4/20/2015
Effect on pH for Different Starting Conditions
acid
Fundamental
Acid
Ionization:
Chapter 15:
K a→
HNO2 (aq) + H2O
c. base
NO2– (aq) + H3O+
Acid + water
base
‒
Applications of
Aqueous Equilibria
‒
HNO2 (aq) + OH
Salts:
NO2 (aq) + H2O
← Kb
Salt + water
Common
Ions:
Part A: Acids and Bases
HNO2 (aq) + H2O
Ka
Acid + water
NO2– (aq) + H3O+
+ Common ion
Kn
Neutralization:
Chem 112
Dr. Gentry
HNO2 (aq) + NaOH (aq)
NaNO2– (aq) + H2O
Acid + Neutralization Base
Part A: Soluble Salts in Water
But What Happens When Salt is Placed in Water?
Do they affect the pH?
Answer: Depends on the cation and anion
NaCl
KNO3
KNO2
NH4Br
1) Soluble salt in water → dissociates into ions
B+ (aq) + A‒(aq)
BA (s)
?
base
NO2– (aq) + H2O
HA (aq) + OH‒
Yes:
acid
B+ (aq) + OH‒
?
acid
Left to right
will decrease pH
NO2‒ / HNO2 are reversible weak base / acid pair
K+ (aq) + 2 H2O
B + H+ (aq)
acid
base
or
HNO2 (aq) + OH‒
2b) Does cation K+ act like an acid
2b) Does cation B+ act like an acid?
?
?
base
acid
BH (aq)
NO2– (aq) + K+(aq)
KNO2 (s)
2a) Does anion NO2‒ act like a base
2a) Does anion A‒ act like a base?
A– (aq) + H2O
1) Soluble salt in water → dissociates into ions
No:
BOH (aq)
?
KOH (aq) + H3O+
base
Left to right
not possible
- no effect on pH
KOH is strong base… only goes in one direction
towards the K+
base
Soluble Salts in Water
Conjugate Bases in Water
(adding only NO2‒ to water… if it were possible)
K+ (aq) + NO2– (aq)
KNO2 (s)
does K+
affect pH?
Acid + water
does NO2–
affect pH?
NO
YES
(no reverse rxn
since KOH is strong base)
(NO2‒ produces OH‒
since HNO2 is weak acid)
HNO2 (aq) + H2O
NO2– (aq) + H3O+
acid
base
−
NO2– (aq) + H2O
HNO2 (aq) + OH‒
base
acid
acid (K+) in water… no rxn
K+(aq) + OH– (aq)
KOH (aq)
acid
base
raise pH
Kb,NO − =
2
[OH ][HNO3 ]
[NO2 − ]
[A − ][H3O + ]
[HA]
Kb =
[OH− ][HA]
[A − ]
increase pH
Conjugate base + water
NO2– (aq) + H2O
base (NO2‒) in water
Ka =
base
HNO2 (aq) + OH‒
acid
decrease pH
Equations are NOT simply the reverse of each other
In each case, water is added to the left side reactant
Kb =
Kw
Ka
1
4/20/2015
Calculate the pH of a 0.15 M solution of
potassium nitrite (KNO2 ).
Acid–Base Properties of Salts
(i.e., what happens when NO2- is added to water?)
If given a soluble salt in water…
NH4Br(aq)
1) Break the salt into its two sets of ions
NH4+ (aq) + Br − (aq)
Acid ionization reaction for nitrous acid
NO2‒ + H3O+
HNO2 + H2O
Ka = 4.5×10–4
Conjugate base ionization reaction for nitrite ion
NO2‒ + H2O
HNO2 + OH‒
Kb = Kw / Ka
K
10 −14
Kb = w =
= 2.2 * 10 −11
K a 4.5 * 10−4
Kb = 2.2 * 10−10 =
[HNO2 ][OH ]
x
≅
[NO2 − ]
.15
- Is the new base a strong base or a weak base
NH3 is a weak base
Br − → HBr
pOH = 5.74
x = [HNO2 ] = [OH− ] = 1.8 * 10−6 M
- Identify what its base is after losing a H+ or adding a OH‒
NH4+ (aq) → NH3
3) A negative anion can gain a H+ (therefore it is a base)
- Identify what its acid is after gaining the H+
2
−
2) A positive cation can lose a H+ or gain a OH‒ (so it is an acid)
pH
- Is the new acid a strong acid or a weak acid
HBr is a strong acid
= 8.26
Acid–Base Properties of Salts
Acid–Base Properties of Salts
(summary)
• Salts made from strong acids and strong bases
(NaCl)
⇒ neutral solutions
no effect on pH
• Salts made from weak acids and strong bases
⇒ basic solutions
(NaNO2)
pH > 7
• Salts from strong acids and weak bases
⇒ acidic solutions
(NH4Cl)
pH < 7
• Salts from weak acids and weak bases
⇒ depends on relative
strengths of acid & base
(NH4NO2)
pH = ?
Effect on pH for Different Starting Conditions
acid
Fundamental
Acid
Ionization:
K a→
HNO2 (aq) + H2O
c. base
NO2– (aq) + H3O+
Acid + water
• Predict whether the following solutions will be
acidic, basic, neutral, or ambiguous:
(a) NH4Br
NH3
HBr
= weak base
= strong acid
Final pH < 7
(b) CaCl2
CaOH
HCl
= strong base
= strong acid
Final pH = 7
(c) KCN
KOH
HCN
= strong base
= weak acid
Final pH > 7
(d) NH4CH3CO2
NH3
= weak base
HC2H3O2 = weak acid
Final pH ???
Part B:
Acid/Base Common Ions
and Buffer Solutions
base
‒
Salts:
‒
HNO2 (aq) + OH
NO2 (aq) + H2O
← Kb
Salt + water
Common
Ions:
HNO2 (aq) + H2O
Ka
Acid + water
NO2– (aq) + H3O+
+ Common ion
Both HNO2 AND NO2– Are Initially Present
0.01M
HNO2 (aq) + H2O
0.02M
NO2– (aq) + H3O+
Ka =
[A − ][H3O + ]
[HA]
Kn
Neutralization:
HNO2 (aq) + NaOH (aq)
NaNO2– (aq) + H2O
Acid + Neutralization Base
2
4/20/2015
Common-Ion Effect
0.10 M Acetic Acid +
Ka = 1.8*10-5
• Calculate the pH of a 0.10 M HC2H3O2 solution
with no salt added.
Using ICE table: [H3O+] = 1.3*10-3
0.12 M Sodium Acetate
H C2H3O2
Na C2H3O2
pH = 2.87
H3O+ (aq) + C2H3O2‒ (aq)
HC2H3O2 (aq) + H2O
H3O+ (aq) + C2H3O2‒ (aq)
HC2H3O2 (aq) + H2O
Shift in reaction caused by addition of a second compound
having an ion in common with system at equilibrium
• Calculate the pH of a 0.10 M HC2H3O2 solution
with 0.12 M NaC2H3O2 also present.
Ka = 1.8*10-5
Using ICE table on next slide:
[H3O+] = 1.5*10-5
pH = 4.82
not as much H3O+
from
previous
page
Common-Ion Effect
Calculate the pH of a 0.10 M HC2H3O2 solution
with 0.12 M NaC2H3O2 also present.
0.10 M HC2H3O2 (acetic acid)
+
0.12 M Na C2H3O2 (sodium acetate)
Ka =
HC2H3O2 + H2O
pH
Ka = 1.8*10-5
Ka =
[H3O+]
[C2H3O2 − ] [H+ ]
= 1.8 * 10 −5
[HC2H3O2 ]
Change
Equilibrium
HC2H3O2
H+
0.10
0
‒x
+x
+x
0.10 ‒ x
+x
0.12 + x
+
Addition of
CH3CO2‒
0.10M HCH3CO2
+ 0.12M CH3CO2‒
C 2H 3O 2‒
0.12
10-5M
5.0
[H3O+] = 1.5*10-5
0
0.040
12
Titration Curve
10
add strong base to HNO2
8
H C2H3O2
Na C2H3O2
pH
H3O+ (aq) + C2H3O2‒ (aq)
Ka = 1.8*10-5
A) If equal amount
weak acid and its
conj. base,
pH = pKa
buffer region
6
pKa
4
HC2H3O2 (aq) + H2O
0.120
Buffer Solutions
Shift in reaction caused by addition of a second compound
having an ion in common with system at equilibrium
0.12 M Sodium Acetate
0.080
Concentration of Added CH3CO2- (M)
pH = 4.82
Common-Ion Effect
0.10 M Acetic Acid +
[C2H3O2 − ] [H3 O+ ]
[HC2H3 O2 ]
10-4M
4.0
Initial
0.10M HCH3CO2
10-3M
3.0
H3O+ + C2H3O2‒
(HNO2
2
0
NO2–)
B) Subsequently
adding small
amount acid or
base has only
small effect on
pH
20
40
mL of 0.100M NaOH added
3
4/20/2015
 [ Conj. Base ]
pH = pK a + log 
[ Acid]

Buffer Solutions



0.20 M Acetic Acid +
If equal amount
weak acid and
its conj. base,
pH = pKa
add
base
pH
Buffer Solutions
(pH slow to respond if add more acid or base)
0.30 M Sodium Acetate
H C2H3O2
Na C2H3O2
• A buffer solution is a solution of:
1) A weak acid or base
2) AND the salt of the same weak acid or base
add
acid
• Both acid and base must be present!
a) base may be added outright,
b) or can be created by adding strong base to wk. acid
pKa
• Buffer solutions play a critically important role in
controlling pH in our bodies
• A buffer solution must have both:
1) a weak acid or base
2) AND the salt of the same weak acid or base
Which of the following are buffer systems?
(a) 1 mol KF / 1 mol HF
(c) 1mol HNO2 / 0.5 mole NaOH
(b) HBr is a strong acid
⇒ not a buffer solution
(d) HNO2 is weak acid, and 1.0 mol NaOH will create ALL NO2−
⇒ not a buffer solution
CONJ. BASE
HF
7.1 x 10 –4
3.15
F–
HNO2
4.5 x 10 –4
3.35
NO2 –
C9H8O4 (aspirin)
3.0 x 10 –4
3.52
C 9H 7O 4 –
HCO2H (formic)
1.7 x 10 –4
3.77
HCO2 –
C6H8O6 (ascorbic)
8.0 x 10
–5
4.10
C 6H 7O 6 –
C6H5CO2H (benzoic)
6.5 x 10 –5
4.19
C6H5CO2 –
CH3CO2H (acetic)
1.8 x 10 –5
4.74
CH3CO2 –
HCN
4.9 x 10 –10
9.31
CN –
C6H5OH (phenol)
1.3 x 10
–10
10.1



with
pK a = − log(K a )
C 6H 5O
pK a = − log(1.8 * 10−5 ) = + 4.74
 0.12 
pH = + 4.74 + log 
 =
 0.10 
4.82
(same result as
earlier ICE slide)
• How would you prepare an “acetate buffer” with a pH of
about 5.50.
Ka = 1.8*10–5
Acid Ionization Constants
pKa
 [ Conj. Base ]
pH = pK a + log 
[ Acid]

Calculate the pH of a solution containing:
0.10 M HC2H3O2 and 0.12 M NaC2H3O2.
Ka = 1.8*10-5
(c) HNO2 is weak acid, and 0.5 mol NaOH will create some NO2−
⇒ buffer solution
Ka
Simplified alternative to ICE table for acid + salt
If [HA] and [salt] are much greater than Ka value:
(b) 1 mol HBr / 1 mol KBr
(d) 1 mol HNO2 / 1 mol NaOH
(a) HF is a weak acid and F- is its conjugate base
⇒ buffer solution
ACID
Henderson–Hasselbalch equation
 base 
pH = pK a + log 

 acid 
pK a = − log ( K a ) = 4.74
 base 
5.50 = 4.74 + log 

 1.00 
set acid = 1.0 mole
base
5.8 mole base
= 10 ( 5.50− 4.74 ) =
1.00
1.0 mole acid
Mix solution with ratio
of 5.8 mole Acetate
salt for every 1.0 mole
Acetic Acid
H-H equation will give the desired mole ratio (which is
equal to molarity ratio) of conjugate base to acid.
–
Volume does not matter since base and acid in same
volume of sol’n. All that matters is ratio of amount of
acid & base
4
4/20/2015
H3O+ (aq) + C6H5CO2H ‒ (aq)
C6H5CO2H (aq) + H2O
Benzoic Acid
( Ka = 6.5*10-5 , pKa = 4.187 )
Using Henderson-Hasselbalch Equation
with Weak Bases
Using ICE table:
6.5 * 10−5 =
x = [H3O+] = 4.6*10-3
(x) (x)
x2
≅
(0.32 − x) 0.32
Using Henderson-Hasselbalch equation
 [ Conj. Base]
pH = pK a + log 
[ Acid]

Ammonia

 0.56 
 = 4.187 + log 

 0.32 

pH = 4.43
( Kb = 1.8*10-5 ,
pKb = 4.75)
• H-H equation is written for weak acids and their pKa’s
 [Base ] 
pH = pK a + log 

 [ Acid] 
pH = 2.34
• Calculate the pH of a 0.32 M C6H5CO2H solution
with 0.56 NaC6H5CO2 added.
NH4+ (aq) + OH ‒ (aq)
NH3 (aq) + H2O
• Calculate the pH of a 0.32 M C6H5CO2H solution.
• Must convert Kb to Ka
Ka =
K w 1* 10 −14
=
Kb
Kb
Calculate pH of 0.18 M NH3 solution with 0.23M NH4Cl
Ka =
Kw
1* 10 −14
= 5.56 * 10−10
=
Kb 1.8 * 10−5
pK a = − log(5.56 * 10−10 ) = 9.25
 NH3 
pH = 9.25 + log  
 NH4 + 






pH = 9.14
Test 4: Monday, April 27
Cover up to this point in Chapters 14 & 15 including:
- Calculations involving weak acids and pKa’s
(on previous test, but still need for this material)
- Calculations involving weak bases and pKb’s
- % Dissociation
- Polyprotic Acids (general trends)
- Addition of salts to water
- Buffer solutions
Will NOT include:
- Titration curves
- Ksp’s and solubility
5