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
AcidsandBases
Acid-Base Theories
Arrhenius (1883)
Brønsted–Lowry (1923)
Lewis (1923-38)
HCl NaOH
NH3
BF3
H2O
H2SO4
AlI3
CO32-
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
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.
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+]
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
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)
AcidsandBases
Weak Acid Dissociation
HX
HA
Strong Acid Dissociation
HXH3O
+ X-
Concentrated 10M Strong Acid
Concentrated 10M Very Weak Acid
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.
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.
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
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.
+
+
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
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
+
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
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
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
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
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
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
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
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
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
AcidsandBases
pH
These are the pH values for several common substances.
pH + pOH = 14
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)
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
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?
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)
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
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]
=
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?
AcidsandBases
(b) Calculate the pH of 0.50 M Lactic Acid.
2002B Q1
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.
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.
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
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)
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?
AcidsandBases
2002A Q1(a-b)
AcidsandBases
2005 A acid base equilibria
AcidsandBases
2005 A acid base equilibria ANSWERS
AcidsandBases
c) ii
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2005 B acid base equilibria
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2005 B acid base equilibria ANSWERS
a)
b)
c)
AcidsandBases
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d)ii
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2001 a/b equilibria titration
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2001 answers
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2002 B
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2003 A
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2003 B
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2005 A
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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).
