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Chapter 7 Acids and Bases
7.1
7.2
7.3
7.4
7.5
7.6
7.7
7.8
7.9
The Nature of Acids and Bases
Acid Strength
The pH Scale
Calculating the pH of Strong Acid Solutions
Calculating the pH of weak Acid Solutions
Bases
Polyprotic Acids
Acid-Base Properties of Salts
Acid Solutions in Which Water Contributes to the
H+ Concentration
7.10 Strong Acid Solutions in Which Water Contributes
to the H+ Concentration
7.11 Strategy for solving Acid-Base Problems: A Summary
From Chapter 4: Acids and Bases
An Acid is a substance that produces H+ (H3O+) ions when dissolved
in water, and is a proton donor
A Base is a substance that produces OH - ions when dissolved in water:
Example: NaOH(aq)  Na+(aq) + OH-(aq)
The OH- ions react with the H+ ions (if an acid is present) to produce
water, H2O, and are therefore proton acceptors.
Acids and Bases are electrolytes. Their strength is categorized in
terms of their degree of dissociation in water to make hydronium or
hydroxide ions. Strong acids and bases dissociate completely, and
are strong electrolytes. Weak acids dissociate partially (some small
% of the molecules dissociate) and are weak electrolytes.
Selected Acids and Bases
Acids
Bases
H+(aq)
A-(aq)
Strong:
+
Hydrochloric, HCl
Hydrobromic, HBr
Hydroiodoic, HI
Nitric acid, HNO3
Sulfuric acid, H2SO4
Perchloric acid, HClO4
Strong: M+(aq) + OH-(aq)
Sodium hydroxide, NaOH
Potassium hydroxide, KOH
Calcium hydroxide, Ca(OH)2
Strontium hydroxide, Sr(OH)2
Barium hydroxide, Ba(OH)2
Weak
Hydrofluoric, HF
Phosphoric acid, H3PO4
Acetic acid, CH3COOH
(or HC2H3O2)
Weak
Ammonia, NH3
accepts proton from water to make
NH4+(aq) and OH-(aq)
Arrhenius (or Classical) Acid-Base Definition
An acid is a substance that contains hydrogen and
dissociates in water to yield a hydronium ion : H3O+
A base is a substance that contains the hydroxyl group
and dissociates in water to yield : OH Neutralization is the reaction of an H+ (H3O+) ion from the
acid and the OH - ion from the base to form water, H2O.
The neutralization reaction is exothermic and releases
approximately 56 kJ per mole as written:
H+(aq) + OH-(aq)
→
H2O(l) + 56 kJ / mol
Brønsted-Lowry Acid-Base Definition
An acid is a proton donor, any species that donates an H+ ion.
An acid must contain H in its formula; HNO3 and H2PO4- are two
examples, all Arrhenius acids are Brønsted-Lowry acids.
A base is a proton acceptor, any species that accepts an H+ ion.
A base must like to bind the H+ ion;
a few examples are NH3, CO32-, F -, as well as OH -.
Brønsted-Lowry bases are not Arrhenius bases, but all Arrhenius
bases produce the Brønsted-Lowry base OH-.
Therefore in the Brønsted-Lowry perspective, one species donates a
proton and another species accepts it: an acid-base reaction is a
proton transfer process.
Acids donate a proton to water
Bases accept a proton from water
Strong Acids and the Molarity of H+ Ions in
Aqueous Solutions of Acids
Problem: In aqueous solutions, each molecule of sulfuric acid will
loose two protons to yield two H+ ions, and one sulfate ion.
What is the molarity of the sulfate and Hydronium ions in a solution
prepared by dissolving 155g of concentrate sulfuric acid into sufficient
water to produce 2.30 Liters of acid solution?
Plan:
Construct Equation
Calculate moles of
Sulfuric Acid
Divide moles by
volume to get
concentration of
sulfate
Use equation to
calculate H+
concentration
Strong Acids and the Molarity of H+ Ions in
Aqueous Solutions of Acids
Solution: Two moles of H+ are released for every mole of acid:
2 H3O+(aq) + SO4- 2(aq)
H2SO4 (l) + 2 H2O(l)
Moles H2SO4 = 155 g H2SO4 x
Molarity of SO4- 2 =
1 mole H2SO4
= 1.58 moles H2SO4
98.09 g H2SO4
1.58 mol SO4-2
= 0.687 Molar in SO4- 2
2.30 L solution
Molarity of H+ = 2 x 0.687 M = 1.37 Molar in H+ (or H3O+)
The Acid-Dissociation Constant (Ka)
Strong acids dissociate completely into ions in water:
HA(g or l) + H2O(l)
H3O+(aq) + A-(aq)
In a dilute solution of a strong acid, almost no HA molecules
exist: [H3O+] = [HA]init or [HA]eq = 0
+
Ka = K = [H3O ][A ] at equilibrium, and Ka >> 1
[HA][H2O]
Nitric acid is an example: HNO3 (l) + H2O(l)
H3O+(aq) + NO3-(aq)
Weak acids dissociate very slightly into ions in water:
HA(aq) + H2O(aq)
H3O+(aq) + A-(aq)
In a dilute solution of a weak acid, the great majority of HA
molecules are not dissociated: [H3O+] << [HA]init or [HA]eq = [HA]init
Ka = K =
[H3O+][A-]
[HA][H2O]
at equilibrium, and Ka << 1
The Meaning of Ka, the Acid Dissociation Constant
For the ionization of an acid, HA:
HA(aq) + H2O(l)
O+]
[A-]
Ka = [H3
[HA] [H2O]
H3O+(aq) + A-(aq)
Since the liquid water is nearly pure (>98%) in all the
aqueous solutions we will discuss, we will treat it like
a pure liquid, so [H2O] = 1.0 in all K expressions.
Therefore:
Ka =
[H3O+] [A-]
[HA]
Stronger acid
The stronger the acid, the higher the [H3O+]
at equilibrium, and the larger the Ka:
higher [H3O+]
larger Ka
•For a weak acid with a relative high Ka
a 1 M solution
has ~10% of the HA molecules dissociated.
•For a weak acid with a moderate Ka (~10-5 ), a 1 M solution has ~
0.3% of the HA molecules dissociated.
•For a weak acid with a relatively low Ka (~10-10 ), a 1 M solution
has ~ 0.001% of the HA molecules dissociated.
(~10-2 ),
The Extent
of
Dissociation
for Strong
and Weak
Acids
pH
What is pH?
How is it important?
The definition of pH
pH = -log10[H3O+] = -log10[H+]
Significant Figures for pH
The number of decimal places in the log is
equal to the number of significant figures in
the original number.
Calculating pH from its definition
pH = - log[H+]
What is the pH of a solution that is 10-12 M in hydronium ion ?
pH = -log[H3O+] = -log 10-12 = -(-12.0) = +12.0
What is the pH of a solution that is 7.3 x 10-9 M in H+ ?
pH = -log(7.3 x 10-9) = -(-8.14) = +8.14
What is the pH of a 10-1 M HNO3(aq) solution ?
Since it’s a strong acid: [H3O+] = [HNO3]initial = 0.1
pH = -log[H3O+] = -log 10-1 = -(-1.0) = +1.0
pH in the body
Body Section
pH
Stomach Acid
0.7
Urine
6.0
Brain Fluid
7.3
Blood
7.45
Pancreas
8.1
Mitochondria
7.5
Lyosomes
4.5
The human body is constantly balancing pH, incorrect
balancing and manipulation of these pH’s can lead to
acidosis and alkalosis.
Figure 7.3: The pH
scale and pH values
of some common
substances
pH of a neutral solution = 7.00
pH of an acidic solution < 7.00
pH of a basic solution > 7.00
Water itself is a weak acid and weak base:
Two water molecules react to form H3O+ and OH(but only slightly)
K = Ka = Kw = 1.0 x 10-14
Autoionization of Water
H2O(l) + H2O(l)
Ka =
[H3O+][OH-]
[H2O]2
H3O+(aq) + OH-(aq)
= [H+][OH-] = Kw
The ion-product for water, Kw:
Set = 1.0 since nearly pure liquid
water even when salts, acids or
bases present.
Ka = Kw = [H3O+][OH-] = 1.0 x 10-14 (at 25°C)
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 25°C)
The molarity of pure water is:
1000 g/L
18.02 g/mol
= 55.5 M
Two New Definitions
Remember: Kw = [H+][OH-] so that [OH-] = Kw / [H+]
1)
pOH = -log10[OH-]
= - log10(Kw/[H+]) = - log10Kw - log10[H+])
= - log10 (1.0 x 10-14) – pH = 14.00 - pH
pOH = 14.00 – pH
2) It is convenient to express equilibrium
constants for acid dissociation in the form
pKa = - log10Ka
The Relationship Between Ka and pKa
Acid Name (Formula)
Hydrogen sulfate ion (HSO4-)
Ka at 25oC
1.02 x 10-2
pKa
1.991
Nitrous acid (HNO2)
7.1 x 10-4
3.15
Acetic acid (CH3COOH)
1.8 x 10-5
4.74
Hypobromous acid (HBrO)
2.3 x 10-9
8.64
Phenol (C6H5OH)
1.0 x 10-10
10.00
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 25°C.
Plan:
HCl is Strong Acid
Complete dissociation:
[H+] = initial concentration
Use [H+] to calculate pH, OHand pOH
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 25°C.
Solution:
(a) [H+] = 3.0 M
[OH-] =
Kw
[H+]
pH = -log[H+] = -log(3.0) = -0.477
=
1 x 10-14
= 3.333 x 10-15 M
3.0
pOH = - log(3.333 x 10-15) = 14.523
(b) [H+] = 0.0024 M
pH = -log[H+] = -log(0.0024) = 2.62
Kw
1 x 10-14
[OH-] =
=
[H+]
0.0024
= 4.167 x 10-12 M
pOH = -log(4.167 x 10-12) = 11.38
Determining Concentrations from Ka and Initial [HA]
Problem: Hypochlorous acid is a weak acid formed in laundry bleach.
What is the [H+] 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-]
= 3.5 x 10-8
[HClO]
Concentration (M)
Initial
Change
Equilibrium
Ka =
HClO
H2O
0.125
-x
0.125 - x
----------
(x)(x)
= 3.5 x 10-8
0.125-x
H+
0
+x
x
ClO0
+x
x
Set 0.125 - x = 0.125 since x must be tiny.
x2 = 0.4375 x 10-8
One More Step With pH
• We calculated x which was equal to the [H+]
concentration.
• We can then calculate the pH by:
-log10(0.661x10-4)
pH = 4.180
x = 0.661 x 10-4
Finding the Ka of a Weak Acid from the pH
of its Solution–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, hence, 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 = x
H3O+(aq) + ClO -(aq)
Concentration (M) HClO(aq) + H2O(l)
Initial
Change
Equilibrium
0.12
-x
0.12 -x
Substituted
----------
0.12
0
+x
+x
0
+x
+x
6.46 x 10-5
6.46 x 10-5
Finding the Ka of a Weak Acid from the pH of
its Solution–II
HClO(aq) + H2O(l)
H3O+(aq) + ClO -(aq)
x = [H3O+] = 6.46 x 10-5 M = [ClO-]
Ka =
[H3O+] [ClO-]
=
[HClO]
Ka = 3.48 x 10-8
(6.46 x 10-5 M) (6.46 x 10-5 M)
= 3.48 x 10-8
0.12 M - 6.46 x 10-5 M
In text books it is found to be: 3.5 x 10-8
Solving Problems Involving Weak-Acid Equilibria–I
There are two general types of equilibrium problems involving weak
acids and their conjugate bases:
• Given equilibrium concentrations, find Ka.
• Given Ka and some concentration information, find the other equilibrium
concentrations.
The problem-solving approach.
•Write the balanced equation and Ka expression; these will tell you what to find.
•Define x as the unknown concentration that changes during the reaction. Frequently,
x = [HA]dissoc., the concentration of HA that dissociates which, through the use of
certain assumptions, also equals [H3O+] and [A-] at equilibrium.
•Construct a reaction table that incorporates the unknown.
•Make assumptions that simplify the calculation, usually that x isvery small relative
to the initial concentration.
Solving Problems Involving Weak-Acid Equilibria–II
•Substitute the values into the Ka expression and solve for x.
•Check that the assumptions are justified.
•We normally apply the 5% rule; if the value of x is greater than 5% of the value
it is compared with, you must use the quadratic formula to find x.
•The notation system. Molar concentrations of species are indicated by using square
brackets around the species of interest. Brackets with no subscript refer to the molar
concentration of the species at equilibrium.
•The assumptions. The two common good assumptions to simplify the arithmetic
are:
•The [H3O+] from the autoionization of water is negligible. In fact, the presence
of acid from whatever is put into solution will hinder the autoionization of water,
and make it even less important.
•A weak acid has a small Ka. Therefore, it dissociates to such a small extent that
we can neglect the change in its concentration to find its equilibrium
concentration.
Calculate the pH of a 1.00 M HNO2 Solution
Problem: Calculate the pH of a 1.00 M Solution of Nitrous acid HNO2.
Ka = 4.0 x 10-4 M
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:
H+(aq) + NO2-(aq)
Ka = 4.0 x 10-4 (Table 7.2)
HNO2 (aq)
Initial concentrations = [H+] = 0 , [NO2-] = 0 , [HNO2] = 1.00 M
Final concentrations = [H+] = x , [NO2-] = x , [HNO2] = 1.00 M - x
[H+] [NO2-]
(x) (x)
= 4.0 x 10-4 =
[HNO2]
1.00 - x
Assume 1.00 – x = 1.00 since left side so tiny.
Ka =
x2
= 4.0 x 10-4 or x2 = 4.0 x 10-4
1.00
x = 2.0 x 10-2 = 0.02 M = [H+] = [NO2-]
pH = - log[H+] = - log(2.0 x 10-2) = 2.00 – 0.30 = 1.70
Molecular model: Nitrous acid
Figure 7.5: Effect of dilution on the percent
dissociation and [H+]
Ka =
[H+] [A-]
[HA]
Although Ka is independent of molarity, the
percent dissociation depends on molarity
fraction dissociated =
([H+]/[HNO2]0 ) ≈ Ka/[NO2-]
Problem: Calculate the Percent dissociation of a 1.00 M Hydrocyanic
acid solution, Ka = 6.20 x 10-10.
H3O+(aq) + CN- (aq)
HCN(aq) + H2O(l)
H3O+
CN-
0
+x
x
0
+x
x
HCN
Initial 1.00 M
Change
-x
Final
1.00 –x
Assume 0.0100-x = 0.0100
Ka =
(x)(x)
Ka = (1.00-x)
x2
1.00
Ka =
[H3O+][CN-]
[HCN]
= 6.20 x 10-10
= 6.2 x 10-10
x = 2.49 x 10-5
% dissociation =
2.49 x 10-5
1.00
x 100% = 0.00249%
Acid - Base Reactions : Neutralization Rxns.
The generalized reaction between an Acid and a Base is:
HA(aq) + MOH(aq)
Acid
+
MA(aq) + H2O(l)

Base
Salt
+ Water
The salt product can either be dissolved as ions or form a precipitate.
Writing Balanced Equations for
Neutralization Reactions - I
Problem: Write balanced chemical reactions (molecular, total ionic, and
net ionic) for the following Chemical reaction:
Calcium Hydroxide(aq) and Hydroiodic acid(aq)
Plan: Strong acid and base, therefore they will make water and the
corresponding salts.
Solution:
Ca(OH)2 (aq) + 2HI(aq)
Ca2+
(aq)
+2
OH -(aq)
+2
H+
(aq)
+2I
CaI2 (aq) + 2H2O(l)
(aq)
2 OH -(aq) + 2 H+(aq)
Ca2+(aq) + 2 I -(aq) + 2 H2O(l)
2 H2O(l)
Molecular Model
The reaction of an acid
HA with water to form H3O+ and a conjugate base
A
A
Acid
Base
Conjugate
acid
Conjugate
base
The Conjugate Pairs in Some Acid-Base Reactions
Conjugate Pair
Acid
+
Base
Conjugate Base + Conjugate Acid
Conjugate Pair
Reaction 1
Reaction 2

HF

+
H2O
HCOOH +
CN–



Reaction 3
NH4+
+
CO3
Reaction 4
H2PO4–
+
OH–
Reaction 5
H2SO4


+

2–


N2H5+
F–


H3O+
+
HCOO–
NH3


+
+
HCN
HCO3–
HPO42– +
H2O
HSO4–
N2H62+

+





Identifying Conjugate Acid-Base Pairs
Problem: The following chemical reactions are important for industrial
processes. Identify the conjugate acid-base pairs.
SO42-(aq) + HCN(aq)
(a) HSO4-(aq) + CN-(aq)
(b) ClO-(aq) + H2O(l)
HClO(aq) + OH-(aq)
(c) S2-(aq) + H2O(aq)
HS-(aq) + OH-(aq)
Plan: To find the conjugate acid-base pairs, we find the species that
donate H+ and those that accept it. The acid (or base) on the left
becomes its conjugate base (or acid) on the right.
Solution:
(a) The proton is transferred from the sulfate to the cyanide so:
HSO4-(aq)/SO42-(aq) and CN-(aq)/HCN(aq ) are the two acid-base pairs.
(b) The water gives up one proton to the hypochlorite anion so:
ClO-(aq)/HClO(aq) and H2O(l) / OH-(aq ) are the two acid-base pairs.
(c) One of water’s protons is transferred to the sulfide ion so:
S2-(aq)/HS-(aq) and H2O(l)/OH-(aq) are the two acid-base pairs.
BASES
Classifying the Relative Strengths of Bases
Strong bases
Soluble compounds containing O2- or OH- ions are strong bases. The cations
are usually those of the most active metals:
1) M2O or MOH, where M= Group 1A(1) metals (Li, Na, K, Rb, Cs)
2) MO or M(OH)2, where M = Group 2A(2) metals (Ca, Sr, Ba)
[MgO and Mg(OH)2 are only slightly soluble, but the soluble
portion dissociates completely.]
Weak bases
Many compounds with an electron-rich nitrogen are weak bases (none are
Arrhenius bases). The common structural feature is an N atom bonded to H
and C atoms.
1) Ammonia (:NH3)
2) Amines (general formula RNH2, R2NH, R3N), such as
CH3CH2NH2, (CH3)2NH, (C3H7)3N, and C5H5N
Molecular Model
Na+ and OH- and H2O
pH of a Base Calculation
Calculate the pH of a 2.0 x 10-3 M solution of NaOH.
Since NaOH is a strong base, it will dissociate 100% in water.
NaOH(aq) →
Na+(aq) + OH-(aq)
Since [NaOH] = 2.0 x 10-3 M , [OH-] = 2.0 x 10-3 M
The concentration of [H+] can be calculated from Kw:
[H+] =
Kw
=
[OH-]
1.0 x 10-14
2.0 x 10-3
= 5.0 x 10-12 M
pH = - log [H+] = - log( 5.0 x 10-12) = 11.30
The Meaning of Kb, the Base Association Constant
For the generalized reaction between a base, B(aq), and water:
B(aq) + H2O(l)
Equilibrium constant:
BH+(aq) + OH-(aq)
BH OH 

K  Kb 
B
The stronger the base, the higher the [OH-] at
equilibrium, and the larger the Kb
Determining pH from Kb and Initial [B] – I
Problem: Ammonia is commonly used cleaning agent in households
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-] and then calculate
[H3O+] and the pH. The balanced equation and Kb expression are:
NH4+(aq) + OH-(aq)
NH3 (aq) + H2O(l)
Kb =
+]
[NH4
[OH-]
[NH3]
Concentration (M)
NH3
Initial
Change
Equilibrium
H2O
1.5
-x
1.5 - x
----------
NH4+
OH-
0
+x
x
0
+x
x
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-x
= 1.8 x 10-5
x2 = 2.7 x 10-5 = 27 x 10-6
x = 5.20 x 10-3 = [OH-] = [NH4+]
Calculating pH:
pOH = -log[OH-] = - log (5.20 x 10-3) = 2.28
pH = 14.00 – pOH = 14.00 – 2.28
pH = 11.72
The Relation Between Ka and Kb of a Conjugate
Acid-Base Pair
Acid
HA + H2O
H3O+ + A-
Base
A-
+ H2O
HA + OH-
2 H2O
H3O+ + OH-
[H3O+] [A-]
x
[HA] [OH-]
[A-]
[HA]
Ka
x
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
Molecular Model
HCN, HNO2, and H2O
Summary: 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-that is, by assuming that [HA]0 – x = [HA]0
•Verify whether the approximation is valid ( the 5% rule is the test in this
case).
•Calculate [H+] and pH.
Mixtures of Several Acids
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)
HOCl(aq)
H+
H2O(aq)
H+(aq) + OH-(aq)
(aq)
+
OCl-(aq)
Ka = 7.2 x 10-4
Ka = 3.5 x 10-8
Ka = 1.0 x 10-14
Even though HF is a weak acid, it has by far the greatest Ka,
therefore it will be the dominant producer of H+.
Solve for its I.C.E. table first, and then use those concentrations in next most
important equilibrium. Move down the list, in each case setting
Ka =
[H+] [A-]
= specific value
[HA]
Mixtures of Several Acids
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+
Ka = 1.0 x 10-14
(aq)
+
OH-(aq)
Even though HF is a weak acid, it has by far the greatest Ka,
therefore it will be the dominant producer of H+.
Ka =
[H+] [F-]
= 7.2 x 10-4
[HF]
Mixtures of Several Acids
Initial Concentration
(mol/L)
[HF]0 = 1.00
[F-] = 0
[H+] = ~ 0
Ka =
Equilibrium Concentration
(mol/L)
x mol HF
dissociates
[HF] = 1.00 – x
[F-] = x
[H+] = x
[H+] [F-]
(x) (x)
= 7.2 x 10-4 =
[HF]
1.00-x
=
x2
1.00
x = 2.7 x 10-2
Therefore,
[F- ] = [H+] = x = 2.7 x 10-2 and pH = 1.56
Mixtures of Several Acids
Ka =
[H+] [OCl-]
= 3.5 x 10-8
[HOCl]
The concentration of H+ comes
from the first part of this
problem = 2.7 x 10-2 M
[HOCl]initial = 5.00 M ; [OCl-]eqbm = x
3.5 x 10-8 =
x
=
(2.7 x 10-2)( x)
(5.00 - x)
5.00 ( 3.5 x 10-8)
2.7 x 10-2
[HOCl]eqbm = 5.00 - x
Left is tiny so x must be tiny.
Assume: 5.00 – x = 5.00
= 6.48 x 10-6 M = [OCl-]
pH = -log [H+] = -log [2.7x10-2] = 1.56
[F-] = 2.7 x 10-2 M ; [OCl-] = 6.48 x 10-6 M
Polyprotic Acids
Molecular Model
H2SO4 :
Strong acid
so Ka is huge
(100% dissociated)
Weak acid
Ka = 1.2x10-2
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)
HSO4-(aq) + H+(aq)
SO4-(aq) + H+(aq)
Ka1 = LARGE
Ka2 = 1.2x10-2
Because first dissociation has a very large Ka it
can be treated as a strong acid.
Calculate the pH of 3.00 x 10-3 M Sulfuric Acid
SO4-(aq) + H+(aq)
HSO4-(aq)
Initial Concentration (mol/L)
[HSO4-]0 = 0.00300
[SO42-]0 = 0
[H+]0= 0.00300
Ka = 1.2x10-2
Equilibrium Concentration (mol/L)
x mol/L HSO4dissociates
to reach
equilibrium
[HSO4-] = 0.00300 – x
[SO42-] = x
[H+] = 0.00300 + x
From dissociation of H2SO4
Ka2 = 1.2 x 10-2 =
[H+][SO42-]
=
[HSO4-]
(0.00300 + x)(x)
(0.00300 – x)
No obvious approximation looks valid, and so we must solve with
the quadratic formula. Multiplying the expression out we get:
0 = x2 + 0.015x – 3.6 x 10-5
a = 1, b = 0.015
c = -3.6 x 10-5
x=
-b +-
b2 – 4ac
2a
[H+] = 0.00300 + x = 0.00510
x = 2.10 x 10-3
pH = 2.29
Note: Change in [H+] does not violate huge Ka for 1st dissociation: [H2SO4] = 0 still.
The Stepwise Dissociation of Phosphoric Acid
Phosphoric acid is a weak acid, and normally only looses one proton
in solution, but it will lose all three when reacted with a strong base
with heat. The equilibrium constants are reflected by arrow sizes.
H3PO4 (aq) + H2O(l)
H2PO4-(aq) + H3O+(aq)
Ka = 7.5x10-3
H2PO4-(aq) + H2O(l)
HPO42-(aq) + H3O+(aq)
Ka = 6.2x10-8
HPO42-(aq) + H2O(l)
PO43-(aq) + H3O+(aq)
Ka = 4.8x10-13
Net: H3PO4 (aq) + 3 H2O(l)
PO43-(aq) + 3 H3O+(aq)
Molecular Model
H3PO4 and H2O
The Stepwise Dissociation of Phosphoric Acid
Calculating ions concentrations in a certain molarity of phosphoric acid
H3PO4 (aq) + H2O(l)
H2PO4-(aq) + H3O+(aq)
Ka = 7.5x10-3
H2PO4-(aq) + H2O(l)
HPO42-(aq) + H3O+(aq)
Ka = 6.2x10-8
HPO42-(aq) + H2O(l)
PO43-(aq) + H3O+(aq)
Ka = 4.8x10-13
Net: H3PO4 (aq) + 3 H2O(l)
PO43-(aq) + 3 H3O+(aq)
Order from biggest Ka to smallest:
Just like in the case with several different acids, each Ka is so much smaller
then the one above it that its contribution can be neglected wrt changing any
concentrations of previous products!!
Like Example 7.8 (P257)-I
Calculate the pH of a 5.0 M H3PO4 solution and determine
equilibrium concentrations of the species:
H3PO4 , H2PO4-, HPO4-2, and PO4-3.
Solution:
H3PO4 (aq)
Ka = 7.5 x 10-3 =
Initial Concentration (mol/L)
[H3PO4]0 = 5.0
[H2PO4-]0 = 0
[H+]0 = 0
Ka1 = 7.5 x 10-3 =
H+(aq) + H2PO4-(aq)
[H+] [H2PO4-]
[H3PO4]
Equilibrium Concentration (mol/L)
[H3PO4] = 5.0 - x
[H2PO4-] =x
[H+] = x
[H+] [H2PO4-]
[H3PO4] =
(x)(x)
= ~
5.0-x
x2
5.0
Like Example 7.8 (P257) - II
Solving for x 1.9 x 10-1 Since 1.9 x 10-1 is less than 5% of 5.0, the
approximation is acceptable (but you could also solve w/ calc.) and:
[H+] = x = 0.19 M = [H2PO4-] ,
pH = 0.72 [H3PO4] = 5.0 – x = 4.8 M
The concentration of HPO42- can be obtained from Ka2:
Ka2 = 6.2 x 10-8 =
where:
[H+] [HPO42-]
[H2PO4-]
[H+] = [H2PO4-] = 0.19 M ; [HPO42-] = Ka2 = 6.2 x 10-8 M
To calculate [PO43-], we use the expression for Ka3 , and the
Values obtained from the other calculations:
[H+] [PO43-]
0.19[PO43-]
Ka3 = 4.8 x 10-13 =
=
[HPO42-]
6.2 x 10-8
(4.8 x 10-13)(6.2 x 10-8)
3[PO4 ] =
= 1.6 x 10-19 M
0.19
A plot of the effect of pH on the fractions of
H2CO3, HCO-3 and CO32-
You can control the
equilibrium by
adding a base,
consuming H+!
The Conjugate Pair for Acids and Bases
Conjugate Pair
Acid
+
Base
Conjugate Base + Conjugate Acid
Conjugate Pair
Reaction 1

HF

+
H2O
Reaction 2
HCOOH +
CN–
Reaction 3
+
NH4
+
CO3
Reaction 4
H2PO4–
+
OH–
Reaction 5
H2SO4



+


2–


N2H5+
F–


H3O+
+
HCOO–
NH3
+


+
HCN
H2O
HSO4–
N2H62+
+

HCO3–
HPO42– +





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.
For any 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 cation is the conjugate acid of a weak base, or small
and highly charged, produces an acidic solution in water.
Examples: NH4Cl, AlCl3, Fe(NO3)3, etc.
Molecular Model
Na+, C2H3O2-, and H2O
The Relation Between Ka and Kb of a Conjugate
Acid-Base Pair
Acid
HA + H2O
H3O+ + A-
Base
A- + H2O
HA + OH-
2 H2O
H3O+ + OH-
[H3O+] [A-]
[HA] [OH-]
x
Ka
x
= [H3O+] [OH-]
[A-]
[HA]
Kb
=
Kw
and for A- =NO2- : Kb = 2.2 x 10-11
HA = HNO2 : Ka = 4.5 x 10-4
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
Like Example 7.12 (P264) - I
Calculate the pH of a 0.45M 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)
HCN(aq) + OH-(aq)
[OH-]
[HCN]
The value of Kb can be calculated
[CN-]
from Kw and the Ka value for HCN.
1.0 x 10-14
Kw
Kb =
=
= 1.61 x 10-5
Ka (for HCN)
6.2 x 10-10
Kb =
Initial Concentration (mol/L)
[CN-]0 = 0.45
[HCN]0 = 0
[OH-]0 = 0
Equilibrium Concentration (mol/L)
X mol/L CN- reacts with
H2O to reach equilibrium
[CN-] = 0.45 – x
[HCN] = x
[OH-] = x
Like Example 7.12 (P264) - II
Thus:
Kb = 1.61 x 10-5 =
~ 2.69 x 10-3
x=
[HCN] [OH-]
=
[CN-]
(x)(x)
=
0.45 - x
~x2
0.45
Although this is not exactly valid by the
5% rule, it is only off by 1%, so we will
use it for now! (Or solve w/ calculator.)
x = [OH-] = 2.69 x 10-3 M
pOH = -log [OH-] = 2.57
pH = 14.00 – pOH = 14.00 - 2.57 = 11.43
Small or highly charged cations produce acidic solutions in water.
Examples: Al3+, Fe3+, etc..
Molecular Model
Cl-, Al(H2O)63+, H2O
Like Example 7.14 (P 267) - I
Calculate the pH of a 0.010 M AlCl3 solution.
The Ka value for the Al(H2O)63+ ion is 1.4 x 10-5.
Solution:
Since the Al(H2O)63+ ion is a stronger acid than water, the
Dominate equilibrium will be:
Al(H2O)63+(aq)
Al(OH)(H2O)52+(aq) + H+(aq)
[Al(OH)(H2O)52+][H+]
1.4 x 10-5 = Ka =
[Al(H2O)63+]
Initial Concentration (mol/L)
[Al(H2O)63+]0 = 0.010
[Al(OH)(H2O)52+] = 0
[H+]0 =~ 0
Equilibrium Concentration (mol/L)
x mol/L
3+
Al(H2O)6
dissociates
to reach
equilibrium
[Al(H2O)63+] = 0.010 – x
[Al(OH)(H2O)52+] = x
[H+] = x
Like Example 7.14 (P 267) - II
Thus:
[Al(OH)(H2O)52+] [H+]
= 1.4 x 10-5
[Al(H2O)63+]
Ka =
Ka =
(x) (x)
=
0.010 - x
x2
= 1.4x10-5
0.010
x = 3.7 x 10-4
Since the approximation is valid by the 5% rule:
[H+] = x = 3.7 x 10-4 M
and pH = 3.43
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.
For any 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 cation is the conjugate acid of a weak base, or small
and highly charged, produces an acidic solution in water.
Examples: NH4Cl, AlCl3, Fe(NO3)3, etc.
THE KEY to recognizing the acid/base properties of a salt:
The Conjugate Pair for Acids and Bases
Conjugate Pair
Acid
+
Base
Conjugate Base + Conjugate Acid
Conjugate Pair
Reaction 1
Reaction 2

HF

+
H2O
HCOOH +
CN–

+

Reaction 3
NH4
+
CO3
Reaction 4
H2PO4–
+
OH–
Reaction 5
H2SO4


+

2–


N2H5+
F–


H3O+
+
HCOO–
NH3


+
+
HCN
HCO3–
HPO42– +
H2O
HSO4–
N2H62+

+





The Relation Between Ka and Kb of a Conjugate
Acid-Base Pair
Acid
HA + H2O
H3O+ + A-
Base
A- + H2O
HA + OH-
2 H2O
H3O+ + OH-
[H3O+] [A-]
x
[HA] [OH-]
[A-]
[HA]
Ka
x
Kb
pKa + pKb
=
= [H3O+] [OH-]
Kw
= 14
= 10-14
Example 7.15 (P268): More complex salts
Predict whether an aqueous solution of each of the following
salts will be acidic, basic, or neutral.
b) NH4CN
c) Al2(SO4)3
a) NH4C2H3O2
Solution: get needed K from Ka = Kw/Kb, or Kb = Kw/Ka
a) The ions are the ammonium and acetate ions, Ka for NH4+ is
5.6 x 10-10, and Kb for C2H3O2- is 5.6 x 10-10 . Since the are equal
the solution will be neutral and the pH close to 7.
b) The solution will contain the ammonium and cyanide ions, the
Ka value for NH4+ is 5.6 x 10-10, and
Kw
= 1.6 x 10-5
Kb (for CN-) =
Ka(for HCN)
Since Kb for CN- is much larger than Ka for NH4+, this solution
will be basic.
c) This solution contains the hydrated Aluminum ion and the sulfate
ion. Ka for Al(H2O)63+ = 1.4 x 10-5, for sulfate, Kb = 8.3 x 10-13 ;
therefore this solution will be acidic.
General reaction:
NR3(aq) + H2O(l)
+]
Kb = [HNR3
[NR3]
HNR3+(aq) + OH-(aq)
[OH-]
Table 7.6 (P 268) - I
Acid – Base Properties of Aqueous Solutions of Various Types of Salts
Type of Salt
Cation is from
strong base;
anion is from
strong acid
Examples
KCl, KNO3
NaCl, NaNO3
Cation is from
strong base;
anion is from
weak acid
Anion acts as
a base; cation
has no effect
on pH
NaC2H3O2
KCN, NaF
Cation is conjugate
acid of weak base;
anion is from
strong acid
Comment
pH of Soln
Neither acts as
an acid or a
Neutral
base
Basic
Cation acts as
an acid; anion
has no effect
on pH
NH4Cl,
NH4NO3
Acidic
Table 7.6 (P 268) - II
Acid – Base Properties of Aqueous Solutions of Various Types of Salts
Type of Salt
Examples
Comment
Cation is conjugate
acid of weak base
anion is conjugate
base of weak acid
NH4C2H3O2
NH4CN
Cation is highly
charged metal ion;
anion is from
strong acid
Al(NO3)3,
FeCl3
pH of Soln
Cation acts as
an acid; anion
acts as a base
Acidic if :
K a > Kb
Basic if :
K b > Ka
Neutral if :
Ka = Kb
Hydrated cation
acts as an acid;
anion has no
effect on pH
Acidic
Like Example 7.3 (P 245)-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+:
H+(aq) + F-(aq)
HF(aq)
HOCl(aq)
H+
H2O(aq)
H+(aq) + OH-(aq)
(aq)
+
OCl-(aq)
Ka = 7.2 x 10-4
Ka = 3.5 x 10-8
Ka = 1.0 x 10-14
Even though HF is a weak acid, it has by far the greatest Ka,
therefore it will be the dominant producer of H+.
Ka =
[H+] [F-]
= 7.2 x 10-4
[HF]
Our assumptions earlier allowed us a short-cut solution to what was
really a complex system of n equations in n unknowns.
Very weak acids: water contributes to [H+]
- I
We start with the expression for the value of Ka , for the weak acid
HA:
From the conservation of charge equation:
Ka =
[H+] [A-]
[H+] = [A-] + [OH-]
[HA]
From the Kw expression for water:
K
[OH-] = [Hw+]
Substitute this into the charge balance equation:
[H+] = [A-] +
Kw
[H+]
or:
[A-] = [H+] -
[Kw]
[H+]
The material balance equation in A is:
[HA]0 = [HA] + [A-]
Since:
[A-] = [H+] -
or
Kw
[H+]
[HA] = [HA]0 – [A-]
We have:
K
[HA] = [HA]0 - ([H+] - w+ )
[H ]
Very weak acids - II
Now substitute these expressions for [A-] and [HA] into Ka:
[H+] [A-]
=
Ka =
[HA]
Ka =
[H+]( [H+] [HA]0 – ( [H+] -
[H+]2 - Kw
[HA]0 -
[H+]2 - Kw
Kw
)
[H+]
Kw
)
[H+]
=
Full Equation!
[H+]
When [H+]2 >> Kw (i.e., when [H+] > 10-6), this simplifies to:
[H+]2
Ka = [HA] – [H+] , which is the usual eqn.
0
Like Example 7.17 (P273) -I
Calculate the [H+] in: a) 1.0 M HOCl and b) 1 x 10-5 M HOCl
For hypochlorous acid HOCl , Ka = 3.5 x 10-8
a) First do the weak acid problem in the normal way.
~x2
x2
= 3.5 x 10-8 =
1.0 – x
1.0
x = 1.87 x 10-4 M = [H+]
b) First we do the weak acid problem in the normal way.
x2
x2
-8 ~
x = 5.92 x 10-7 M = [H+]
1.0 x 10-5 – x = 3.5 x 10 =1.0 x 10-5
In this very dilute solution of HOCl, [H+] < 10-6, so we need to use the full
equation to obtain the correct H+ concentration:
Ka = 3.5 x 10-8 =
[H+]2 – 10-14
[H+]2 – 10-14
1.0 x 10-5 [H+]
One equation in
one unknown, so
definitely soluble.
But hard to solve!
Use calculator!
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
acid 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) Compute the initial concentrations (assuming that the dominant equilibrium
has not yet occurred-for example, there has been no acid dissociation).
c) Define x.
d) Compute the equilibrium concentrations in terms of x.
e) Substitute the concentrations into the equilibrium expression, and solve for x.
f) Check the validity of the approximation.
g) Calculate the pH and other concentrations as required.
Summary:
(P 233)
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 eqbm.
eqbm. constant equation, list what you can calculate, and list what’
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.