Acid Base Theory, stronger Lewis acid and base

Acid base Theory

1. Updates
•Assignment 05 is is due today (in class)
•Midterm 2 is Thurs., March 15
–Huggins 10, 7-8pm
–For conflicts: ELL 221, 6-7pm (must arrange at
least one week in advance)

2. Acids and Bases
Chapter 16

3. SAMPLE EXERCISE Lewis Acids and Bases
Describe this reaction according to the Lewis theory of acids
and bases: AlCl
3 (s)
+ Cl
-

(aq)
AlCl
4
-
(aq)
.
AlCl
3
would be the Lewis acid because it has a vacant orbital
that accepts an electron pair from Cl
-
. Therefore, Cl
-
is the Lewis
base.

4. SAMPLE EXERCISE Lewis Acids and Bases
Which would be considered a stronger Lewis acid:
(a) BF
3
or BCl
3
(b) Fe
2+
or Fe
3+
? Explain.
The more electron deficient the central atom with the vacant
orbital, the better the Lewis acid. Three fluorines would pull
electron density away from boron more strongly than three
chlorines so BF
3
is the better Lewis acid in (a). Fe
3+
is more
electron deficient than Fe
2+
so Fe
3+
will be the better Lewis acid.

5. Acid-Base Equilibria and
Solubility Equilibria
Chapter 17

6. The Common-Ion Effect
•Consider a solution of acetic acid:
•If acetate ion is added to the solution,
Le Châtelier says the equilibrium will
shift to the left.
CH
3
CO
2
H(aq) + H
2
O(l) H
3
O
+
(aq) + CH
3
CO
2
−
(aq)

7. The Common-Ion Effect
“The extent of ionization of a weak
electrolyte is decreased by adding to
the solution a strong electrolyte that has
an ion in common with the weak
electrolyte.”

8. The Common-Ion Effect
Calculate the fluoride ion concentration and pH
of a solution that is 0.20 M in HF and 0.10 M in
HCl.
K
a
for HF is 6.8  10
−4
.
[H
3
O
+
] [F
−
]
[HF]
K
a = = 6.8  10
-4

9. The Common-Ion Effect
Because HCl, a strong acid, is also present,
the initial [H
3
O
+
] is not 0, but rather 0.10 M.
[HF], M [H
3
O
+
], M [F
−
], M
Initially 0.20 0.10 0
Change −x +x +x
At Equilibrium 0.20 − x  0.200.10 + x  0.10 x
HF(aq) + H
2
O(l) H
3
O
+
(aq) + F
−
(aq)

10. The Common-Ion Effect
= x
1.4  10
−3
= x
(0.10) (x)
(0.20)
6.8  10
−4
=
(0.20) (6.8  10
−4
)
(0.10)

11. The Common-Ion Effect
•Therefore, [F
−
] = x = 1.4  10
−3
[H
3O
+
] = 0.10 + x = 0.10 + 1.4  10
−3
= 0.10 M
•So,pH = −log (0.10)
pH = 1.00

12. We say that the second solution is able to
“buffer” against changes in pH. Why does the
solution containing carbononic acid and its
conjugate base resist a change in pH?
•When 0.01 mmol of HCl is added to 1 mL of
water at pH 7, the pH drops to 2.00.
•When 0.01 mmol of HCl is added to 1 mL of
water containing 0.16 mmol of H
2CO
3/HCO
3
-

at pH 7, the pH drops to 6.92.

13. HCl H
+
+ Cl
-
HCl + H
3CCO
2
-
H
3CCO
2H

+ Cl
-
17.3

14. Buffers:
•Solutions of a weak
conjugate acid-base
pair.
•They are particularly
resistant to pH
changes, even when
strong acid or base is
added.

15. Buffers
If a small amount of hydroxide is added to an
equimolar solution of HF in NaF, for example, the HF
reacts with the OH
−
to make F
−
and water.

16. Buffers
If acid is added, the F
−
reacts to form HF and water.

17. Buffer Calculations
Consider the equilibrium constant
expression for the dissociation of a
generic acid, HA:
[H
3
O
+
] [A
−
]
[HA]
K
a =
HA + H
2
O H
3
O
+
+ A
−

18. Buffer Calculations
Rearranging slightly, this becomes
[A
−
]
[HA]
K
a =[H
3
O
+
]
Taking the negative log of both side, we get
[A
−
]
[HA]
−log K
a
=−log [H
3O
+
] + −log
pK
a
pH
acid
base

19. Buffer Calculations
•So
pK
a = pH − log
[base]
[acid]
•Rearranging, this becomes
pH = pK
a + log
[base]
[acid]
•This is the Henderson–Hasselbalch equation.
•When [base] = [acid], pH = pK
a

20. Henderson–Hasselbalch
Equation
What is the pH of a buffer that is 0.12 M
in lactic acid, HC
3
H
5
O
3
, and 0.10 M in
sodium lactate? K
a for lactic acid is
1.4  10
−4
.

21. Henderson–Hasselbalch
Equation
pH = pK
a
+ log
[base]
[acid]
pH = −log (1.4  10
−4
) + log
(0.10)
(0.12)
pH = 3.85 + (−0.08)
pH = 3.77

22. pH Range
•The pH range is the range of pH values
over which a buffer system works
effectively.
•It is best to choose an acid with a pK
a
close to the desired pH.

23. When Strong Acids or Bases
Are Added to a Buffer…
…it is safe to assume that all of the strong acid
or base is consumed in the reaction.

24. Addition of Strong Acid or
Base to a Buffer
1.Determine how the neutralization
reaction affects the amounts of
the weak acid and its conjugate
base in solution.
2.Use the Henderson–Hasselbalch
equation to determine the new
pH of the solution.