(aq)
Whole (Initial)
Part (@ Equi)
[HCOOH], M [H3O+], M [HCOO−], M
I 0.10 0 0
C −4.2 10-3 +4.2 10-3 +4.2 10−3
E 0.0958 = 0.10 4.2 10−3 4.2 10−3
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AcidsandBases
Calculating pH from Ka
Calculate the pH of a 0.30 M solution of acetic acid, HC2H3O2, at 25°C.
Problem: For acetic acid at 25°C is Ka =1.8 10−5
HC2H3O2(aq) + H2O(l) H3O+(aq) + C2H3O2−(aq)
[H3O+] [C2H3O2−]
[HC2H3O2]Ka =
[C2H3O2], M [H3O+], M [C2H3O2−], M
Initially 0.30 0 0
Change −x +x +x
At Equilibrium 0.30 − x 0.30 x x
We are assuming that x will be very small compared to 0.30 and can, therefore, be ignored.
[x] [x][0.3]
=
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AcidsandBases
Calculating pH from Ka
(x)2
(0.30)1.8 10−5 =
(1.8 10−5) (0.30) = x2
5.4 10−6 = x2
[H3O+] = 2.3 10−3 = x
[H3O+] [C2H3O2−]
[HC2H3O2]Ka =
• pH = −log [H3O+] = − log x
• pH = −log (2.3 10−3)
• pH = 2.64
What is the pH?
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AcidsandBases
(b) Calculate the pH of 0.50 M Lactic Acid.
2002B Q1
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AcidsandBases
Polyprotic Acids• Have more than one acidic proton.
• If the difference between the Ka for the first dissociation and subsequent Ka values is 103 or more, the pH generally depends only on the first dissociation.
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AcidsandBases
Bases react with water to produce hydroxide ion.
Dissociation Constants for Weak Bases
The generalized equilibrium constant expression for these reactions is
[HB+] [OH−][B]
Kb =
where Kb is the base-dissociation constant.
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AcidsandBases
Weak Bases
Kb can be used to find [OH−] and, through it, pH.
pKb
4.7
8.8
8.0
3.4
6.7
3.7
6.5
pKb
Salts of weak bases are strong acids
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AcidsandBases
NH3(aq) + H2O(l) NH4+
(aq) + OH−(aq)
pH & pOH for Weak Acids & Bases
Kb = 1.8 x 10-5
[NH3], M [NH4+], M [OH−], M
Initially 0.15 0 0
Change - x + x + x
[NH4+] [OH−]
[NH3]Kb = = 1.8 10−5 =
At Equilibrium 0.15 - x 0.15
x x
What is the pH of a 0.15 M solution of NH3?
(x)2
(0.15)
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AcidsandBases
pH of Basic Solutions
(1.8 10−5) (0.15) = x2
2.7 10−6 = x2
1.6 10−3 = x2
(x)2
(0.15) =
[NH4+] [OH−]
[NH3]Kb = = 1.8 10−5
[OH−] = 1.6 10−3 M
pOH = −log (1.6 10−3)
pOH = 2.80
pH = 14.00 − 2.80
pH = 11.20
What is pH of soln?What is x?
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AcidsandBases
2002A Q1(a-b)
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AcidsandBases
2005 A acid base equilibria
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AcidsandBases
2005 A acid base equilibria ANSWERS
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AcidsandBases
c) ii
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AcidsandBases
2005 B acid base equilibria
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AcidsandBases
2005 B acid base equilibria ANSWERS
a)
b)
c)
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AcidsandBases
d)i
d)ii
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AcidsandBases
d)iii
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AcidsandBases
2001 a/b equilibria titration
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AcidsandBases
2001 answers
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AcidsandBases
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AcidsandBases
2002
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AcidsandBases
a)
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AcidsandBases
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AcidsandBases
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AcidsandBases
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AcidsandBases
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AcidsandBases
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AcidsandBases
2002 B
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AcidsandBases
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AcidsandBases
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AcidsandBases
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AcidsandBases
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AcidsandBases
2003 A
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AcidsandBases
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AcidsandBases
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AcidsandBases
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AcidsandBases
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AcidsandBases
2003 B
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AcidsandBases
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AcidsandBases
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AcidsandBases
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AcidsandBases
2005 A
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AcidsandBases
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AcidsandBases
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AcidsandBases
H2O(l) + H2O(l) H3O+(aq) + OH−(aq)
Hydroxide is a much stronger base than H2O, so the equilibrium favors the left side (K<<<1).