Download Percent Ionization

Solutions of a Weak Acid or Base
Acid-Ionization Equilibria
Acid-Base Equilibria
Chapter 16.1 and 16.3 16.6
Essentials of General Chemistry, 2nd Ed.,
Ebbing and Gammon
Acid-ionization Equilibria
Consider an equilibrium involving a weak acid in H2O:
HA (aq) + H2O (l)
H3O+ (aq) + A- (aq)
Acid-Ionization Constant, Ka (acid-dissociation constant)
- equilibrium constant for the ionization of a weak
acid
Ka = [H3O+][A-]
[H2O]
Acid dissociation
- acid reacts with water to produce hydronium ion
and the conjugate base ion
Consider acetic acid
HC2H3O2 (aq) + H2O (l)
H3O+ + C2H3O2- (aq)
- since acetic acid is weak electrolyte, it ionizes only to
small amount (about 1% or less/partially ionized)
Note: this is not the same thing as a dilute sol n of a strong
acid (100% dissociated)
AcidIonization
Constants @
25oC
Table 16.1
AcidIonization
Constants @
25oC
Calculations of and using Ka
Experimental determination of Ka
- using pH of a solution of a weak acid
- degree of ionization
- percent ionization
Calculations with Ka
- calculate the equilibrium concentrations of HA, Aand H3O+ for solutions of different molarities
- use simplifying assumption or quadratic formula
Table 16.1
1
Percent Ionization (Degree of ionization)
Use of Percent Ionization
Degree of Ionization
- the concentration of the acid that dissolves divided
by the initial concentration of the acid
Consider
HA (aq) + H2O (l)
H3O+ (aq) + A- (aq)
- degree of ionization of weak acid depends of both Ka
and the concentration of the acid solution
i.) for given concentration
- the larger the Ka, the greater the degree of
ionization
ii.) for a given value of Ka
- the more dilute the solution, the greater the
degree of ionization
How do you know when to use the simplifying assumption?
- it can be shown that this simplifying assumption
gives an error of less than 5% if the concentration
of acid, Ca, divided by Ka equals 100 of more
Degree of Ionization = [HA]dissociated = [H3O+]eq
[HA]initial
[HA]initial
Percent Ionization = degree of ionization x
100%
Base-Ionization Equilibria
Consider the reaction
NH3 (aq) + H2O (l)
Base-Ionization Constants @ 25oC
NH4+ (aq) + OH- (aq)
Kb = [NH4+][OH-]
[NH3]
Base-Ionization Constant, Kb (base-dissociation constant
- equilibrium constant for the ionization of a weak
acid
B (aq) + H2O (l)
HB+ (aq) + OH- (aq)
Kb = [BH+][OH-]
[B]
amines
- derivatives of ammonia in which one or more hydrogen
atoms are replaced by another group
CH3NH2 (aq) + H2O (l)
CH3NH3+
(aq)
+ OH- (aq)
Acid-Base Properties of Salt Solutions
Salt
- formed when an acid neutralizes a base
- salt solutions can be neutral, basic or acidic (with acidity
or basicity dependent upon individual ions of the salt)
Eg. NaCN
- gives Na+ and CN- ions
- Na+ is unreactive with water (neutral)
- CN- reacts with water
CN (aq) + H2O (l)
HCN (aq) + OH- (aq)
base
acid
(Kb)
to form OH- (basic solution)
2
Hydrolysis (of an ion)
- the reaction of an ion with water to produce the
conjugate acid and hydroxide ion or the conjugate
base and hydronium ion
Eg. NH4Cl
- gives NH4+ and Cl- ions
- Cl- is unreactive with water (neutral)
- NH4+ reacts with water
NH4+ (aq) + H2O (l)
NH3 (aq) + H3O+ (aq)
acid
base
(Ka)
Predicting the Acidity/Basicity of a Salt
Consider hydrolysis of CN- to produce HCN
- here, HCN is a weak acid and CN- acts as a base
CN- (aq) + H2O (l)
HCN (aq) + OH- (aq)
generally:
- the anions of weak acids are basic
- the anions of strong acid have hardly any basic
character
eg. HCl
Cl- (aq) + H2O (l)
no reaction
to form H3O+ (acidic solution)
Consider NH4+ coming from NH3
- a cation conjugate of a weak base
- NH4+ behaves as an acid with NH3 acting as a base
NH4+ (aq) + H2O (l)
NH3 (aq) + H3O+ (aq)
generally:
- the cations of weak bases are acidic
- the cations of strong bases (metal ions of Groups
IA and IIA elements except Be) have hardly any
acidic character
eg. NaOH
Na+ (aq) + H2O (l)
no reaction
Predicting the Acidity/Basicity of a Salt
These rules apply to normal salts (those in which the anion
has no acidic hydrogen)
1.) A salt of a strong base and a strong acid.
- the salt has no hydrolyzable ions and so gives a
neutral aqueous solution
- eg. NaCl
2.) A salt of a strong base and a weak acid.
- the anion of the salt is the conjugate of the weak
acid. It hydrolyzes to give a basic solution
- eg. NaCN
- aqueous metal ions other than the cations of strong bases
usually hydrolyze by acting as acids
eg. hydrated aluminum ion, Al(H2O)63+ (aq)
- e- are drawn away from O atoms by Al atom so O
atoms draw e- from O-H bonds, making them
highly polar
- Al(H2O)63+ hydrolyzes in aqueous sol n by donating
proton (from acidic H2O molecules on ion) to water
Predicting the Acidity/Basicity of a Salt
3.) A salt of a weak base and a strong acid.
- the cation of the salt is the conjugate of the weak
base. It hydrolyzes to give an acidic solution
- eg. NH4Cl
4.) A salt of a weak base and a weak acid.
- both ions hydrolyze. You must compare the Ka of
the cation with the Kb of the anion
- if the Ka of the cation is larger the solution is
acidic
- if the Kb of the anion is larger, the solution is
basic
3
Predicting the Acidity/Basicity of a Salt
- to predict the acidity or basicity of a salt sol n, examine
acidity or basicity of ions composing salt
Consider KC2H3O2 (potassium acetate)
- potassium is Group IA element does not hydrolyze
- acetate ion is conjugate of acetic acid (weak acid);
therefore, acetate ion is basic
K+ (aq) + H2O (l)
no reaction
C2H3O2 (aq) + H2O (l)
HC2H3O2 (aq) + OH- (aq)
The pH of a Salt Solution
Consider pH of 0.10 M NaCN
- already seen that sol n should be basic due to the
hydrolysis of CN- ion
- require the Kb for CN- (related to Ka for HCN)
Relation between Ka and Kb
- relationship between Ka and Kb for conjugate acidbase pairs
HCN (aq) + H2O (l)
H3O+ (aq) + CN- (aq)
Ka
CN- (aq) + H2O (l)
HCN (aq) + OH- (aq)
Kb
___________________________________________________
2 H2O (l)
H3O+ (aq) + OH- (aq)
Kw
- by the rule of multiple equilibria (chpt. 14)
KaKb = Kw
Consider three possible cases:
a.) Ka > Kb
- if Ka for cation is greater than Kb for anion,
the solution will contain excess of H3O+ ions
(pH < 7)
b.) Ka < Kb
- if Ka for cation is less than Kb for anion, the
solution will contain excess of OH- ions
(pH > 7)
c.) Ka Kb
- if Ka for cation and Kb for anion are
comparable, the solution contains
approximately equal concentrations of H3O+
and OH- ions
(pH 7)
Buffer Solutions
- a solution characterized by the ability to resist changes in
pH when limited amounts of acid or base are added to it
- contain either a weak acid and its conjugate base or a
base and it conjugate acid
- so, buffer solution contains both an acid species and a
base species in equilibrium
Consider buffer system containing approximately equal
amounts of weak acid, HA, and conjugate base, A-
The Common-Ion Effect
- the shift in an ionic equilibrium caused by the addition of a
solute that provides an ion that takes part in the
equilibrium
Note: the common-ion effect is just another example of
Le Chatelier s principle (applies to weak acids and
bases
Consider a solution of HC2H3O2
HC2H3O2 (aq) + H2O (l)
C2H3O2- (aq) + H3O+ (aq)
- now add HCl or C2H3O2HC2H3O2 (aq) + H2O (l)
C2H3O2- (aq) + H3O+ (aq)
added
or
added
*** degree of ionization of acetic acid is decreased by the
addition of a strong acid
OH- added to buffer (strong base)
HA (aq) + OH- (aq)
A- (aq) + H2O (l)
- the pH increases slightly
H3O+ added to buffer (strong acid)
A- (aq) + H3O+ (aq)
HA (aq) + H2O (l)
- the pH decreases slightly
4
Two important characteristics of a buffer:
1.) pH
2.) buffer capacity
- the amount of acid or base the buffer can react
with before giving a significant pH change
- depends on the amount of acid and conjugate
base in the solution
Figure 16.10
- now, use the preceding equation to derive an equation for
the pH of a buffer
- take negative logarithm of both sides of equation
so, - log [H3O+] = - log Ka x [HA] = - log Ka - log [HA]
[A-]
[A-]
Note: left side equals pH and can simplify right side pKa of
weak acid
- defined in same manner as pH and pOH
pKa = -log Ka
Now,
pH = pKa log [HA] = pKa + [A-]
[A-]
[HA]
Henderson-Hasselbalch Equation
How do you prepare a buffer of given pH?
- show that buffer must be prepared from conjugate acidbase pair in which acid-ionization constant is
approximately equal to desired H3O+ concentration
Consider a buffer made of weak acid, HA, and conjugate
base, A-:
HA (aq) + H2O (l)
H3O+ (aq) + A- (aq)
+
to give
Ka = [H3O ][A ]
[HA]
and rearrange to give [H3O+] = Ka x [HA]
[A-]
Henderson-Hasselbaslch equation
- an equation relating the pH of a buffer for
different concentrations of conjugate acid and base
Generally,
pH = pKa + log [base]
[acid]
- by substituting the value of pKa for the conjugate
acid and the ratio [base]/[acid], you obtain the pH
of the buffer
5
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