Transcript Slide 1

“A” students work
(without solutions manual)
~ 10 problems/night.
Dr. Alanah Fitch
Flanner Hall 402
508-3119
[email protected]
Office Hours Th&F 2-3:30 pm
Module #17B:
Acid Base Ionization
Computations
To determine if our protein unfolds or not is it
sufficient to know the Ka and Kb
Values of the functional groups?
2
1
3D structure of ALAD directs
Reactants into proper orientation
3D structure controlled by proper hydrogen and ionic bonds, pH dependent!!!!!!!
http://www.jenner.ac.uk/PPD/
http://www.biology.arizona.edu/biochemistry/problem_sets/aa/aa.htm
Ionization constants for proteins
O C
S-
S-
Zn2+
-
O
H
S

RCOOHaq  H2 O 
RCOO

H
O

aq
3 aq


RSHaq  H2 O 
RS

H
O

aq
3 aq
R refers to rest of protein
cysteine
Ka ,COOH  10
. x10 2
Ka,SH  6.3x10 9
Phosphoenolate carboxylase, human, cys
Ka  7.94 x10 9
Control of protein shape is due to fraction of sites charged
Protein folding due to
FRACTION of sites charged
Ka,Val , NH3  2.5x1010
Ka ,Val ,COOH  501
. x10 3
Ka,tyr ,ROH  10
. x1010
Hemeglobin
% Ionized (dissociated)

HAaq 
H
O

3 aq , from
HA
 Aaq
 amount dissociated ( M ) 
%dissociation  100

initial
concentration
(
M
)


 

Aaq
% dissociation  100
 HA
aq ,initial







This seems pretty straight forward
Calculating [A-]
This will require knowing [H3Oaq+]

aq 
HA
Ka



aq , from HA
H3 O

Aaq ,eq H 3 Oaq ,eq
 HA 
eq ,aq


aq
 A
AND a new vocabulary for
comparing the solution
acidity from experiment to
experiment
Ka and Kb tell us about the possibility
Of donating protons, not what the solution
Acidity is
Define another comparison number: pH
“A” students work
(without solutions manual)
~ 10 problems/night.
Dr. Alanah Fitch
Flanner Hall 402
508-3119
[email protected]
Office Hours Th&F 2-3:30 pm
Module #17B:
Acid Base Ionization
Computations
pH, pOH, pKa
LAZY
Chemists are _____
To shorten calculations use log
log ab  log a  logb
 a
log   log a   log b p x   log x
 b
x
log a  x log a
1
x
log x  loga    x log a
a 
K



OH
H
O
 eq  3 eq
 H O

2
2
Assumption that 55.5 molar is relative unchanged

K H2 O

2


H O 
 OH  eq  H 3 O  eq
2



K 5555
.
 OH eq

3
eq
K5555
.   KW  1014
2


KW  10 14  OHeq H3Oeq



KW  10 14  OHaq H3 Oaq
log KW  log10

 14  log OH

14

aq

  logOH

aq
  log H O 
14   log OH
3

aq
  logH O 
3

aq

aq
   log H O 
3

aq
p x   log x
14  pOH  pH
pH scale runs from 0 to 14
Which is more
Acidic?
acid/base ave [H+]
base blood
5.01x10-8
saliva
1x10-7
acid urine
2.51x10-7
cow’s milk 3.54x10-7
cheese
7.94x10-6
pH
pOH ave [OH-]
7.3 6.7
7
You do the rest
pH   log501
. x108   7.3
# is slightly larger than 10-8, so I know it is
7….. something
14-7.3=6.7
1.99x10-7
[OH  ]  10 6.7  1995
. x10 7
p x   log x
14  pH  pOH
x  10
 pX
What are the two
pKas?
px    1log x
O C


pKa    1 log6.3x10  9 
S-
-
O
H
pKa    1  8.2  8.2


RCOOHaq  H2 O 
RCOO

H
O

aq
3 a
cysteine
What is the Ka of a compound
Whose pKa is 3.7?
x  10  px
 3.7
4
Ka  10  199
. x10
Ka ,COOH  10
. x10 2
Ka,SH  6.3x10 9
“A” students work
(without solutions manual)
~ 10 problems/night.
Dr. Alanah Fitch
Flanner Hall 402
508-3119
[email protected]
Office Hours Th&F 2-3:30 pm
Module #17B:
Acid Base Ionization
Computations
Calculating [A-]
This will require an equilibrium calculation

aq 
HA
Ka



aq , from HA
H3 O

Aaq ,eq H 3 Oaq ,eq
 HA 
eq ,aq


aq
 A
Generalized Strategy involves comparing Kas
1. Write down ALL possible reactions involving a proton
2. Excluding water, identify all the proton donors as
1. Strong acid
a. Strong electrolyte: HNO3, HCl, H2SO4 (No Clean Socks)
b. Give all strong acid protons to water or alpha dog
c. Calculate hydronium conc.
d. Calculate pH
2. Weak Acid
a. Identify strongest acid (omega dog, can not hold protons)
b. Has largest Ka; smallest charge density anion
c. Calculate how many protons omega gives up (equil)
d. Calculate pH
e. Use to determine what alpha gets
Example Calculations
1.HCl
2.Acetic acid (vinegar)
2.HF
3.B(OH)3 (Boric acid (eye wash))
4.Mixture (HF and phenol)
5.Mixture (H2SO4, HSO4-)
6.Triethylamine
7.NaAcetate
8.Our heme example
Calculate the pH of 0.004 M HCl
Generalized Strategy involves comparing Kas
1. Write down ALL possible reactions involving a proton


HClaq 
H

Cl

aq
aq


H2 O 
H

OH

aq
aq
Kw  10 14
2. Excluding water, identify all the proton donors as
1. Strong acid
a. Strong electrolyte: HNO3, HCl, H2SO4 (No Clean Socks)
Cl  low ch arg e density Omega dog
aq
b. Give all strong acid protons to water or alpha present
H2 O  HClaq
give all bone to water

complete reaction
H3Oaq  Claq
c. Calculate hydronium conc.
d. Calculate pH
H3Oaq  0.004 M
H3Oaq  0.004 M
Scientific notation allows you to quickly check if
Your answer is in the right “ballpark”
pH   log4x10 3 
3


pH   log 4   log10 
pH   log 4     3
pH    0.602  3
pH  2.39
pH has to be slightly less
Than 3
Example Calculations
1.HCl
2.Acetic acid (vinegar)
2.HF
3.B(OH)3 (Boric acid (eye wash))
4.Mixture (HF and phenol)
5.Mixture (H2SO4, HSO4-)
6.Triethylamine
7.NaAcetate
8.Our heme example
Example: What is the % ionization of commercial
vinegar? The label reads 5% acidity (by weight).
Vinegar is acetic acid which has the formula
HC2H3O2. (CH3COOH) Ka = 1.8x10-5
Density of 5% acetic acid 1.0023 g/mL
1. Write down ALL possible reactions involving a proton


CH3COOHaq  H2 O 
H
O

CH
COO

3 aq ,
3
aq


H2 O 
H

OH

aq
aq
Ka  18
. x10  5
Kw  10 14
2. Excluding water, identify all the proton donors as
1. Strong acid
a. Strong electrolyte: HNO3, HCl, H2SO4 (None)
2. Weak Acid:


CH3COOHaq  H2 O 
 H3 Oaq ,  CH3 COOaq
Ka  18
. x10  5
a. Calculate how many protons omega gives up (equil)
Example: What is the % ionization of commercial
vinegar? The label reads 5% acidity (by weight).
Vinegar is acetic acid which has the formula
HC2H3O2. (CH3COOH) Ka = 1.8x10-5
Density of 5% acetic acid 1.0023g/mL
Don’t Know
Know
Need the initial molarity
need the final dissociation
% by wt.
Ka = 1.8x10-5
 5g acetic acid 

 100 g solution  5g acetic acid
 100 g solution 
 1moleacetic acid 

5gacetic acid 
5g acetic acid
60
g

acetic acid 
0.083mole


 0.835 M
0.0997 L
0.0997 L
 1mLsolution   1L 


100g solution  10023
.
g solution   10 3 mL 

 1moleacetic acid



 212  41  216 



.
M
HC H O   08355
2
3
2
Why complicate this situation by adding in 10-7
When we get rid off it with an assumption?
Because it creates a habit necessary of multiple rx
H2O
OH55.5
10-7
HC2H3O2
C2H3O2stoic
1
1
conc. init
5%
0
[Init]
0.8355
0
Change
-x
+x
Assume
0.8355>>x
+x
[Equil]
0.8355
+x
What is the % ionization of commercial
vinegar? The label reads 5% acidity (by
weight). Density of 5% acetic acid is 1.0023
g/mL. Vinegar is acetic acid which has the
formula HC2H3O2. (CH3COOH) Ka = 1.8x10-5
Ka 
 x  x  10  7 
H+
10-7
H+
1
10-7
10-7
+x
10-7 << x
+x

x2
 HA   x  HA 
init
init
18
. x10  5
x2

0.8355
HA
0.8355>>x
Assume
08355
.
 Ka   x
A
+x
2
0.8355 Ka   x
x
18. x105  0.8355  0.003878
7
10
  0.003878
Check:
0.0038780
+
0.0000001
0.0039779
Sig figs
H+
10-7 << x
08355
.
 0003878
.
?
0.835500
0.003878
0.831622
Original sig figs
were = 0.83
So if we round to
2 sig fig, have
Same answer
pH =-log(0.003878)= 2.41


 A
aq ,eq

% dissociation  100
 HA
aq ,initial







 0.003878 

 100  0.46%
 0.835 
What is the % ionization of commercial
vinegar? The label reads 5% acidity (by
weight). Density of 5% acetic acid is 1.0023
g/mL. Vinegar is acetic acid which has the
formula HC2H3O2. (CH3COOH) Ka = 1.8x10-5
How does % dissociation or ionization
vary with concentration?
[Acetic Acid]
1.00 M
0.835M
0.1 M
% ionization
0.42%
0.46%
1.3% Observations?
% ionization increases with the lower molarity.
Why should this be so?
What is the % ionization of commercial
vinegar? The label reads 5% acidity (by
weight). Density of 5% acetic acid is 1.0023
g/mL. Vinegar is acetic acid which has the
formula HC2H3O2. (CH3COOH) Ka = 1.8x10-5
H  A 



Ka

 HA
Dilute by 10 (make less concentrated):
  H      A  

 

 10   10   H   A  
Q

 , , K ?
 HA
 HA10
10
What does this tell us, if anything?
We have too many reactants, need to shift to
the right, or dissociate some more.
general rule of thumb: dilution gives
more dissociation.
Example Calculations
1.HCl
2.Acetic acid (vinegar)
3.HF
4.B(OH)3 (Boric acid (eye wash)) (students
Do this one yourself)
5.Mixture (HF and phenol)
6.Mixture (H2SO4, HSO4-)
7.Triethylamine
8.NaAcetate
9.Our heme example
Calculate the F- of a solution of 1.00 M HF.
Ka= 7.2x10-4
1. Write down ALL possible reactions involving a proton


HFaq 
H

F

aq , F
aq


H2 O 
H

OH

aq
aq
Ka  7.2 x10  4
Kw  10 14
2. Excluding water, identify all the proton donors as
1. Strong acid
a. Strong electrolyte: HNO3, HCl, H2SO4 (No Clean Socks)
No Strong Acids (SA)
2. Weak Acid
a. Identify strongest acid (omega dog, can not hold protons)
b. Has largest Ka; smallest charge density anion


HFaq 
H

F

aq , F
aq
Ka  7.2 x10  4
c. Calculate how many protons omega gives up (equil)
stoic.
Init
Change
Assum
Equil
H 2O
55.5
HF(aq)
1
1.0
-x
1>>x
1
H+
10-7
H+from HF
1
10-7
+x
10-7 <<x
x
OH10-7
F1
0
+x
x
Init
Change
Assum
Equil
 x x
Ka 
 HA 
1 K   x
init
2
a
x  2.7 x10
H+from HF F10-7
0
+x
+x
10-7 <<x
x
x
HF(aq)
1.0
-x
1>>x
1
Check assumptions
1   2.7 x10  2
1
10  7   2.7 x10  2
0.027
 0.027
 0.0000001
0.983
0.0270001
2
Calculate the pH of a solution of 1.00 M HF.
K1 = 7.2x10-4
Sig fig is here
0983
.
1
00270001
.
 0027
.
Init
Change
Assum
Equil
HF(aq)
1.0
-x
1>>x
1
x  2.7 x10  2
x   F    2.7x10 2
pH   log2.7x10 2   1568
.
 16
.
Calculate the F- of a solution of 1.00 M HF.
K1 = 7.2x10-4
H+from HF F10-7
0
+x
+x
10-7 <<x
x
x
Example: Boric acid is commonly used in eyewash
solutions to neutralize bases splashed in the eye. It
acts as a monoprotic acid, but the dissociation reaction
looks different. Calculate the pH of a 0.75 M solution
of boric acid, and the concentration of B(OH)4-.



B OH  3,aq  H2 O 
B
OH

H
O

4 ,aq
3 aq

Ka  58
. x10 10
1. Write down ALL possible reactions involving a proton

 



B OH 3,aq  H2 O  B OH 4 ,aq  H3Oaq


H2 O 
H

OH

aq
aq
Ka  58
. x10 10
Kw  10 14
2. Identify proton donors
1. strong acids: No Clean Socks?
2. Weak acids:
No SA
Students do
B(OH)3
This on your own
Set up ICAE chart
stoic
[Init]
Change
Assume
Equil
Ka 
x2
 HA 
Students do
This on your own
OHH+
10-7
10-7
B(OH)4- + H+
1
1
0
10-7
+x
+x
x
10-7<<x
x
x
H 2O
55.5
B(OH)3 + H2O
1
n.a.
0.75
-x
0.75>>x
0.75
58. x1010 0.75 
x  2.1x10  5
init
2
x
58
. xcheck??
10 10  two assumptions.
B(OH) + H O
B(OH) + H
K = 5.8x10
Calculate the pH of a 0.75 M solution of boric acid.
0.75
3
2
4
-
+
a
-10
B(OH)4- + H+
0
10-7
+x
+x
x
10-7<<x
x
x
B(OH)3 + H2O
0.75
-x
0.75>>x
0.75
[Init]
Change
Assume
Equil
58. x1010 0.75 
075
.  x ?
yes
x    10  7 ?
0.000021
x  2.1x10  5

5


x   B OH 4   2.1x10
x   H    21
. x105
pH   log21
. x105   4.68  4.7
 0.0000001
0.0000211two assumptions.
check??
B(OH)
Sig fig
Students do
This on your own
+ H2O
B(OH)4- + H+
Ka = 5.8x10-10
Calculate the pH of a 0.75 M solution of boric acid.
3
Example Calculations
1.HCl
2.Acetic acid (vinegar)
3.HF
4.B(OH)3 (Boric acid (eye wash)) (students
Do this one yourself)
5.Mixture (HF and phenol)
6.Mixture (H2SO4, HSO4-)
7.Triethylamine
8.NaAcetate
9.Our heme example
Mixtures of Acids Calculate the pH of a solution
that contains 1.0 M HF and 1.0 M HOC6 H5.
Calculate the conc. of -OC6 H5 at this concentration.
1. Write down ALL possible reactions involving a proton


HFaq 
H

F

aq , F
aq


C6 H5OH 
H

C
H
O

aq
6 5 aq


H2 O 
H

OH

aq
aq
Ka  7.2 x10  4
Ka  18
. x10  5
Kw  10 14
2. Excluding water, identify all the proton donors as
1. Strong acid NONE
2. Weak Acid
Identify strongest acid (omega dog, can not hold protons)
Has largest Ka; smallest charge density anion
Calculate how many protons omega gives up (equil)
Calculate pH (Use to determine what alpha gets)
HF will control the proton concentration, but
Should include all possible sources to remind ourselves.
stoic
[Init]
Change
Assume
Equil
Ka 
OH10-7
F- +
1
0
+x
x
x
H 2O
55.5
HF + H2O
1
n.a.
1.0
-x
1.0>>x
1.0
x2
x2

1
7.2 x10  4
 HA 
init
7.2 x10 4  
check?? H two
assumptions.
+F
K = 7.2x10
HF
HOC6 H5
+
-
H+ + -OC6 H5
a
-4
Ka = 1.8x10-5
H+
10-7
H+
1
10-7
+x
10-7<<x
x
x  2.7 x10  2
Calculate the pH of a solution that contains 1.0 M
HF and 1.0 M HOC6 H5. Calculate the conc.
of -OC6 H5 at this concentration.
stoic
[Init]
Change
Assume
Equil
HF + H2O
1
n.a.
1.0
-x
1.0>>x
1.0
7.2 x10 4  
x  2.7 x10  2
Check assumptions:
2.7 x10  2   1
10
.
 0.027
F- +
1
0
+x
x
x
H+
1
10-7
+x
10-7<<x
x
2.7 x10  2   107
0.027
 0.0000001
0.0270001
0.983
Sig fig = 1.0
Sig fig = 0.027
F- +
H+
1
1
0
10-7
+x
+x
x
10-7<<x
x
x
 2.7 x10 C H O 
HF + H2O
1
n.a.
1.0
-x
1.0>>x
1.0
stoic
[Init]
Change
Assume
Equil
7.2 x10  
4
x  2.7 x10
 H   x  2.7x10

2
2
2
Ka  18
. x10  5 
C H O 
pH  157
.
6
5

aq


C6 H5 OH  H2 O 
H
O

C
H
O

3 aq
6
5 aq
Ka  18
. x10
5
H O C H O 


3

aq
6
5
C H OH
6
5

aq
6
5

aq
1
18
. x105
4


6
.
66
x
10
 2.7 x10 2 
Ka  18
. x10  5
Calculate the pH of a solution that contains 1.0 M
HF and 1.0 M HOC6 H5. Calculate the conc.
of -OC6 H5 at this concentration.
Example Calculations
1.HCl
2.Acetic acid (vinegar)
3.HF
4.B(OH)3 (Boric acid (eye wash)) (students
Do this one yourself)
5.Mixture (HF and phenol)
6.Mixture (H2SO4, HSO4-)
7.Triethylamine
8.NaAcetate
9.Our heme example
Example: calculate the pH of 0.0010 M sulfuric acid
1. Write down ALL possible reactions involving a proton


H2 SO4 ,aq  H2 O 
HSO

H
O

4 aq , F
3 aq

2
HSO4 ,aq   H2 O 
H
O

SO

3 aq
4.aq


H2 O 
H

OH

aq
aq
Ka  l arg e
Ka  12
. x10  2
Kw  10 14
2. Excluding water, identify all the proton donors as
1. Strong acid
a. Strong electrolyte: HNO3, HCl, H2SO4 (No Clean Socks)


H2 SO4 ,aq  H2 O 
HSO

H
O

4 aq , F
3 aq
Ka  l arg e
b. Give all strong acid protons to water or alpha dog
1
2
3
Pure Water
Control/complete
stoic.
[init]
complete
stoic.
H2O
55.5
H2SO4
1
.001
0
[Init]
Change
Assume?
[Equil]
Ka 2  12
. x10
2
H O SO 


3

aq
2
4.aq
HSO4,aq
OH10-7
HSO41
0
0.001
HSO41
0.0010
-x
0.001>x
0.001
H+
10-7
H+
1
10-7
0.001+10-7
H+
SO421
1
0. 0010 0
+x
+x
0.001>x +x
0.001
x
 0.001 x
Ka2  12
. x10 
 0.001
2

Ka 2  12
. x10  2  x  SO42,aq
Example: calculate the pH of 0.0010
M sulfuric acid; Ka2 = 1.2x10-2

Ka2  12
. x10 2  x  10
. x10 3
NO!
HSO41
0.0010
-x
0.001>x
0.001-x
stoic.
[Init]
Change
Assume?
[Equil]
H+
1
0. 0010
+x
0.001>x
0.001+x
SO421
0
+x
+x
x
Here is our first example in which we can not
Make assumptions
Ka 2  12
. x10
2
H O SO 


3

aq
2
4.aq
HSO4,aq
 0.001  x x
Ka2  12
. x10 
 0.001  x
2
. x10 2  0.001  x   0.001  x x
12
12
. x10  5  12
. x10  2 x  0.001x  x 2
x 2  0.001x  12
. x10  2 x  12
. x10  5  0
Example: calculate the pH of 0.0010
M sulfuric acid; Ka2 = 1.2x10-2
x 2  0.013x  12
. x10  5  0
x 2  0.013x  12
. x10  5  0
2
 0.013   0.013  41  0.000012
x
21
x
 0.013 
2.17 x10
2
4
 0.013  0.0147
x
2
x  8.65x10
[SO42-]=x
4
ax 2  bx  c  0
x
 b
b 2  4ac
2a
-Solution gives a neg
Number which is not allowed
[H+]=0.001
+0.000865
0.001865
pH=-log(0.001865)=2.73
Successive Approximations
(iterations)
Alternative Strategy to
going to “exact equil. Expression”
ITERATIVE SOLUTIONS
Why? – because the real body or real world
Is much too complex to always be able to
Find an exact equilibrium expression
Calculate proton concentration of 0.100 M HNO2 using the iterative
method (Ka=6.0x10-4)
Pure Water
H2O
OHH+
1
55.5
10-7
10-7
HNO2
NO2
H+
stoic.
1
1
1
0.1
0
10-7
2 [Init]
Change
-x
+x
+x
Assum
0.1>>x
x
x>>10-7
[Equil]
0.1
x
x
Calc
7.7x10-3
New Equil
0.1-7.7x10-3 x’’
x’’
3
New calc
7.44x10-3
-3 x’’’
New
New
Equil
0.1-7.44x10
x’’’
4
New new Calc
7.45x10-3
Ka  6.0 x10
x
4
x2

01
.
.   7.7 x10  3
6.0x10  4  01
x' ' 
x'' 
.  7.7 x10  3   7.7 x10  3
6.0x10  4 01
5538
.
x105  7.44 x10 3
Converging, plausible answer for iterative
method: 0.100 M HNO2, Ka=6.0x10-4
Can Skip This for BLB
“A” students work
(without solutions manual)
~ 10 problems/night.
Dr. Alanah Fitch
Flanner Hall 402
508-3119
[email protected]
Office Hours Th&F 2-3:30 pm
Module #17B:
Acid Base Ionization
Computations
Weak Bases
Example Calculations
1.HCl
2.Acetic acid (vinegar)
3.HF
4.B(OH)3 (Boric acid (eye wash)) (students
Do this one yourself)
5.Mixture (HF and phenol)
6.Mixture (H2SO4, HSO4-)
7.Triethylamine
8.NaAcetate
9.Our heme example
Calculation with Weak Base
Calc. the [OH], [H], and pH of 0.20 M solns
of triethylamine, Kb = 4.0x10-4
1
stoic
2 [Init]
Change
Assum
Equil
B
1
0.20
-x
0.20>>x
0.20
H 2O
55.5
H 2O
1
Calc. the [OH], [H], and pH of 0.20 M solns
of triethylamine, Kb = 4.0x10-4
H+
10-7
BH+
1
0
+x
x
x
OH10-7
OH1
10-7
+x
10-7 < x
x
1
stoic
2 [Init]
Change
Assum
Equil
B
1
0.20
-x
0.20>>x
0.20
7
x
10
 x x 2

4
Kb  4.0x10 

0.20  x
0.2
4.0x10  4 0.2  x  8.94 x10  3
8.94 x10  3   10  7 ?
H+
10-7
BH+
1
0
+x
x
x
H2O
55.5
H2O
1
0.00894
OH10-7
OH1
10-7
+x
10-7 < x
x
0.2
 0.00000001
 0.00894
0.00895001
019106
.
Rounds to 0.2
Rounds to 0.0089
8.94 x10  3   0.2 ?
Calc. the [OH], [H], and pH of 0.20 M solns
of triethylamine, Kb = 4.0x10-4
x   OH    8.94x10 3
pOH  2.048
pH  14  2.048  1195
.
“A” students work
(without solutions manual)
~ 10 problems/night.
Dr. Alanah Fitch
Flanner Hall 402
508-3119
[email protected]
Office Hours Th&F 2-3:30 pm
Module #17B:
Acid Base Ionization
Computations
Salts
Example Calculations
1.HCl
2.Acetic acid (vinegar)
3.HF
4.B(OH)3 (Boric acid (eye wash)) (students
Do this one yourself)
5.Mixture (HF and phenol)
6.Mixture (H2SO4, HSO4-)
7.Triethylamine
8.NaAcetate
9.Our heme example
Write all reactions involving protons, hydroxides


NaCH3COO  H2 O 
Na

CH
COO

aq
3
aq

CH3COOaq  H2 O 
OH

aq  CH3 COOHaq

aq
H2 O  Na 

no reaction


H2 O 
H

OH

aq
aq
Kw  10 14
Determine who is omega and will donate

CH3COOaq  H2 O 
OH

aq  CH3 COOHaq
Hmm, a slight problem – we don’t know Kb
If we place Na acetate in solution (to make a 0.1 M
solution) what are the main species present? What
will be the pH of the solution? Ka = 1.8x10-5


CH3COOHaq  H2 O 
CH
COO

H
O

3
aq
3 aq

aq
 
aq 
CH3COO  H3O
+
1
Ka
CH3COOHaq  H2 O


H2 O 
H

OH

aq
aq

aq
Ka  18
. x10  5
CH3COO  H2 O

 
Kw  10 14
CH3COOHaq  OH

aq
1
Kw
Ka
Kw  Ka Kb
10 14
10

K

555
.
x
10
b
18
. x10  5
If we place Na acetate in solution (to make a 0.1 M
solution) what are the main species present? What
will be the pH of the solution? Ka = 1.8x10-5
14  pKa  pKb
Kb  555
. x1010
1
2 stoich
[Init]
Change
Sum
Assume
[Equil]
H2O
H+
OH55.5
10-7
10-7
CH3COO- + H2O = CH3COOH + OH1
1
1
0.1
0
10-7
-x
+x
10-7 +x
0.1-x
0+x
10-7 +x
x<<<0.1
x>>>10-7
0.1
x
x
If we place Na acetate in solution (to make a 0.1 M
solution) what are the main species present? What
will be the pH of the solution? Ka = 1.8x10-5
CH3COO- + H2O = CH3COOH + OH1
1
1
0.1
0
10-7
-x
+x
10-7 +x
0.1-x
0+x
10-7 +x
x<<<0.1
x>>>10-7
0.1
x
x
stoich
[Init]
Change
Sum
Assume
[Equil]
Kb  555
. x10
555
. x10
 10



OH aq CH 3 COOH aq
CH COO 
3
555
. x10 10
x  7.45x10  6
10

aq
x10  7  x x 2


01
. x
01
.
. x10 10 01
. 
555
x

7.45x10 6   01
. ?
7.45x10  6   10  7 ??
yes
X100 rule
No
If we place Na acetate in solution (to make a 0.1 M
solution) what are the main species present? What
will be the pH of the solution? Ka = 1.8x10-5
CH3COO- + H2O = CH3COOH + OHstoich
1
1
1
[Init]
0.1
0
10-7
Change -x
+x
10-7 +x
Sum
0.1-x
0+x
10-7 +x
Assume x<<<0.1
x>>>10-7
[Equil]
0.1
x
x
Equil new 0.1-x
x
x+10-7
7


x
10
 x

10
Kb  555
. x10 
 01
.  x
. x1010  01
.  x   x107  x  107 x  x 2
555
If we place Na acetate in solution (to make a 0.1 M
solution) what are the main species present? What
will be the pH of the solution? Ka = 1.8x10-5
555
. x1011  555
. x1010 x  10 7 x  x 2
x 2  107 x  555
. x1010 x  555
. x1011  0
x 2  100555
.
x107 x  555
. x1011  0 ax 2  bx  c  0
x
 100555
.
x10 7 
2
.
x10 7   41555
. x10 11 
100555
21
 100555
.
x10 7  101103
.
x10 14  2.22 x10 10
x
2
 100555
.
x10  7  149
. x10 5
x
 7.399 x10 6
2
Before we got 7.45x10-6
 b  b 2  4ac
x
2a
“A” students work
(without solutions manual)
~ 10 problems/night.
Dr. Alanah Fitch
Flanner Hall 402
508-3119
[email protected]
Office Hours Th&F 2-3:30 pm
Module #17B:
Acid Base Ionization
Computations
Biological Chemistry
Example Calculations
1.HCl
2.Acetic acid (vinegar)
3.HF
4.B(OH)3 (Boric acid (eye wash)) (students
Do this one yourself)
5.Mixture (HF and phenol)
6.Mixture (H2SO4, HSO4-)
7.Triethylamine
8.NaAcetate
9.Our heme example
Protein folding due to
FRACTION of sites charged
Ka,Val , NH3  2.5x1010
Ka ,Val ,COOH  501
. x10 3
Ka,tyr ,ROH  10
. x1010
Hemeglobin
EXAMPLE Which pH (2, 7, 11) is most favorable for the
formation of a hydrogen bond between Val and tyr in
hemoglobin
It would be useful to be able to predict fraction
ionized at all pH values
Without knowing the concentration of the
protein acid sites.
Skill would be useful in other equilibrium
Calculations.
Need to arrange equations to get
Rid of equilibrium concentrations

HAaq  H2 O 
H
O

3 aq , from
 HA 
 HA 
H 
H 
init
init
x

init
A 

init
0
x
H
 
 HA    A   H
 HA    A   H
 HA    A   H
 Aeq
Ka 

init

init
init

eq

init
init

eq

init

  A 
 HA
init


init
x
 A  A 
 A  A 
  A  A 

A 

init

eq


Hinit
A
 
HA
 Aaq

eq

eq

eq

eq

eq

eq

eq
f ionized 
 x
HA 
init
Ka
H  A 


 HA    A 

init
f ionized 

init

 

Aeq
 HA 
init

  
  K  A    H  A 
   H  A   K  A 
   A  H   K 

Ka HAinit   A    Hinit
 A 

K  HA
K  HA
Ka HAinit
a
init
a
init


init
a

init


init
A 



a

Ka

a
Ka


A
 
 H   K   HA 

init
a
init
 H   K   HA 

init
a
init
An alternative method
To this equation is on next
Slide (can be skipped)
Here is an alternative way to get the equation
HAaq  H2 O

 
f dissociated

aq , from HA
H3 O
A 


 HA 
 
f dissociated
3
eq ,aq
f dissociated 

A 
 H O    A 
K

aq ,eq
A

aq ,eq
3

aq ,eq

aq ,eq
a
f dissociated 
1
H O   1
3
A 


 HA    A 

aq ,eq
eq

aq ,eq

aq ,eq
a
Mass balance
 
3
eq ,aq

aq ,eq
init

 A

aq ,eq
 A H O   HA 
K

aq ,eq
HAinit  HAeq ,aq  Aaq ,eq
Ka

aq
A
HO




 HA 

aq ,eq
Ka

aq ,eq
f dissociated 
Ka
H O   K
3

aq ,eq
a
1
0.9
f ionized ,val
RCOO 

aq ,val
 RCOOH 
f ionized ,tyr
init
0.8
RO 

aq ,tyr
 RCOH 
Fraction Ionized
init
0.7
0.6
0.5
0.4
0.3
0.2
0.1
0
0
2
4
6
8
10
12
14
pH
Which pH will allow best H-bonding?
16
“A” students work
(without solutions manual)
~ 10 problems/night.
Dr. Alanah Fitch
Flanner Hall 402
508-3119
[email protected]
Office Hours Th&F 2-3:30 pm
Module #17B:
Acid Base Ionization
Computations
When should we
Be making assumptions?
Example on Using Simplifications
Module 17A
If 1.0 mol NOCl is placed in a 2.0 L flask what are the equilibrium
concentrations of NO and Cl2 given that at 35 oC the equilibrium
constant, Kc, is 1.6x10-5 mol/L?
Red herrings:
35 oC is a red herring
Clues?
K is “small” compared to others (<<< 1) we have worked with !!!!!
Example 2
EXAMPLE 3:
Example 4:
CO2 , g  H2 , g 
 COg  H2 Og
N 2 O4 
 2 NO2 ( g )
H2 , g  I 2 , g 
 2 HI ( g )
Kc is 0.64.
Kc = 0.36M
Kp is 1x10-2.
We will define
Small in the
Next chapter!
The issue is small Ka with respect
To the initial concentration!!
Considering a simple system


HAeq 
A

H
 eq
eq
The issue is small Ka with respect
To the initial concentration!!
Considering a simple system


HAeq 
A

H
 eq
eq
 A   HA K
Mass balance

aq ,eq
 HA    HA    A 
init

aq ,eq
eq ,aq




A
aq ,eq
% dissociated  

HA

A

eq
aq ,eq

Ka
 
A
HO




 HA 

aq ,eq
3

aq ,eq

aq ,eq
3

aq ,eq
eq ,aq

aq ,eq


 100


eq ,aq

a
HA K
eq ,aq
a



Ka HAeq

%dissociated  
 HAeq  Ka HAeq







 100


OK this is not a nice equation
Do you need to know it?
eq ,aq
 A H O    HA K
A 
2
a
“A” students work
(without solutions manual)
~ 10 problems/night.
Dr. Alanah Fitch
Flanner Hall 402
508-3119
[email protected]
Office Hours Th&F 2-3:30 pm
Module #17B:
Acid Base Ionization
Computations
What you need
To know
1.pH, pOH, pKa, pKb
2.Fraction Ionized
3.ICE charts for equilibrium calculation
4.Recognize a fraction ionized chart and interpret it
5.Recognize when you can make “assumptions”
“A” students work
(without solutions manual)
~ 10 problems/night.
Dr. Alanah Fitch
Flanner Hall 402
508-3119
[email protected]
Office Hours Th&F 2-3:30 pm
Module #17B:
Acid Base Ionization
Computations
END
Which pH (2, 7, 11) is most favorable for the formation of
a hydrogen bond between Val and tyr in hemoglobin
Ka ,Val ,COOH  501
. x10 3
% dissociated 


aq ,eq
100 A
Ka,tyr ,ROH  10
. x1010

 A  H O    A 
K

aq ,eq

aq ,eq
3

aq ,eq
a
% dissociated 
100
H O   1
3

aq ,eq
Ka
pH
2
7
11
[H+]
1.00E-02
1.00E-07
1.00E-11
pH2  10 2


100


100
% dissociatedCOOH  
  33.33
% dissociatedCOOH  10  2 2
 10 3  1 
5.01x10  3  1
 5.01x10

KaCOOH Ka ROH
5.00E-03 1.00E-10
%diss
%diss
33.33333
1E-06
99.998
0.0999
100
90.90909
Repeat procedure
With tyrosine
Ka ,Val ,COOH  501
. x10 3
Ka,tyr ,ROH  10
. x1010
Which pH is best?
pH
2
7
11
[H+]
1.00E-02
1.00E-07
1.00E-11
KaCOOH Ka ROH
5.00E-03 1.00E-10
%diss
%diss
33.33333
1E-06
99.998
0.0999
100
90.90909