11/20/2011
Water itself is a weak acid and weak base:
Two water molecules react to form H3O+ and OH(but only slightly)
H2O
H2O
H3O+
OH-
Autoionization of Water
H2O(l) + H2O(l) ⇌
K=
[H3O+][OH-]
[H2O]2
The ion-product for water, Kw:
H3O+(aq) + OH-(aq)
= [H+][OH-] = Kw
Set = 1.0 since nearly pure liquid
water even when salts, acids, or
bases present.
K = Kw = [H3O+][OH-] = 1.0 x 10-14 (at 25oC)
For pure water the concentration of hydroxyl and hydronium ions
must be equal:
[H3O+] = [OH-] =
√1.0 x 10-14 = 1.0 x 10 -7 M (at 25oC)
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11/20/2011
The Meaning of Kb, the Base Dissociation Constant
For the generalized reaction between a base, B(aq), and water:
⇌
B(aq) + H2O(l)
BH+(aq) + OH-(aq)
Equilibrium constant:
K = Kb = [BH+][OH-]
[B]
The stronger the base, the higher the [OH-] at
equilibrium and the larger the Kb.
The Relation Between Ka and Kb for
a Conjugate Acid-Base Pair
Acid
HA + H2O
A- + H2O
Base
x
[HA]
Ka
x
HA + OHH3O+ + OH-
2 H2O
[H3O+] [A-]
H3O+ + A-
[HA] [OH-]
[A-]
Kb
= [H3O+] [OH-]
=
Kw
For HNO2: Ka = 4.5 x 10-4 and for NO2- : Kb = 2.2 x 10-11
Ka x Kb = (4.5 x 10-4)(2.2 x 10-11) = 9.9 x 10-15
or ~ 10 x 10-15 = 1 x 10 -14 = Kw
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pH of a Strong Acid
Calculate pH of 1.0 M HClO4
• major species : H+, ClO4- and H2O
• perchloric acid is a strong acid: FULLY dissociated
Therefore, 1.0 M HClO4  1.0 M H+
pH = - log [H+] = - log [1.0] = 0.00
Calculate pH of 0.01 M HClO4
pH = - log [H+] = - log [0.01] = 2.0
Calculate pH of 0.02 M HClO4
pH = - log [H+] = - log [0.02] = 1.7
Two New Definitions
Remember: Kw = [H+][OH-] so that [OH-] = Kw / [H+]
1)
pOH = -log [OH-]
= - log (Kw/[H+]) = - log Kw – (- log [H+])
= pKw – pH
= - log (1.0 x 10-14) – pH
pOH = 14.00 – pH
2) It is convenient to express equilibrium constants for acid
dissociation in the form
pKa = - log Ka
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pH of a Strong Base
Calculate pH of 1.0 M NaOH
• major species : Na+, OH- and H2O
• sodium hydroxide is a strong base: FULLY dissociated
Therefore, 1.0 M NaOH  1.0 M [OH-]
pH = - log [H+]; pOH= - log [OH-]; pKw= pH + pOH = 14
pH = 14 – pOH
pH = 14 – (-log 1.0) = 14.0
Calculate pH of 0.01 M NaOH
pH= 14 – (-log 0.01) = 14 – 2 = 12.0
Calculate pH of 0.02 M NaOH
pH= 14 – (-log 0.02) = 14 - 1.70 = 12.3
Calculating [H3O+], pH, [OH-], and pOH
Problem: A chemist dilutes concentrated hydrochloric acid to
make two solutions: (a) 3.0 M and (b) 0.0024 M. Calculate the
[H3O+], pH, [OH-], and pOH of the two solutions at 25oC.
Solution:
(a) [H+] = 3.0 M
pH = -log[H+] = -log(3.0) = -0.48
-14
[OH-] = Kw+ = 1 x 10
= 3.3 x 10-15 M
[H ]
3.0
pOH = - log(3.3 x 10-15) = 14.52
(b) [H+] = 0.0024 M
pH = -log[H+] = -log(0.0024) = 2.62
-14
[OH-] = Kw+ = 1 x 10
[H ]
0.0024
= 4.2 x 10-12 M
pOH = -log(4.2 x 10-12) = 11.38
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pH= -log[H+]
pOH= - log [OH-]
pKw= 14 = pH + pOH
Kw= [H+][OH-]
The Acid-Dissociation Constant (Ka)
Remember
Acids dissociate into ions in water:
H3O+(aq) + A-(aq)
HA(aq) + H2O(l)
H+(aq) + A-(aq)
HA(aq)
Ka =
[H3O+][A-]
[HA]
=
[H+][A-]
[HA]
In a dilute solution of a WEAK acid, most of HA remains
undissociated and therefore
[HA]init = [HA]eq
Ka = [H+][A-]/[HA] = [H+]2 / [HA]init
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Calculate the pH of a 1.00 M HNO2 Solution
Problem: Calculate the pH of a 1.00 M solution of nitrous acid HNO2.
Solution:
HNO2 (aq)
H+(aq) + NO2-(aq)
Ka = 4.0 x 10-4
Initial concentrations: [H+] = 0, [NO2-] = 0, [HNO2] = 1.00 M
Final concentrations: [H+] = x , [NO2-] = x , [HNO2] = 1.00 M - x
Ka =
[H+] [NO2-]
[HNO2]
(x) (x)
= 4.0 x 10-4 =
1.00 - x
Assume 1.00 – x ≈ 1.00 to simplify the problem.
x2
1.00
≈ 4.0 x 10-4
or x2 ≈ 4.0 x 10-4
x = 2.0 x 10-2 = 0.02 M = [H+] = [NO2-]
pH = - log[H+] = - log(2.0 x 10-2) = 1.70
Determining Concentrations from Ka and Initial [HA]
Problem: Hypochlorous acid is a weak acid formed in laundry bleach.
What is the [H+] and pH of a 0.125 M HClO solution? Ka = 3.5 x 10-8
Plan: We need to find [H+]. First we write the balanced equation and
the expression for Ka and solve for the hydronium ion concentration.
Solution:
HClO(aq) + H2O(l)
H3O+(aq) + ClO-(aq)
Ka =
[H+] [ClO-]
[HClO]
Concentration (M)
Initial
Change
Equilibrium
Ka =
= 3.5 x 10-8
HClO
H 2O
0.125
-x
0.125 - x
----------
(x)(x)
= 3.5 x 10-8
0.125-x
H+
0
+x
x
ClO0
+x
x
Assume (0.125 – x) ≈ 0.125
x2 = 4.375 x 10-9
x = 6.61 x 10-5
pH = -log(H+) = -log(6.61 x 10-5) = 4.18
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Finding Ka for a Weak Acid from the pH – I
Problem: The weak acid hypochlorous acid is formed in bleach
solutions. If the pH of a 0.12 M solution of HClO is 4.19, what is the
value of the Ka of this weak acid?
Plan: We are given [HClO]initial and the pH, which will allow us to find
[H3O+] and then the hypochlorite anion concentration, so we can
write the reaction and expression for Ka and solve directly.
Solution:
Calculating [H3O+] : [H3O+] = 10-pH = 10-4.19 = 6.46 x 10-5 M
Concentration (M) HClO(aq)
Initial
Change
Equilibrium
0.12
-x
0.12 -x
+
H2O(l)
H3O+(aq)
1 x 10 -7
+x
1 x 10 -7 + x
----------
+
ClO -(aq)
0
+x
x
Assumptions: [H3O+] = [H3O+]HClO
since HClO is a weak acid, we assume (0.12 M – x) ≈ 0.12 M
Finding the Ka for a Weak Acid from the pH – II
H3O+(aq) + ClO -(aq)
HClO(aq) + H2O(l)
x = [H3O+] = [ClO-] = 6.46 x 10-5 M
Ka =
[H3O+] [ClO-]
=
(6.46 x 10-5 M) (6.46 x 10-5 M)
[HClO]
Ka = 3.48 x 10-8
= 3.48 x 10-8
0.12 M
In text books it is found to be: 3.5 x 10-8
Checking:
1. For [H3O+]from water :
1 x 10-7 M
x 100 = 0.155%
6.46 x 10-5 M
assumptions are OK
2. For [HClO]dissoc :
6.46 x 10-5 M
0.12 M
x 100 = 0.0538 %
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Overview: Solving Weak Acid Equilibrium Problems
• List the major species in the solution.
• Choose the species that can produce H+ and write balanced equations for
the reactions producing H+.
• Comparing the values of the equilibrium constants for the reactions you
have written, decide which reaction will dominate in the production of H+.
• Write the equilibrium expression for the dominant reaction.
• List the initial concentrations of the species participating in the dominate
reaction.
• Define the change needed to achieve equilibrium; that is, define x.
• Write the equilibrium concentrations in terms of x.
• Substitute the equilibrium concentrations into the equilibrium expression.
• Solve for x the “easy” way… assume that [HA]0 – x ≈ [HA]0
• Verify the approximation is valid (5% rule).
• Calculate [H+] and pH.
Mixtures of Several Acids - I
Calculate the pH of a solution that contains 1.00 M HF
(Ka = 7.2 x 10-4) and 5.00 M HOCl (Ka = 3.5 x 10-8). Also calculate the
concentrations of the F- and OCl- ions at equilibrium.
Three components produce H+:
HF(aq)
⇌
H+(aq) + F-(aq)
Ka = 7.2 x 10-4
HOCl(aq)
⇌
H+(aq) + OCl-(aq)
Ka = 3.5 x 10-8
H2O(aq)
⇌
H+(aq) + OH-(aq)
Ka = 1.0 x 10-14
Even though HF is a weak acid, it has by far the greatest Ka, so it will be
the dominant producer of H+. Solve for its I.C.E. table first, then use those
concentrations in the second most important equilibrium. Move down the
list, in each case setting
Ka =
[H+] [A-]
[HA]
= specific value
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Mixtures of Several Acids - II
Initial Concentration
(mol/L)
Equilibrium Concentration
(mol/L)
[HF]0 = 1.00
[F-] = 0
[H+] = ~ 0
Ka =
x mol HF
dissociates
[H+] [F-]
[HF]
= 7.2 x 10-4 =
[HF] = 1.00 – x
[F-] = x
[H+] = x
(x) (x)
1.00-x
≈
x2
1.00
x = 2.7 x 10-2
Therefore,
[F- ] = [H+] = x = 2.7 x 10-2
and pH = 1.57
Mixtures of Several Acids - III
The concentration of H+ comes from the first part of this problem:
Initial Concentration
(mol/L)
[HOCl]0 = 5.00
[OCl-] = 0
[H+] = 2.7 x 10-2
Ka =
Equilibrium Concentration
(mol/L)
x mol HOCl
dissociates
[HOCl] = 5.00 – x
[OCl-] = x
[H+] = 2.7 x 10-2 + x
[H+] [OCl-]
(2.7 x 10-2 + x) (x)
= 3.5 x 10-8 =
[HOCl]
5.00-x
≈
(2.7 x 10-2) x
5.00
x = 6.5 x 10-6 M = [OCl-]
Therefore,
pH = 1.56
[F- ] = 2.7 x 10-2 M
[OCl-] = 6.5 x 10-6 M
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11/20/2011
PERCENT DISSOCIATION
% dissociation =
amount dissociated (mol/L) x 100
initial concentration (mol/L)
x
% dissociation = [HA] x 100
o
•for a given acid, % dissociation increases as the acid is diluted
•for a weak acid, [H+] decreases with increasing dilution as [HA]
does, but % dissociation increases as dilution increases
Figure 7.5: Effect of dilution on the percent
dissociation and [H+]
Ka =
[H+] [A-]
[HA]
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Problem: Calculate the percent dissociation of a 1.00 M
hydrocyanic acid solution, Ka = 6.20 x 10-10
⇌
HCN(aq) + H2O(l)
HCN
Initial 1.00 M
Change
-x
Final
1.00 –x
H 3O +
0
+x
x
H3O+(aq) + CN- (aq)
CN0
+x
x
[H3O+][CN-]
[HCN]
Ka =
Ka =
(x)(x)
= 6.20 x 10-10
(1.00-x)
Assume 1.00-x ≈ 1.00
Ka =
x2
1.00
= 6.2 x 10-10
x = 2.49 x 10-5
% dissociation =
2.49 x 10-5
x 100 = 0.00249%
1.00
Weak Bases
• Many compounds with an electron-rich nitrogen are
weak bases.
•The common structural feature is an N atom that
has a lone electron pair in its Lewis structure.
EXAMPLES:
1) Ammonia (:NH3)
2) Amines (general formula RNH2, R2NH, R3N)
..
N
..
N
H3C H CH3
H3C CH CH3
3
Dimethylamine
Trimethylamine
..
N
H3C H H
Methylamine
..
N
H H C2H5
Ethylamine
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General reaction (R is a hydrocarbon group or a H):
NR3(aq) + H2O(l) ⇌ HNR3+(aq) + OH-(aq)
+
Kb = [HNR3 ] [OH ]
[NR3]
Determining pH from Kb and Initial [B]–I
Problem: Ammonia is commonly used cleaning agent and is a weak
base, with a Kb of 1.8 x 10-5. What is the pH of a 1.5 M NH3 solution?
Plan: Ammonia reacts with water to form [OH-]. Calculate [H3O+] and the
pH. The balanced equation and Kb expression are:
NH4+(aq) + OH-(aq)
NH3 (aq) + H2O(l)
Kb =
Concentration (M)
Initial
Change
Equilibrium
[NH4+] [OH-]
[NH3]
NH3
H2O
NH4+
OH-
1.5
-x
1.5 - x
----------
0
+x
x
0
+x
x
making the assumption: since Kb is small, (1.5 M – x) ≈ 1.5 M
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Determining pH from Kb and Initial [B]–II
Substituting into the Kb expression and solving for x:
Kb =
[NH4+] [OH-]
=
[NH3]
(x)(x)
1.5
= 1.8 x 10-5
x2 = 2.7 x 10-5
x = 5.20 x 10-3 = [OH-] = [NH4+]
Calculating pH:
[H3O+] =
Kw
=
[OH-]
1.0 x 10-14
5.20 x 10-3
= 1.92 x 10-12
pH = -log[H3O+] = - log (1.92 x 10-12)
pH = 11.72
Polyprotic Acids
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Calculate the pH of 3.00 x 10-3 M Sulfuric Acid
Polyprotic acid, more than one proton to lose!!! Multiple Ka’s
H2SO4 (aq)
⇌ HSO4-(aq) + H+(aq)
Ka1 = LARGE
HSO4-(aq) ⇌ SO4-(aq) + H+(aq)
Ka2 = 1.2 x 10-2
Because first dissociation has a very large Ka it can be treated
as a strong acid (~100% dissociated).
Calculate the pH of 3.00 x 10-3 M Sulfuric Acid
HSO4-(aq) ⇌ SO4-(aq) + H+(aq)
Initial Concentration (mol/L)
[HSO4-]0 = 0.00300
[SO42-]0 = 0
[H+]0= 0.00300
Ka = 1.2x10-2
Equilibrium Concentration (mol/L)
[HSO4-] = 0.00300 – x
[SO42-] = x
[H+] = 0.00300 + x
x mol/L HSO4dissociates
to reach
equilibrium
From dissociation of H2SO4
[H+][SO42-]
(0.00300 + x)(x)
=
[HSO4 ]
(0.00300 – x)
No obvious approximation looks valid, so we must solve with the
quadratic formula.
Ka2 = 1.2 x 10-2 =
0 = x2 + 0.015x – 3.6 x 10-5
x=
x = 2.10 x 10-3
[H+] = 0.00300 + x = 0.00510
-b +-
b2 – 4ac
2a
a=1
b = 0.015
c = -3.6 x 10-5
pH = 2.29
Note: Change in [H+] does not violate huge Ka for 1st dissociation: [H2SO4] = 0.
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SALTS and their pH
sodium acetate (CH3COONa) solution is basic
CH3COO- + H2O ⇌ CH3COOH + OHammonium chloride (NH4Cl) solution is acidic
NH4+ + H2O ⇌
NH3 + H3O+
sodium chloride solution is neutral
NaCl + H2O ⇌ Na+ + Cl- + H2O
Effects of Salts on pH and Acidity
Salts that consist of cations of strong bases and the anions of strong
acids have no effect on the [H+] when dissolved in water.
Examples: NaCl, KNO3, Na2SO4, NaClO4, KBr, etc.
A salt whose cation alone has neutral properties (such as Na+ or K+)
and whose anion is the conjugate base of a weak acid, produces a
basic solution in water, since anion captures some H+.
Examples: NaF, KCN, NaC2H3O2, Na3PO4, Na2CO3, K2S, Na2C2O4, etc.
A salt whose anion alone has neutral properties and whose cation is
the conjugate acid of a weak base OR a small and highly charged
metal ion will produce an acidic solution in water.
Examples: NH4Cl, AlCl3, Fe(NO3)3, etc.
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See
Table 7.6
Like Example 7.12 - I
Calculate the pH of a 0.45 M NaCN solution. The Ka value for HCN is
6.2 x 10-10.
Solution: Since HCN is a weak acid, the cyanide ion must
have significant affinity for protons.
CN-(aq) +
H2O(l)
Kb = [HCN] [OH ]
[CN ]
Kb =
Kw
=
Ka (for HCN)
Initial Concentration (mol/L)
[CN-]0 = 0.45
[HCN]0 = 0
[OH-]0 = 0
⇌
HCN(aq) +
OH-(aq)
The value of Kb can be calculated
from Kw and the Ka value for HCN.
1.0 x 10-14 = 1.61 x 10-5
6.2 x 10-10
Equilibrium Concentration (mol/L)
X mol/L CN- reacts with
H2O to reach equilibrium
[CN-] = 0.45 – x
[HCN] = x
[OH -] = x
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Like Example 7.12 - II
Thus:
Kb = 1.61 x 10-5 =
[HCN] [OH-]
=
[CN-]
(x)(x)
0.45 - x
The approximation is not valid by the 5% rule, so
you have to use the quadratic equation.
x = 2.68 x 10-3
x = [OH-] = 2.68 x 10-3 M
pOH = -log [OH-] = 2.57
pH = 14.00 – pOH = 14.00 - 2.57 = 11.43
Small and highly charged cations produce acidic solutions in water.
Examples: Al3+, Fe3+, etc..
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Predicting the Relative Acidity of Salt Solutions
Problem: Determine whether an aqueous solution of iron(III) nitrite,
Fe(NO2)3, is acidic, basic, or neutral.
Plan: The formula consists of the small, highly charged, and therefore
weakly acidic, Fe3+ cation and the weakly basic NO2- (anion of the
weak acid HNO2). To determine the relative acidity of the solution, we
write equations that show the reactions of the ions with water, and then
find Ka and Kb of the ions to see which ion reacts to form H+ or OH- to a
greater extent.
Solution: Writing the reactions with water:
Fe(H2O)63+(aq) + H2O(l)
Fe(H2O)5OH2+(aq) + H3O+(aq)
NO2-(aq) + H2O(l)
HNO2(aq) + OH -(aq)
Obtaining Ka and Kb of the ions: for Fe3+(aq) Ka = 6 x 10-3
for NO2-(aq), Kb must be determined:
Kw
1.0 x 10-14
Kb of NO2- =
=
= 2.5 x 10-11
Ka of HNO2 4.1 x 10-4
Since Ka of Fe3+ > Kb of NO2-, the solution is acidic.
Summary: Solving Acid-Base Equilibria Problems
• List the major species in solution.
• Look for reactions that can be assumed to go to completion, such as a strong
acids/bases dissociating or H+ reacting with OH-.
• For a reaction that can be assumed to go to completion:
a) Determine the concentrations of the products.
b) Write down the major species in solution after the reaction.
• Look at each major component of the solution and decide whether it is an acid
or a base.
• Pick the equilibrium that will control the pH. Use known values of the
dissociation constants for the various species to determine the dominant
equilibrium.
a) Write the equation for the reaction and the equilibrium expression.
b) Set up the I.C.E. table and define x.
c) Compute the equilibrium concentrations in terms of x.
d) Substitute the [ ]eq into the equilibrium expression and solve for x.
e) Check the validity of the approximation.
f) Calculate the pH and other concentrations as required.
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Summary: General Strategies for Solving Acid-Base Problems
Think Chemistry. Focus on the solution components and their reactions. It
will almost always be possible to choose one reaction that is the most
important.
Be systematic. Write down all the things you know, including things like the
equilibrium expression, list what you can calculate, and list what’s needed.
Be flexible. Although all acid-base problems are similar in many ways,
important differences do occur. Treat each problem as a separate entity. Do
not try to force a given problem to match any you have solved before. Look
for both the similarities and the differences.
Be patient. The complete solution to a complicated problem cannot be seen
immediately in all its detail. Pick the problem apart into its workable steps.
Be confident. Look within the problem for the solution, and let the problem
guide you. Assume that you can think it out. Do not rely on memorizing
solutions to problems. In fact, memorizing solutions is usually detrimental
because you tend to try to force a new problem to be the same as one you
have seen before.
Understand and think - don’t just memorize.
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