Acid Base Equilibria

Report
Acid Base Equilibria
maybe do packet 2 before
packet 3? Combine concepts
from 1 and 3?
Arrhenius Acids & Bases
(1st year chem definition!)
An Arrhenius acid is a substance that, when
dissolved in water, increases the
concentration of H+ (needs H+ in it)
Example: HCl (monoprotic)
H2SO4 (diprotic)
An Arrhenius base is a substance that, when
dissolved in water, increases the
concentration of OH–
Example: NaOH (monobasic)
Ca(OH)2 (dibasic)
Brønsted-Lowry Acids and
Bases
Brønsted-Lowry acid is a species that
donates H+ (proton)
Brønsted-Lowry base is a species that
accepts H+ (proton)
Brønsted-Lowry definition of a base does not
mention OH– and the reaction does not need
to be aqueous.
The H+ Ion in Water
• The H+(aq) ion is
simply a proton
with no
surrounding
valence electrons.
• In water, clusters
of hydrated H+(aq)
ions form.
• The simplest cluster is H3O+(aq)
We call this a hydronium ion.
• Larger clusters are also possible (such as
H5O2+ and H9O4+).
• Generally we use H+(aq) and H3O+(aq)
interchangeably.
Proton-Transfer Reactions
• Consider
NH3(aq) + H2O(l)  NH4+(aq) + OH–(aq)
• H2O donates a proton to ammonia.
– Therefore, water is acting as an acid.
• NH3 accepts a proton from water.
– Therefore, ammonia is acting as a base.
Amphoteric substances
• Can behave as acids and bases.
• Water is an example of an amphoteric
species.
Water as an acid (proton donor)
H2O + NH3  NH4+ + OHWater as a base (proton acceptor)
H2O + HNO2  H30+ + NO2-
Conjugate Acid-Base Pairs
• A conjugate acid is
the substance formed
by adding a proton to
the base.
• A conjugate base is
the substance left
over after the acid
donates a proton.
Within a pair the acid has more hydrogen!
Strong Acids & Bases
Strong Acids
Strong Bases
sulfuric acid
H2SO4
Lithium hydroxide
LiOH
Nitric acid
HNO3
Sodium hydroxide
NaOH
Perchloric acid
HClO4
Potassium hydroxide
KOH
Chloric acid
HClO3
Rubidium hydroxide
RbOH
Hydrochloric acid
HCl
Cesium hydroxide
CsOH
Hydrobromic acid
HBr
Calcium hydroxide
Ca(OH)2
Hydroiodic acid
HI
Strontium hydroxide
Sr(OH)2
Barium hydroxide
Ba(OH)2
Less common:
O2- H-
CH3-
All other acids and bases are weak!
Ions
Most anions are weak bases
Most cations are weak acids
Anions of strong acids and cations of strong bases are neutral
Neutral anions
Neutral cations
Hydrogen sulfate
HSO4-
Lithium
Li+
Nitrate
NO3-
Sodium
Na+
Perchlorate
ClO4-
Potassium
K+
Chlorate
ClO3-
Rubidium
Rb+
Chloride
Cl-
Cesium
Cs+
Bromide
Br-
Calcium
Ca2+
Iodide
I-
Strontium
Sr2+
Barium
Ba2+
Strengths of Acids and Bases
Strong acids completely ionize in water.
HCl + H2O  H3O+ + ClHCl  H+ + Cl• Essentially no un-ionized molecules remain in solution
so the equation usually does not contain equilibrium
arrows. Keq >>1
• Their conjugate bases have negligible tendencies to
become protonated
Cl- + H+  HCl.
The conjugate base of a strong acid is a neutral anion.
Strengths of Acids and Bases
Strong bases completely dissociate in water
NaOH(aq)  Na+(aq) + OH-(aq)
• Essentially no undissociated compound remains in
solution so the equation usually does not contain
equilibrium arrows. Keq >>1
• The ions have negligible tendencies to attract OH- in
solution.
Na+(aq) + OH-(aq)  NaOH(aq)
The cation of a strong base is a neutral cation.
All other acids are Weak acids. They only
partially dissociate in aqueous solution.
HC2H3O2(aq) + H2O(l)  H3O+(aq) + C2H3O2-(aq)
acid
base
conjugate
acid
conjugate
base
• They exist in solution as a mixture of molecules
and component ions. (usually mostly molecules
in equilibrium)
• Their conjugate bases are weak bases.
– Besides non-neutral anions, weak bases tend to be
nitrogen containing organic compounds
• Example: Acetic acid is a weak acid; acetate
ion (conjugate base) is a weak base.
Non-acid compounds with
hydrogen
Not all compounds containing hydrogen are acidic.
These are extremely weak acids…so much that we
don’t consider them acids at all.
Their conjugate bases are strong bases!
Negligible acidity: OHStrong bases: O2-
H2
H-
CH4
CH3-
Page 657 in textbook
The stronger an acid is, the weaker its conjugate
base will be.
In acid-base reactions, the reaction favors the
transfer of a proton from the stronger acid to the
stronger base to form a weaker acid and weaker
base.
We need a more specific way to determine
acid strength!
Where does that acid/base ranking come from?
Keq!!!
Since weak acids and bases are in equilibrium…we can write
equilibrium constant expressions!
When looking at the reaction of a weak acid with water we label
the equilibrium constant Ka:


[
H
O
][
F
]
3
HF(aq) + H2O(l)  H3O+(aq) + F-(aq)
Ka 
[ HF ]
When looking at the reaction of a weak base with water we label
the equilibrium constant Kb:


+
[
NH
][
OH
]
NH3(aq) + H2O(l)  NH4 (aq) + OH (aq) K 
4
b
[ NH3 ]
Ka and Kb
• The value of Ka or Kb indication the extent
to which the weak acid or base
ionizes/dissociates.
– Larger Ka or Kb means more products!
• Weak acids with larger Ka values are
stronger.
• Weak bases with larger Kb values are
stronger.
(Values in Appendix D)
Percent Ionization
• Percent ionization is another method to assess
acid strength.
• For the reaction: HA(aq)  H+(aq) + A–(aq)
 H+ 
  equilibrium
% ionization =
× 100
 HA 
initial
• The higher the percent ionization,
the stronger the acid.
19
Polyprotic Acids
Polyprotic acids have more than one ionizable proton.
• The protons are removed in successive steps.
Consider the weak acid, H2SO3 (sulfurous acid):
H2SO3(aq)  H+(aq) + HSO3–(aq) Ka1 = 1.7 x 10–2
HSO3–(aq)  H+(aq) + SO32–(aq)
Ka2 = 6.4 x 10–8
• It is always easier to remove the first proton in a
polyprotic acid than the second.
Ka1 > Ka2 > Ka3, etc
The Autoionization of Water
In pure water the following equilibrium is established:
H2O(l) + H2O(l)  H3O+(aq) + OH–(aq)
acid
base
acid
base
• This process is called the autoionization of water.
We can write an equilibrium constant expression for
the autoionization of water:
Kw = [H3O+] [OH–] = 1.0*10-14 @25°C
Kw is called the “ion-product constant”
Kw = Ka*Kb for conjugate pair
The Ion Product Constant
• This applies to pure water as well as to aqueous
solutions. (for our purposes…)
• A solution is neutral if [OH–] = [H3O+].
• If the [H3O+] > [OH–], the solution is acidic.
• If the [H3O+] < [OH–], the solution is basic.
In a neutral solution at 25oC,
[H+] = [OH-] = 1.0 x 10-7 M
Give the conjugate base of the following
Bronsted-Lowry acids:
(a) HIO3
(b) NH4+1 (c) H2PO4-1
remove H+
(a)
(b)
(c)
(d)
IO3-1
NH3
HPO4-2
C7H5O2-
(d) HC7H5O2
Designate the Bronsted-Lowry acid and the Bronsted-Lowry base on the
left side of each of the following equations, and also designate the
conjucate acid and base on the right side:
(a) NH4+1(aq)
acid
+
CN-1(aq)
base
(b) (CH3)3N(aq) +
base
(c) HCHO2(aq) +
acid
24

H 2O
acid
PO4-3(aq)
base
HCN(aq)
acid
+
 OH-1(aq) +
base

HPO4-2(aq)
acid
NH3(aq)
base
(CH3)3NH+1(aq)
acid
+
CHO2-1(aq)
base
(a) The hydrogen oxalate ion (HC2O4-1) is amphoteric. Write a balanced
chemical equation showing how it acts as an acid toward water and
another equation showing how it acts as a base towards water.
(b) What is the conjugate acid and base of HC2O4-1?
(a) Acid:
Base:
HC2O4-1(aq) + H2O(l)  C2O4-2(aq) + H3O+1(aq)
acid
base
conj base
conj acid
HC2O4-1(aq) + H2O(l)  H2C2O4(aq) + OH-1(aq)
base
acid
conj acid
conj base
(b)
Label each of the following as being a strong acid, a weak
acid, or a species with negligible acidity. In each case write
the formula of its conjugate base, and indicate whether the
conjugate base is a strong base, a weak base or a species
with negligible basicity: (a) HNO2 (b) H2SO4 (c) HPO4-2
(d) CH4 (e) CH3NH3+1
(a)
(b)
(c)
(d)
(e)
HNO2 , weak acid
H2SO4 , strong acid
HPO4-2 , weak acid
CH4 , negligible acid
CH3NH3+1 , weak acid
IMPORTANT
26
NO2-1 , weak base
HSO4-1 , negligible base
PO4-3 , weak base
CH3-1 , strong base
CH3NH2 , weak base
ACID
strong
weak
negligible
BASE
negligible
weak
strong
(a)
Which of the following is the stronger acid, HBrO or HBr?
(b)
Which is the stronger base, F-1 or Cl-1?
Briefly explain your choices.
(a)
(b)
HBr - It is one of the seven strong acids
F-1
HF is a weak acid so F- is a weak base
HCl is a strong acid so Cl- is neutral
Predict the products of the following acid-base reactions & determine
whether equilibrium lies to the right or left:
(a) O-2(aq) + H2O(l) 
(b) CH3COOH(aq) + HS-1(aq) 
(c) NO3-1(aq) + H2O(l) 
* You will need a chart like this or Ka/Kb values to determine equilibrium!
(a)
O-2(aq)
+
base
H2O(l)

acid
OH-1(aq)
OH-1(aq)
+
acid
base
*OH- is the weaker acid so equilibrium lies to the right
(b)
CH3COOH(aq) = HC2H3O2 (aq)
HC2H3O2 (aq) + HS-1(aq)  C2H3O2-1(aq)
acid
base
+
base
H2S(aq)
acid
*H2S is the weaker acid so equilibrium lies to the right
(c)
NO3-1(aq)
base
+
H2O(l)
acid

HNO3(aq)
acid
+
OH-1(aq)
base
*H2O is the weaker acid so equilibrium lies (far!) to the left
The “p” Function
• p is short for “– log10”
pH = -log10[H+] = -log[H+]
pOH = -log10[OH-] = -log [OH-]
• Note that this is a logarithmic scale.
• Thus a change in [H+] by a factor of 10
causes the pH to change by 1 unit.
• Most pH values fall between 0 and 14.
• In neutral solutions at 25oC, [H+] = 1.0 x 10-7
pH = -log[1.0 x 10-7] = 7.00
[H+]
pH
Acidic
> 1.0 x 10–7
< 7.00
Neutral
1.0 x 10-7
7.00
Basic
< 1.0 x 10–7
> 7.00
Lower pH = more acidic
Higher pH = more basic
Another one: pK
• We can use a similar system to describe the
equilibrium constant
pK = - log[K]
• The value of Kw at 25oC is 1.0 x 10–14
pKw = - log (1.0 x 10–14)= 14.0
Kw=[H+][OH-]
pKw= pH + pOH = 14.0
The pH Loop
[H+]=10-pH
[H+]
pH
pH = -log[H+]
[H+] [OH-]=10-14
[OH-]
pH + pOH=14.0
[OH-]=10-pOH
pOH =
-log[OH-]
pOH
Measuring pH
The most accurate method to measure pH is to use a pH meter.
Acid Base Indicators:
• certain dyes change color as pH changes.
• Indicators are less precise than pH meters.
– Many indicators do not have a sharp color change as a function of pH.
• Most acid-base indicators can exist as either an acid or a
base.
– These two forms have different colors.
– The relative concentration of the two different forms is sensitive to the
pH of the solution.
• Thus, if we know the pH at which the indicator turns color, we
can use this color change to determine whether a solution has
a higher or lower pH than this value.
Example 1: Calculate [H+1] for each of the following solutions and indicate whether
the solution is acidic, basic or neutral. (a) [OH-1] = 0.0007 M (b) a solution
where [OH-1] is 100 times greater than [H+1]
kw  [ H  ][OH  ]
1.0 x1014  [ H  ](7 x104 )
[ H  ]  1x1011 M (basic)
[OH  ]  100[ H  ]


 2
kw  [ H ] *100[ H ]  100[ H ]
14
Kw
1.0 x10
[H ] 

100
100

 1.0 x108 M (basic)
Example 2:
(a) If NaOH is added to water, how does the [H+1] change? How does pH change?
(b) If [H+1] = 0.005 M, what is the pH of the solution? Is the solution acidic or basic?
(c) If pH = 6.3, what are the molar concentrations of H+1(aq) and OH-1(aq) in the
solution?
(a) Kw = [H+] [OH-].
the [OH-] will increase and the [H+] will decrease.
[H+] decreases, pH increases.
(b) pH = - log [H+] = - log (0.005) = 2.3
(c) pH = 6.3
[H+] = 10-pH = 10-6.3 = 5 x 10-7 M
pOH = 14 – pH = 14 – 6.3 = 7.7
[OH-] = 10-pOH = 10-7.7 = 2 x 10-8 M
acidic
Strong Acid Calculations
• In solution the strong acid is usually the only
source of H+
• The pH of a solution of a monoprotic acid
may usually be calculated directly from the
initial molarity of the acid.
Caution: If the molarity of the acid is less than
10–6 M then the autoionization of water needs
to be taken into account.
Example 3: Calculate the pH of each of the following strong acid solutions: (a) 1.8  10-4 M HBr
(b) 1.02 g HNO3 in 250 mL of solution (c) 2.00 mL of 0.500 M HClO4 diluted to 50.0 mL (d) a
solution formed by mixing 10.0 mL of 0.0100 M HBr with 20.0 mL of 2.5  10-3 M HCl
(a)
(b)
1.8 x 10-4M HBr = 1.8 x 10-4M H+
1.02 g HNO3 1 mol HNO3
*
 0.0647 M HNO3
0.250 L
63.0 g HNO3
0.0647 M HNO3 = 0.0647 M H+
(c) M1V1 = M2V2
(0.500 M)(.00200 L) = (x M)(0.0500 L)
x = .0200 M HCl
(d)
pH = -log(1.8 x 10-4) = 3.74
[ H  total ] 
pH = -log (.0647) = 1.19
[H+] = .0200 M
pH = -log(.0200) = 1.70
(.0100M *.0100 L)  (.00250M *.0200 L)
 .00500M
.0300 L
pH = -log(.0050) = 2.30
Strong Base Calculations
• Strong bases are strong electrolytes and
dissociate completely in solution.
• For example:
NaOH(aq)  Na+(aq) + OH–(aq)
The pOH (and thus the pH) of a strong base
may be calculated using the initial molarity of
the base.
Example 4: Calculate [OH-1] and pH for (a) 3.5  10-4 M Sr(OH)2 (b) 1.50 g LiOH in 250 mL of
solution (c) 1.00 mL of 0.095 M NaOH diluted to 2.00 L (d) a solution formed by adding 5.00
mL of 0.0105 M KOH to 15.0 mL of 3.5  10-3 M Ca(OH)2
(a) [OH-] = 2[Sr(OH)2] = 2(0.00035 M) = .00070 M OHpOH = -log(.00070) = 3.15
pH = 14 – 3.15 = 10.85
(b)
1.50 g LiOH 1 mol LiOH
*
 0.251M LiOH  [OH- ]
0.250L
24.0g LiOH
pOH = -log (.251) = 0.601
pH = 14 - .601 = 13.399
(c) M1V1 = M2V2
(0.095 M)(.00100 L) = (x M)(2.00 L)
x = .000048 M NaOH = [OH-]
pOH = -log(.000048) = 4.32
pH = 14 – 4.32 = 9.68
(d)
[OH  total ] 
(. 0105 M * .00500 L)  2(. 0035 M * .0150 L)
 .0079 M
.0200 L
pOH = -log(.0079) = 2.1
pH = 14 – 2.1 = 11.9
Weak Acid Calculations
• Weak acids are only partially ionized in
aqueous solution.
• Therefore, weak acids are in equilibrium:
HA(aq) + H2O(l)  H3O+(aq) + A–(aq)
OR
HA(aq)  H+(aq) + A–(aq)
• We can write Ka for this dissociation:
Ka


H O A 
H A 

or K 
3

HA


a
HA

Calculating Ka from pH for weak
acids
• In order to find the value of Ka, we need to know all
of the equilibrium concentrations. (ICE Chart)
• The pH gives the equilibrium concentration of H+.
• We then substitute these equilibrium
concentrations into the equilibrium constant
expression and solve for Ka.
Example 4: A 0.20 M solution of niacin (a
monoprotic weak acid) has a pH of 3.26.
What is the Ka for niacin?
HA

H+
A-
I
0.20
0
0
C
-.000550
+.000550
+.000550
E
.019945
10-3.26=.000550
.000550
(.000550)(.000550)
Ka 
 1.5*105
(.019945)
Using Ka to Calculate pH for
weak acids
• Write the balanced chemical equation clearly showing the
equilibrium.
• Write the equilibrium expression. Look up the value for Ka
(in a table).
• Write down the initial and equilibrium concentrations for
everything except pure water. (ICE table)
• We usually assume that the equilibrium concentration of H+
is x.
• Substitute into the equilibrium constant expression and
solve.
• Remember to convert x to pH if necessary.
Polyprotic Acids
• Polyprotic acids have more than one ionizable
proton.
H2SO3(aq)  H+(aq) + HSO3–(aq) Ka1 = 1.7 x 10–2
HSO3–(aq)  H+(aq) + SO32–(aq) Ka2 = 6.4 x 10–8
• The majority of the H+(aq) at equilibrium usually
comes from the first ionization
• If the successive Ka values differ by a factor of
103, we can usually get a good approximation of
the pH of a solution of a polyprotic acid by
considering the first ionization only.
• If not, then we have to account for the successive
ionizations
Example 5: The acid dissociation constant for benzoic acid, HC7H5O2 is 6.5 x 10-5. Calculate
the equilibrium concentrations of H3O+, C7H5O2-, and HC7H5O2 . The initial
concentration of HC7H5O2 is 0.050M.
HC7H5O2 (aq)  H+(aq) + C7H5O2- (aq)
I
C
E
0.050
-x
(.050-x)
0
x
x
0
x
x
[ H  ][C7 H5O2  ]
x2
Ka 

 6.5*105
[ HC7 H 5O2 ]
(.050  x)
x2 + (6.5*10-5)x – (3.25*10-6) = 0
x= .0018M = [H+]=[C7H5O2- ]
[HC7H5O2] = .050 - .0018 = .048 M
What if I don’t have a quadratic equation program?
Algebra Shortcut:
Assume x is much smaller (less than 5% of .050)
To simplify: .050-x = .050
Now you don’t need the quadratic equation!
[ H  ][C7 H5O2 ]
x2
x2
Ka 


 6.5*105
[ HC7 H5O2 ]
(.050  x) .050
x= .0018M = [H+]=[C7H5O2- ]
[HC7H5O2] = .050 - .0018 = .048 M
If you made this assumption you need to check and make
sure it’s valid – if this answer isn’t less than 5% use the
quadratic equation!
.0018
.050
*100%  3.6%
Example 6: Calculate the pH of the following solution (Ka and Kb values are
in Appendix D). 0.175 M hydrazoic acid, HN3
(a)
I
C
E
HN3(aq) 
0.175M
-x
0.175-x
H+(aq) + N3-(aq)
0
0
x
x
x
x
[ H  ][ N3 ]
x2
Ka 

 1.9*105
HN3
(0.175  x)
x2 + (1.9*10-5)x – (3.325*10-6) = 0
x=.0018M H+
pH = -log(0.0018) = 2.74
Example 7: Calculate the percent ionization of 0.400 M hydrazoic acid, HN3,
solution.
I
C
E
HN3(aq) 
0.400M
-x
0.400-x
H+(aq) +
0
x
x
N3-(aq)
0
x
x
[ H  ][ N3 ]
x2
Ka 

 1.9*105
HN3
(0.400  x)
x2 + (1.9*10-5)x – (7.6*10-6) = 0
x=.0028M H+
%ionization 
[ H  ]eq
[ HA]initial
0.0028

*100%  0.70%
0.400
Example 8: Citric acid, which is present in citrus fruits, is a triprotic acid. Calculate
the pH of a 0.050 M solution of citric acid.
H3C6H5O7(aq)  H+(aq)
H2C6H5O7-1(aq)  H+(aq)
HC6H5O7-2(aq)  H+(aq)
+
+
+
H2C6H5O7-1(aq) Ka = 7.4*10-4
HC6H5O7-2(aq) Ka = 1.7*10-5
C6H5O7-3(aq) Ka = 4.0*10-7
To calculate the pH of a .050M solution, assume initially that only the first
ionization is important
H3C6H5O7(aq)
H+(aq)
+
H2C6H5O7-1(aq)
I
0.050
0
0
C
-x
x
x
E
0.050-x
x
x
[ H  ][ H 2C6 H5O7  ]
x2
Ka 

 7.4*104
HN3
(0.050  x)
x = 0.0057
0.0057=[H+] = [H2C6H5O7-1]
Does the second ionization have any effect?
H2C6H5O7-1(aq)  H+(aq) + HC6H5O7-2(aq)
I
0.0057
0.0057
0
C
-y
y
y
E
0.0057-y 0.0057+y
y
Ka = 1.7*10-5
2
Ka2
[ H  ][HC6 H 5O7 ] (.0057 y ) y
5



1
.
7
*
10
1
0.0057 y
[ H 2C6 H 5O7 ]
y = 0.000017
This value is small compared to 0.0057 (think SDs)
Total [H+] = 0.0057 + .000017 = 0.0057
which indicates the 2nd (and any subsequent) ionizations can be ignored
pH = -log(0.0057) = 2.24
Weak Base Calculations
• Weak bases remove protons from substances.
• There is an equilibrium between the base and the
resulting ions:
Example:
NH3(aq) + H2O(l) 
NH4+(aq) + OH–(aq).
The base-dissociation constant, Kb, is
 NH 4   OH  
Kb 
 NH 3
The larger Kb, the stronger the base.
16.8 Relationship Between Ka and Kb
• Generally only either Ka or Kb for a conjugate pair is
reported in tables. If you know one you can find the
other!
• Consider the following equilibria:
NH4+(aq)  NH3(aq) + H+(aq)
NH3(aq) + H2O(l)  NH4+(aq) + OH–(aq)
• We can write equilibrium expressions for these
reactions:
[NH3 ][H + ]
Ka =
+
[NH 4 ]
[NH 4 + ][OH - ]
Kb =
[NH3 ]
• If we add these equations together:
NH4+(aq)  NH3(aq) + H+(aq)
NH3(aq) + H2O(l)  NH4+(aq) + OH–(aq)
• The net reaction is the autoionization of water.
H2O(l)  H+(aq) + OH–(aq)
• When we add equations we multiply K.
Kw = Ka x Kb
Alternatively, we can express this as:
pKa + pKb = pKw = 14.00 (at 25oC)
Ka
Kb
Kw
Example 9: Calculate the molar concentration of OH-1 ions
in a 0.050 M solution of hydrazine, H2NNH2,
Kb = 1.3  10-6. What is the pH of this solution?
H 2 NNH 2 (aq) + H 2 O (l)




H 2 NNH 3+1 (aq)
+
OH1 (aq)
initial
0.050
0
0
change
-x
x
x
equil.
0.050 - x
x
x
[H2 NNH3+1 ][OH-1 ]
x2
Kb =
=
= 1.3 x 10-6
[H2 NNH2 ]
0.050 - x
x = 0.00025 M
pOH = - log [OH-1 ] = - log (.00025) = 3.60
pH = 14 - pOH = 14 - 3.60 = 10.40
Example 10: Although the acid dissociation constant for phenol,
C6H5OH, is listed in Appendix D, the base dissociation
constant for the phenolate ion, C6H5O-1 is not. (a) Explain
why it is not necessary to list both. (b) Calculate the Kb for
the phenolate ion. (c) Is the phenolate ion a weaker or
stronger base than ammonia, NH3?
(a) For a conjugate acid/conjugate base pair Ka x Kb = Kw
K w 1.0 x 10-14
-5
(b) K b =
=
=
7.7
x
10
Ka
1.3 x 10-10
Appendix D
(c) K b for C6H5O-1 (7.7 x 10-5 ) > K b for NH3 (1.8 x 10-5 )
C6H5O-1 is a stronger base
16.9 Acid-Base Properties of Salt Solutions
• All soluble salts are strong electrolytes.
- In solution, they exist nearly entirely of ions.
- Acid-base properties of salts are due to the
reactions of their ions in solution.
• Many ions can react with water to form OH– or H+.
• This process is called hydrolysis.
59
Remember this?
Most anions are weak bases
Most cations are weak acids
Anions of strong acids and cations of strong bases are neutral
Neutral anions
Neutral cations
Hydrogen sulfate
HSO4-
Lithium
Li+
Nitrate
NO3-
Sodium
Na+
Perchlorate
ClO4-
Potassium
K+
Chlorate
ClO3-
Rubidium
Rb+
Chloride
Cl-
Cesium
Cs+
Bromide
Br-
Calcium
Ca2+
Iodide
I-
Strontium
Sr2+
Barium
Ba2+
Summary
Anions
– Of strong acids are neutral.
• Example: Cl- of HCl
– Cl- + H2O  X
– Of weak acids are basic.
• Example: F- of HF
– F- + H2O  HF + OH-
– With ionizable protons are amphoteric.
• Example: HSO4–
Summary
Cations
– Of strong bases are neutral.
• Example: Na+ of NaOH
– Na+ + H2O  X
–All other cations are weak acids
• Example: Fe3+
• Cation hydrolysis reaction:
Smaller and more highly charged ions =
stronger (weak) acids
(All are 1.00 M solutions)
The pH of a solution may be qualitatively predicted:
Cation
Anion
Solution is:
Example
Neutral
(from strong base)
Neutral
(from strong acid)
Neutral
NaCl
Neutral
(from strong base)
Basic
(from weak acid)
Basic
NaF
Acidic
Neutral
(from a weak base) (from a strong acid)
Acidic
FeCl3
NH4Cl
Acidic
Basic
(from a weak base) (from a weak acid)
Depends!
NH4CN
Salts from a weak acid and weak base can be either acidic or
basic.
Compare Ka of the cation and Kb of the anion
- If the Ka is larger the solution will be acidic
- If the Kb is larger the solution will be basic
For example, consider NH4CN.
The Kb of CN-1 is larger than the Ka of NH4+1
so the solution will be basic.
Example 1: Predict whether aqueous solutions of the following
compounds are acidic, basic or neutral.
(a) NH4Br
(b) FeCl3
(c) Na2CO3
acidic
(d) KClO4
neutral
basic
acidic
(e) NaHC2O4
acidic
Ka for acid HC2O4-1 = 6.410-5
Kb for base HC2O4-1 = 1.710-13
Example 2: Using data from Appendix D, calculate [OH-1] and pH
for the following solution 0.10 M NaCN
NaCN is a soluble salt and will dissociate into Na+1 and CN-1 ions.
Na+1 ion does not interact with water, but CN-1 acts as a weak base:
I
C
E
CN -1
0.10
-x
0.10 - x
+
H2O




HCN
0
x
x
+
OH -1
0
x
x
Ka for HCN (given in Appendix D) = 4.9x10-10
Kw
1 x 10-14
-5
K b for CN =
=
=
2.0
x
10
K a for HCN 4.9 x 10-10
-1
[HCN][OH-1 ]
x2
-5
Kb =
= 2.0 x 10 =
-1
[CN ]
0.10-x
x = 0.0014 M = [OH-1 ]
pOH = - log(0.0014) = 2.85
pH = 14 - 2.85 = 11.15
16.10 Acid-Base Behavior and Chemical Structure
Acidity is directly related to the strength of attraction for
a pair of electrons to a central atom.
4 situations to consider:
1. Ions
Ionic Charge and Size
When comparing ions of similar structure:
More positive ions are stronger acids.
tie breaker: Smaller ion is stronger acid
Acid strength: Na+ < Ca2+ < Cu2+ < Al3+
PO43- < HPO42- < H2PO4- < H3PO4
Example 3: Predict which member of each pair produces the
more acidic aqueous solution:
(a) K+1 or Cu+2
(b) Fe+2 or Fe+3
(c) Al+3 or Ga+3
(a)
Cu+2 has higher charge (and K+1 is neutral)
(b) Fe+3
has higher charge
(c) Al+3
has a smaller size
69
2. Binary Acids:
Bond Polarity (Electronegativity) & Strength
• The H–X bond strength is important in determining
relative acid strength in any group in the periodic
table.
– The weaker the bond the easier it will break
– The H–X bond strength tends to decrease down
a group - acid strength increases down a group
• H–X bond polarity is important in determining
relative acid strength in any period of the periodic
table.
– The H-X bond polarity tends to increase across a period
- acid strength increases (from left to right) across a
period
more polar bond – stronger acid
71
larger atom
weaker bond
stronger acid
3. Oxyacids (Acids with oxygen) with
different central atoms
Generally, the larger the electronegativity of
the central atom the stronger the acid.
– The stronger the pull on electrons the less
tightly the H is held
Acid Strength: H3BO3 < H2CO3 < HNO3
The higher EN of central atom means more electron density shift
away from H - H is easier to remove – stronger acid
4. Oxyacids with the same central atom
Generally, the more oxygens attached to the
central atom the stronger the acid.
– The more atoms pulling on electrons the
less tightly the H is held
Acid Strength: HClO < HClO2 < HClO3 < HClO4
More O atoms means more electron density
shift away from H and H is easier to remove –
stronger acid
Example 4:
Explain the following observations:
(a) HNO3 is a stronger acid than HNO2
(b) H2S is a stronger acid than H2O
(c) H2SO4 is a stronger acid than HSO4-1
(d) H2SO4 is a stronger acid than H2SeO4
(e) CCl3COOH is a stronger acid than CH3COOH
(a)
more oxygen atoms - electron density shifts away from H bond easier to remove H
(b)
bond strength decreases down a group - H easier to remove from
S
(c)
More positive ion is stronger
(d)
S has higher electronegativity than Se – pulls electrons from H –
easier to remove H
(e)
the more electronegative 3 Cl atoms (as opposed to the 3 H
atoms) pull electron density more weakens the O-H bond and
makes H easier to remove
76
16.11 Lewis Acids and Bases
• A Brønsted-Lowry acid is a proton donor.
• Lewis proposed a new definition of acids and
bases that emphasizes the shared electron pair.
• A Lewis acid is an electron pair acceptor.
• A Lewis base is an electron pair donor.
• Note: Lewis acids and bases do not need to
contain protons.
• Therefore, the Lewis definition is the most
general definition of acids and bases.
Typical Lewis acid-base reaction
78
•
What types of compounds can act as Lewis
acids?
• Lewis acids must have a vacant orbital (into
which the electron pair can be donated).
• Lewis acids sometimes have an incomplete
octet (e.g., BF3).
• Transition-metal ions can be Lewis acids
(empty d orbitals)
• Compounds with multiple bonds can act as
Lewis acids.
79
Example 5: Identify the Lewis acid and Lewis base in each
of the following reactions:
(a) Fe(ClO4)3 + 6 H2O  Fe(H2O)6+3 + 3 ClO4-1
(b) CN-1 + H2O  HCN + OH-1
(c) (CH3)3N + BF3  (CH3)NBF3
(d) HIO + NH2-1  NH3 + IO-1
(a)
(b)
(c)
(d)
Acid
Base
H donor
e acceptor
empty orbitals/mult. bonds
incomplete octet/cation
H acceptor
e donator
has lone pairs
often contains N
Fe(ClO4)3 or Fe+3
H2O
BF3
HIO
H2O
CN-1
(CH3)3N
NH2-1

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