Transcript Acid Base Notes

Acid and Base
Equilibrium
Chapter 16 Brown LeMay
Basic Concepts
Acids – sour or tart taste, electrolytes,
described by Arrhenius as sub that inc. the
H+ conc.
 Bases – bitter to the taste, slippery,
electrolytes, describes by Arrhenius as sub
that inc. the OH- conc.

Dissociation of Water
Pure water exists almost entirely of water
molecules. It is essentially a nonelectrolyte.
 Water ionizes to a small extent – autoionization
 The equilibrium expression is
H20(l) <-> H+(aq) + OH-(aq)
Kw = [H+] [OH-]

Since the H+ ion does
not exist alone in
water Kw is often
expressed
Kw = [H3O+] [OH-]
because water conc
is constant it does not
appear in the
expression


The proton in water
Values for Kw and [H+] and [OH-]
Kw = [H3O+] [OH-] = 1.0 x 10-14
1.0 x 10-14 = [X] [X]
1.0 x 10-14 = X2
X = 1.0 X 10-7 = [H3O+] = [OH-]

The Bronsted – Lowry definition



holds true for
situations not
involving water
Acids donate protons
Bases accept protons

HCl(g) + NH3(g)  NH4+(g) + Cl-(aq)\
donates H+ accepts H+
BL-Acid
BL-Base

conj base conj acid
notice that the reaction doesn’t happen in
water and that the OH- concentration has
not increased
Conjugate Acids & Bases and
Amphoteric Substances
HNO2(aq) + H2O(l)  NO2-(aq)+H3O+(aq)
conjugate
acid
base
base
acid
donates - accepts protons
 note HNO2 is considered a Arrenius acid H+ inc in
H20 also a BL because donates a Proton
 all Arrenius acids and bases are also BL acids and
bases – however all BL acids and bases may or
may not be arrenius acids or bases water is not an
Arrenius base in this example


every acid has a
conjugate base
formed from the
removal of a proton
from that acid
every base has a
conjugate acid formed
from the addition of a
proton to that base
Amphotheric Substances
NH3(aq)+H20(l)  HH4+(aq) + OH-(aq)
Base Acid
Conj Acid Conj Base
p-acc p-donar
note water is acting as an acid in this
reaction and a base in the previous one
that makes it a amphortic substance

Strengths of Acid, Bases



The stronger the acid is the weaker its conjugate
base (weaker acid  stronger conj base)
The stronger the base the weaker its conjugate
acid (weaker base  stronger conj acid)
Stronger acids and bases ionize to a greater
extent than do weak acids and bases.
Strong Acids – dissociate
completely into ions
HNO3(aq)  H+(aq) + NO3-(aq)
the production of H+ ions from the acid
dominates – ignore the H+ donated from the
water it is insignificant
 The  equilibrium lies so far to the right
because HNO3 doesn’t reform
 The neg log of the H+ from the acid determines
pH.
 Strong bases also dissociate completely and the
conc of OH- from the base is the only factor
considered when calculating pH.

Acids and Bases
Strong Acids
HClO4 prechloric
H2SO4 sulfuric
HI
hydroiodic
HCl
hydochloric
HBr
hydrobromic
HNO3 nitric

Strong Bases
GI hydroxides ex.
NaOH,KOH
G2 Hydroxides Sr(OH)2
GI Oxides ex.
Na2O, K2O
GI,II Amides ex.
KNH2,Ca(NH2)2

The pH scale
pH – is defined as the neg log (base-10) of
the H+ ion concentration
pH = -log [H+]
 What is the pH of a neutral solution
[H+] = 1.0 x 10^-7
pH = -log [1.0 x 10^-7
pH = 7

Strong acids and the pH scale
An acidic solution must have a [H+] conc
greater than 1.0 X 10-7 ex 1.0 X 10-6
 -log [1.0 X 10-6] = pH 6
 What is the pH of a basic solution?
A basic solution is one in which the [OH-]
is greater than 1X10-7.

Calc. pH of strong basic solutions
Calculate the pH of a sol that has a [OH-]
con. Of 1.0 X10-5
 Kw = 1x10-14 = [H+] [OH-]
Kw = 1x10-14 = [H+] [1.0X10-5]
[H+] = 1x10-14 = 1.0 X 10-9
1.0X10-5
pH = -log 1.0 x 10-9 pH = 9

The “p” Scale
The negative log of a quantity is labeled p
(quantity)
 Ex: we could reference the quantity of
OH- directly: pOH = -log[OH-]
 From the definition of Kw
-log Kw =(-log [H+]) (-log [OH-])= -log 1x10-4
Kw = pH + pOH = 14

Calc. pH using the p scale
Ex. OH- conc = 1.0X10-5
 -log 1.0X10-5 = 5 = pOH
 pH + pOH = 14
 pH + 5 = 14
 pH = 9

Weak Acids
partially ionize in aqueous solution
 mixture of ions and un-ionized acid in sol.
 WA are in equilibrium (H20 is left out
because it a pure liquid)
HA(aq) + H20(l)  H30+(aq) + A-(aq)
 Ka is the acid dissociation constant
 Ka = [H30] [A-] = [H+] [A-]
[HA]
[HA]

Acid Dissociation Constant

The larger the
Ka value the
stronger the
acid is – more
product is in
solution
Weak Bases
Weak bases in water react to release a
hydroxide (OH-) ion and their conjugate
acid:
 Weak Base(aq) + H2O(l) Conjugate
Acid(aq) + OH-(aq)

A common weak base is ammonia
 NH3(aq) + H2O(l)  NH4+(aq) + OH-(aq)
 Since H2O is a pure liquid it is not
expressed in the equilibrium Kb
expression
 Kb = [NH4+][OH-] (base dissociation
[NH3]
constant)
 Kb always refers to the equilibrium in
which a base reacts with H2O to form the
conjugate acid and OH
Lewis Acids and Bases
Review - An Arrhenius acid reacts in water
to release a proton - base reacts in water
to release a hydroxide ion
 In the Bronstead-Lowry description of
acids and bases: acid reacts to donate a
proton - a base accepts a proton

G.N.Lewis defination  Lewis acid is defined as an electronpair acceptor
 Lewis base is defined as an electronpair donor


In the example with ammonia, the
ammonia is acting as a Lewis base
(donates a pair of electrons), and the
proton is a Lewis acid (accepts a pair of
electrons)

Lewis is consistent with
the description by
Arrhenius, and with the
definition by BronsteadLowry. However, the
Lewis description, a base
is not restricted in
donating its electrons to a
proton, it can donate
them to any molecule that
can accept them.
Calculating the pH of a Weak Acid
What is the pH of an aq sol that is
0.0030M pyruvic acid HC2H3P3? Ka =
1.4x10-4 at 25oC
 HC2H3P3  H+ + C2H3P3I 0.0030
0
0
C -X
X
X
E 0.0030-X +X
+X

Ka = [H+] [C2H3P3-] pluggin in the values
[HC2H3P3]
from the table
1.4x10-4 =
X2
(0.0030-X)
1.4x10-4 (0.0030) = X2
4.2x10-7-1.4x10-4x = X2
X2 + 1.4x10-4x-4.2x10-7 = 0 a quadratic
ignore the neg sol x = 5.82 x10-4
pH = -log 5.82 x10-4 pH = 3.24

Learning Check
What is the pH at 25oC of a solution made
by dissolving a 5.00 grain tablet of aspirin
(acetylsalicylic acid) in 0.500 liters of
water? The tablet contains 0.325g of the
acid HC9H7O4. Ka = 3.3x10-4 mm =
180.2g/l
 H+ = 9.4x10-4 pH = 3.03

Buffers
A solution that resists changes in pH when
a limited amount of an acid or base is
added to it.
 Buffers contain either a weak acid and it’s
conj. base or a weak base and it’s conj.
acid.

Examples
Ex. Weak acid and conj. base equal molar
amounts
 Strong Acid added
H+ + A-  HA
conj. base weak acid
the conj base interacts with the H+ ions
from the strong acid changing them to a
weak acid


Strong base added
OH + HA  HOH + Aweak acid
conj base
the weak acid interacts with the OH- ion
from the base to form water and the conj.
base
If the concentration of A- and HA are large
and the amount of H+ or OH- is small the
solution will be buffered or the change in
pH will be minimized.
Buffering capacity – the amount of acid or
base a buffer can react with before a
significant change in pH occurs
 Ratio of acid to conj base – unless the
ratio is close to 1 ( between 1:10 and 10:1)
will be too low to be useful.

Calculating the pH of a buffer
Note: a solution of 0.10 M acedic acid and
its conj base 0.20 M acetate from sodium
acetate is a buffer solution pH = 5.07
 Ex. Calc. the pH of a buffer by mixing 60.0
ml of 0.100 M NH3 with 40.0ml of 0.100 M
NH4Cl.

1St cal the conc. of each species
M = moles/liters
mol of NH3 0.10M = X/0.060 l = 0.0060mol
mol of NH4 0.10M = X/0.040 l = 0.0040mol
[NH3] = 0.0060 mol/ 0.100 l = 0.060 M
[NH4] = 0.0040 mol/ 0.100 l = 0.040 M
NH3 + H2O  NH4+ + OHI 0.060M
O.040 0
C -X
+X
+X
E 0.060-x
0.040+x X

Kb =
[NH4+] [OH-] ( 0.040+X)X
1.8X10-5
[NH3]
(0.060-X)
Ignore X 1.8X10-5 = 0.040X X = 2.7X10-5
0.060
-log (2.7x10-5) = 4.6 pOH pH = 9.4
Or using Henderson - Hasselbalch equation
pOH = pKb + log [conj acid] = 4.74 + log ( [0.04]
[B]
[0.06])
= 4.6 pOH pH= 9.4

What is the pH of a buffer prepared
by adding 30.0ml of .15M HC2H3O2
to 70ml of .2M NaC2H3O2?
[HC2H3O2] = .15M = x/.03 = .0045/.01 = .045
[C2H3O2] = .20M =x/.07 = .0140/.01 = .140
ka = 1.7x10-5
HHeq pH = pKa + log [conj base]
[acid]
pH = 4.77 + log(.140 = 5.3
.045)
Adding an acid or a base to a buffer
Calc the ph of 75ml of the buffer solution
of(0.1M HC2H3O2 and 0.2M NaC2H3O2)
to which 9.5 ml of 0.10M HCl has been
added. Compare the change to that of
adding HCl to pure water.
H+ + C2H3O2 -  HC2H3O2
H+ = 0.10M = n/.0095l = 0.00095 moles
C2H3O2 - = 0.2M = n/.075 = 0.0150 moles
HC2H3O2 = 0.10M = n/.075 = 0.0075 moles

Neutralization Reaction
 C2H3O2 = C2H3O2 moles – H+ moles
0.0150n - .00095n = 0.014
 HC2H3O2 = Orginial Conc. + Conc Contributed
by reaction
0.075 moles + 0.00095 = 0.0085mol
[C2H3O2] = 0.014mol/0.085l = 0.16M
[HC2H3O2] = 0.0085/0.085 = 0.10M
pH = 4.76 + log (.16/.10) = 4.96
