Transcript 4-4-2011 - Professor Monzir Abdel

4-4-2011
1
Choosing a Buffer
• Pick a buffer whose acid has a pKa near the
pH that you want
• Must consider pH and capacity
Buffer Capacity: Not all buffers resist changes
in pH by the same extent. The effectiveness
of a buffer to resist changes in pH is called
the buffer capacity.
Buffer capacity is the measure of the ability of
a buffer to absorb acid or base without
significant change in pH.
2
Buffer capacity increases as the concentration
of acid and conjugate base increases.
1. Larger volumes of buffer solutions have a
larger buffer capacity than smaller volumes
with the same concentration.
2. Buffer solutions of higher concentrations
have a larger buffer capacity than a buffer
solution of the same volume with smaller
concentrations.
3. Buffers with weak acid/conjugate base ratio
close to 1 have larger capacities
3
- Calculate the pH of a solution which is 0.40 M in
acetic acid (HOAc) and 0.20 M in sodium acetate
(NaOAc). Ka=1.7x10-5
Initial M:
Change
Equil M:
HOAc
0.40
-x
0.40-x
0.20 >> x
1.7x10-5 = [H+][OAc-]
[HOAc]
=
H+ + OAc0
0.20
+x
+x
x
0.20+x
[x][0.20]
[0.40]
You should observe that 0.2>>x
pH = - log 3.4x10-5 = 4.47
4
x = [H+] = 3.4x10-5 M
What is the pH of a solution containing 0.30 M HCOOH
and 0.52 M HCOOK?
HCOOH (aq)
Initial (M)
Change (M)
0.30
0.00
0.52
-x
+x
+x
x
0.52 + x
Equilibrium (M) 0.30 - x
0.30 – x  0.30
0.52 + x  0.52
5
H+ (aq) + HCOO- (aq)
Ka = {0.52 * x}/0.30
1.7*10-4 = 0.52*x/0.3
X = 9.4*10-5
Again 0.3 is really much greater than x
X = [H+] = 9.4*10-5 M
pH = 4.02
6
Calculate the pH of a solution that is
0.025 mol/L HCN and 0.010 mol/L NaCN.
(Ka of HCN = 4.9 x 10-10)
HCN D H+ + CNEven if you do not know it is a buffer
you can still work it out easily:
7
(all in mol/L)
Initial conc.
Conc. change
Equil. conc.
+
-
H O (l)
D H O (aq) +
CN (aq)
0.025
N/A
0.0
0.010
-x
N/A
+x
+x
0.025 – x
N/A
+x
HCN(aq) +
2
3
0.010 + x
Ka = {x*(0.01+x)}/(0.025 – x), assume 0.01>>x
However if you observe it is a buffer you
could work it without including the “x”
as it is always very small.
4.9X10-10 = {x*(0.01)}/(0.025)
X = [H+] = 1.2*10-9 M, and pH = 8.91
8
A diprotic acid, H2A, has Ka1 = 1.1*10-3 and Ka2 =
2.5*10-6. To make a buffer at pH = 5.8, which
combination would you choose, H2A/HA- or
HA-/A2-? What is the ratio between the two
buffer components you chose that will give
the required pH?
The weak acid/conjugate base system should
have a pKa as close to the required pH as
possible.
pKa1 = 2.96, while pKa2 = 5.6
It is clear that the second equilibrium should
be used (HA-/A2-).
9
The ratio between HA- and A2- can easily be
found from the equilibrium constant
expression where:
HA- D H+ + A2Ka2 = {[H+][A2-]}/[HA-]
Ka2/[H+] = [A2-]/[HA-]
[H+] = 10-5.8 = 1.6*10-6 M
{2.5*10-6/1.6*10-6} = [A2-]/[HA-]
[A2-]/[HA-] = 1.56
10
Calculate the pH of the 0.30 M NH3/0.36 M NH4Cl
buffer system. What is the pH after the addition
of 20.0 mL of 0.050 M NaOH to 80.0 mL of the
buffer solution? Kb = 1.8*10-5
NH4+ (aq)
H+ (aq) + NH3 (aq)
First, get initial pH before addition of any base
Ka = {[H+][NH3]}/[NH4+]
(10-14/1.8*10-5) = {[H+] * 0.30}/0.36
[H+] = 6.7*10-10
pHinitial = 9.18
11
Now calculate the pH after addition of the base:
Addition of the base will decrease NH4+ and will
increase NH3 by the same amount due to a
1:1 stoichiometry, which is always the case.
NH4+ (aq)
H+ (aq) + NH3 (aq)
mmol of NH4+ = 0.36*80 – 0.05*20 = 27.8
mmol NH3 = 0.30*80 + 0.05 * 20 = 25
Can use mmoles instead of molarity, since both
ammonia and ammonium are present in
same solution
12
NH4+ (aq)
H+ (aq) + NH3 (aq)
Ka = {[H+][NH3]}/[NH4+]
(10-14/1.8*10-5) = {[H+] * 25}/27.8
[H+] = 6.2*10-10
pHfinal = 9.21
DpH = pHfinal – pHinitial
DpH = 9.21 – 9.18 = + 0.03
13
A 0.10 M acetic / 0.10 M acetate mixture has a pH of
4.74 and is a buffer solution! What happens if we add
0.01 mol of NaOH (strong base) to 1.00 L of the acetic
acid – acetate buffer solution?
This reaction goes to completion since OH- is a
strong base and keeps occurring until we run out of
the limiting reagent OH-
CH 3COOH (aq) + OH (aq) g H2 O (l) + CH 3COO (aq)
(all in moles)
Initial
Change
Final (where x = 0.01
-
due tolimiting OH )
14
0.10
0.01
N/A
0.10
-x
-x
N/A
+x
0.10 – x
0.01 – x =
N/A
0.10 + x
= 0.09
0.00
= 0.11
(all in mol/L)
Initial conc.
Conc. change
Equil. conc.
CH COOH (aq) +
3
H O (l)
2
+
D H O (aq) + CH COO (aq)
3
3
0.09
N/A
0.0
0.11
-x
N/A
+x
+x
0.09 – x
N/A
+x
0.11 + x
[H 3O  ][CH 3COO ]
(x)(0.11  x)
5
Ka 
 1.8 x 10 
[CH 3COOH]
(0.09  x)
With the assumption that x is much smaller than
0.09 mol (an assumption we always need to check,
after calculations are done!), we find
pH = - log [H3O+]
pH = - log 1.5 x 10-5
pH = 4.82
15
What happens if we add 0.01 mol of HCl (strong acid)
to 1.00 L of the acetic acid – acetate buffer solution?
CH3COO- (aq) + H3O+ (aq) → H2O (l) + CH3COOH (aq)
This reaction goes to completion and keeps
occurring until we run out of the limiting reagent H3O+
-
+
CH3COO (aq) + H3O (aq) →
(all in moles)
Initial
0.10
0.01
Change
-x
-x
Final (where x = 0.01
0.10 – x
0.01 – x =
due limiting H3O+)
= 0.09
0.00
H2O (l) + CH3COOH (aq)
N/A
0.10
N/A
+x
N/A
0.10 + x
= 0.11
New [CH3COOH] = 0.11 M and new [CH3COO-] = 0.09 M
16
(all in mol/L)
Initial conc.
Conc. change
Equil. conc.
CH 3COOH (aq) +
H 2O (l)
D
+
-
H3O (aq) + CH 3COO (aq)
0.11
N/A
0.0
0.09
-x
N/A
+x
+x
0.11 – x
N/A
+x
0.09 + x
[H 3O  ][CH 3COO ]
(x)(0.09  x)
5
Ka 
 1.8 x 10 
[CH 3COOH]
(0.11  x)
With the assumption that x is much smaller than 0.09
mol (an assumption we always need to check, after
calculations are done!), we find
Note we’ve made the assumption that x << 0.09 M!
pH = - log [H3O+]
pH = - log 2.2 x 10-5
pH = 4.66
17
Calculate the pH of a 0.100 L buffer solution
that is 0.25 mol/L in HF and 0.50 mol/L in NaF.
(all in mol/L)
HF (aq) +
Initial conc.
Conc. change
Equil. conc.

H2 O (l)
D
+
H3 O (aq)
+
-
F (aq)
0.25
N/A
0.0
0.50
-x
N/A
+x
+x
0.25 – x
N/A
+x
0.50 + x

[H 3O ][F ]
(x)(0.50  x)
4
Ka 
 3.5 x 10 
[HF]
(0.25  x)
Assume 0.25>>x, which is always safe to
assume for buffers
18
[HF]
4 [HF]
[H3 O ] Ka  3.5 x 10

[F ]
[F ]
4 0.25
4
3.5 x 10
1.8 x 10 M
0.50

pH = - log [H3O+]
pH = - log 1.8 x 10-4
pH = 3.76
No. mol HF = 0.25*0.1 = 0.025
No. mol F- = 0.50*0.1 = 0.050
19
a) What is the change in pH on addition of 0.002 mol
of HNO3? It is easier here to use moles not molarity
-
+
+ H3 O (aq) g H 2O (l) +
(all in moles)
F (aq)
Initial
0.050
0.002
N/A
0.025
-x
-x
N/A
+x
0.050 – x
0.002 – x
N/A
0.025 + x
= 0.048
= 0.00
Change
Final (where x = 0.002
+
due tolimiting H3 O )
HF (aq)
= 0.027
Use moles: New HF = 0.027 mol and new F- = 0.048 mol
0.027
N/A
0.00
0.048
-x
N/A
+x
+x
0.027 – x
N/A
+x
0.048 + x
Ka 
20


H3O (aq) +
-
Initial conc.
Equil. conc.
H 2O (l)
+
HF (aq)
Conc. change
+
D
(all in mol/L)
[H3 O ][F ]
(x)(0.048 )
4
 3.5 x 10 
[HF]
(0.027 )
F (aq)
[HF]
 4 0.027
4



[H3 O ] Ka 
3.5 x 10
2.0 x 10 M
[F ]
0.048

Notice we’ve made the assumption that
x << 0.027 M.
pH = - log [H3O+]
pH = - log 2.0 x 10-4
pH = 3.71
21
b) What is the change in pH on addition of
0.004 mol of KOH?
(all in moles )
Initial
Change
Final(where x = 0.0 4
due tolimiting OH- )
HF (aq) + OH - (aq) gH 2O (l) + F - (aq)
0.004
0. 025
0.050
N/A
-x
-x
N/A
+x
0.025 – x
0.050 + x
0.04 – x
N/A
= 0.021
= 0. 05
= 0.00
4
-
Use moles: New HF = 0.021 mol and new F = 0.054 mol
(all in mol/L)
Initial conc.
Conc. change
Equil. conc.
HF(aq) +
0.021
-x
0.021 – x

Ka  3.5x10 4
22
H2O(l)
N/A
N/A
N/A


D H 3O+ (aq)
F- - (aq)
0.00 + 0.054
+x
+x
+x
0.054+ x
[H 3O ][F ]
 4 (x)(0.054 )

so 3.5x10 
[HF]
(0.021)
[HF]
4
4 0.21
1.4 x10
[H3 O ]  Ka  3.5 x 10
[F ]
0.54

Notice we’ve made the assumption that
x << 0.21 M. We should check this!
pH = - log [H3O+]
pH = - log 1.4 x 10-4
pH = 3.87
23
M