Download 901 Polyprotic Acids

901
Polyprotic
Acids-Bases Equilibria
Dr. Fred Omega Garces
Analytical Chemistry 251
Miramar College, SDCCD
HW: 9.14, 9.25, 9.34
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Polyfunctional Acids and Bases
Sept 24, 2009
Behavior of Polyprotic Acids
Many acids have more than one ionizable H atom.
Polyprotic acids are acids with multiple ioniziable protons
Consider the following Acids:
1. 
HCl + H2O g H3O+ + Cl-
Ka >>1
Monoprotic
2.
H2SO3 + H2O g H3O+ + HSO3HSO3- + H2O D H3O+ + SO3 2-
Ka1 = 1.7•10 -2
Ka2 = 6.4•10 -8
Diprotic
3.
H3PO4 + H2O D H3O+ + H2PO4H2PO4- + H2O D H3O+ + HPO4 2HPO42- + H2O D H3O+ + PO4 3-
Ka1 =7.3•10 -3
Ka2 = 6.2•10 -8
Ka3 = 4.2•10 -13
Triprotic
Note that the proceeding Ka’s are always smaller than the previous. This makes sense in that the
removal of the next H+ is from a negatively charge specie. The proceeding Ka’s (for successive
H+ ionization) are generally smaller by factors of 1000.
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Polyfunctional Acids and Bases
Sept 24, 2009
Ka’s of Polyprotic Acids
The pH estimate of a polyprotic acid can be estimated by only the pKa1 for
polyprotic acids if pKa1 differs by a factor of 103 or greater to pKa2
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Polyfunctional Acids and Bases
Sept 24, 2009
Ka’s Amino Acids (1)
The pH estimate of a polyprotic acid can be estimated by only the pKa1 for
polyprotic acids if pKa1 differs by a factor of 103 or greater to pKa2
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Polyfunctional Acids and Bases
Sept 24, 2009
Ka’s of Amino Acids (2)
The pH estimate of a polyprotic acid can be estimated by only the pKa1 for
polyprotic acids if pKa1 differs by a factor of 103 or greater to pKa2
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Polyfunctional Acids and Bases
Sept 24, 2009
9.1 Series Equation for the Amino Acid Leucine
Consider the ionization of Leucine,
designated by H2L 0.050 M :
-The reaction is never: H2L+ g
2 H+ + L-
rather it is a two step process to find pH of solution and [L- ]:
step i ) H2L+ g H+ + HL
Ka1 = 4.70 •10-3 , pka1 = 2.328
Ka1 = [H+] [HL] / [H2L+ ]
step ii) HL g H+ + L Ka2 = [H+] [L- ] / [HL] g
g
(SA, 3 iterations)
[HL] = [H+](1) = 1.32•10-2 M g
pH (1) = 1.88
Ka2 = 1.80 • 10 -10
[L-]= 1.80 •10-10 M
step iii) The basic form of L- is treated in the same manner:
H2O + L- g
H2O + HL g
HL + OHH2L+ + OH-
Kb1 =
Kb2 =
Where - Ka • Kb = Kw: Kb1 = Kw / Ka2
Kb2 = Kw / Ka1
Approximation method is valid even if Ka2 is just 10 times less than Ka1.
Approximation would only lead to 4% error or 0.01 pH Units.
Diprotic acids behave like monoprotic acids with Ka = Ka1
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Polyfunctional Acids and Bases
Sept 24, 2009
9.1 The acidic form, H2L
Step i) What is the concentration of the acidic form of Leucine, H2L+ (0.05 M) ?
The acidic form of Leucine, H2L+ is a typical weak acid calculation:
H2L+ g H+ + HL
i
0.5
0
0
Δ  -x
x
x
e
x
x
0.5-x
K a1 =
[HL][H+ ]
[H 2L+ ]
Ka1 = 4.70 •10-3 , pka1 = 2.328
, Remember that F = [HL] + [H 2L+ ]
therefore, K a1 =
[HL][H+ ]
[H 2L+ ]
=
[HL][H+ ]
F-[HL]
=
x2
F-x
→
[H 2L+ ] = F - [HL]
& [HL] = [H+ ] = x
, → K a1 (F-x)=x2 → K a1F - K a1x - x2 = 0
Solving Quadratic, K a1 = 4.7•10−3 , F = 0.050M , x = 1.32•10-2M = [HL] = [H+ ]
[H 2L+ ] = F-[HL] = 0.05M - 1.32•10-2M = 3.68•10-2M
[HL] = [H+ ] = 1.32•10-2M, pH = 1.88
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Polyfunctional Acids and Bases
Sept 24, 2009
9.1 The conjugate base form, LWhat is the concentration of the anionic form of Leucine, L- (0.05 M) ?
The basic form of L- is treated in the same manner:
HL + OH-
H2O + L- g
H2O + HL g
Kb1 = 5.55•10-5
H2L+ + OH-
Kb2 = 2.13•10-12
Where Ka • Kb = Kw: Kb1 = Kw / Ka2
K b1 =
Kw
K a2
=
[HL][OH− ]
[L ]
-
therefore, K b1 =
Kb2 = Kw / Ka1
, Remember that F = [L- ] + [HL] → [L- ] = F - [HL] & [HL] = [OH− ] = x
[HL][OH− ]
=
[HL][OH− ]
=
x2
, → K b1 (F-x)=x2 → x2 + K b1x - K b1F = 0
[L ]
F-[HL]
F-x
−5
Solving Quadratic, K b1 = 5.55•10 , F = 0.050M , x = 1.64•10-3M = [HL] = [OH− ]
-
[L- ] = F-[HL] = 0.05M - 1.64•10-3M = 4.84•10-2M
[HL] = [OH− ] = 1.64•10-3M, pOH = 2.7=9, pH = 11.21
[H 2L+ ] is determine by similar strategy using K b2 .
K b2 =
8
[[H 2L+ ][OH− ]
[HL]
=
[H 2L+ ] x
x
= [H 2L+ ]
Polyfunctional Acids and Bases
Sept 24, 2009
9.1 Amphoteric or Amphiprotic
Specie which can act as either a base or an acid by donating or accepting protons are considered
amphoteric substances. Under certain conditions, chemicals can either give up a proton or accept a
proton.
i.e., H2O, HCO3- , HPO42 -, H2PO4Consider Hydrogen carbonate (bicarbonate) :
Acid process:
H2CO3 + H2O g H3O+ + HCO3-
1st H+, Ka1 =4.3 •10 -7
HCO3- + H2O g H3O+ + CO3-2
2nd H+, Ka2 = 5.6 • 10 -11
Base process:
CO3-2 + H2O g OH - + HCO3HCO3 - + H2O g OH - + H2CO3
1st OH-, Kb1 = 1.79•10
-4
2nd OH-, Kb2 =2.33 •10 -8
Note from the above reaction that Ka1 competes against Kb2
these two equilibrium constant are related by Kw: Kb2 = kw / Ka1
To determine acidity or basicity of an amphoteric solution, the Ka’s and Kb’s must
be weighed against each other.
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Polyfunctional Acids and Bases
Sept 24, 2009
9.1 Amphoteric Behavior of Carbonic Acid
H2CO3
Ka1 = 4.3•10-7
HCO3
Ka2 = 5.6•10-11
-
Kb1 = 1.79•10-4
Kb2 = 2.33•10-8
CO3-2
From the magnitude of the equilibrium constant for bicarbonate ion For the HCO3- : |Kb2| > |Ka2|
In this process, the HCO3- is a better base than HCO3- is an acid.
Therefore is a solution basic or acidic,
if HCO3- is introduced in a solution i.e., NaHCO3 ?
i.e., 42 g NaHCO3 is added to 500 ml of water, What is the pH ?
NaHCO3
OH-
HCO3 Ka
2
Kb2 = 2.33•10-8
-
From inspection, Kb2 >> Ka2,
therefore this solution is basic.
H+
= 5.6•10-11
To determine the pH, solve the equilb. rxn for both reaction.
Kb2 = [OH -] [H2CO3 ]
[HCO3- ]

[OH-]
&
Ka2 = [H3O+] [CO32 ]
[HCO3- ]
 [H3O+]
Excess of OH- and H3O+ will determine the pH of the solution.
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Polyfunctional Acids and Bases
Sept 24, 2009
9.1 The neutral form HL
Systematic approach
Consider the ionization of Leucine,
The basic form of L- is treated in the same manner:
H2O + HL g
L- + H+
Ka2 = 1.80•10-10
H2O + HL g
H2L+ + OH-
Kb2 = 2.13•10-12
Substituting in the charge balance equation-
A systematic approach:
Charge Balance:
[H+ ] + [H 2L+ ] = [OH- ] + [L- ]
multiplying by [H+ ]: [H+ ]2 +
K w = [H+ ] + [OH− ] → [OH− ] =
K a1 =
K b1 =
[L- ] =
[HL][H ]
+
[H 2L+ ]
→ [H 2L ] =
[HL][OH− ]
[L- ]
#K &
[HL] % w (
% [H+ ] (
$
'
K b1
+
=
Kw
[H+ ]
[HL][H+ ]
K a1
[HL][OH− ]
→ [L- ] =
K b1
K w ⋅ [HL]
K b1 ⋅ [H ]
+
[HL][H+ ]
[H+ ] + [H 2L+ ] = [OH- ] + [L- ] → [H+ ] +
= K a2
[H+ ]
K a1
Kw
[H ]
+
+ K a2
[HL]
[H+ ]
,
= K w + K a2[HL]
#
&
K + K a2[HL]
[HL] (
solving for [H+ ]: [H+ ]2 ⋅ %1 +
=K w + K a2[HL] → [H+ ]2 = w
→ [H+ ] =
%
(
K a1 '
[HL]
$
1+
K a1
K a1 ⋅K w + K a1K a2[HL]
K a1 + [HL]
Since K a2 = 1.8 ⋅ 10-10 and K b2 =2.13 ⋅ 10-12 are very small,
[HL] remains unchange = F:
[HL]
[HL][H+ ]2
K a1
=
K a1 ⋅K w + K a1K a2[HL]
[H+ ] =
K a1 + [HL]
= [H+ ] =
K a1 ⋅K w + K a1K a2F
K a1 + F
,
For F = 0.050M, [H+ ] = 8.80 ⋅ 10-7 , pH = 6.06
[HL][H+ ]
0.050M ⋅ 8.80 ⋅ 10-7
[HL]
0.050M
[H 2L+ ] =
=
=9.36 ⋅ 10-6M, [L- ] = K a2
= 1.8 ⋅ 10-10
=1.02 ⋅ 10-5M
-3
+
K a1
4.70 ⋅ 10
[H ]
8.80 ⋅ 10-7
Summary :
[H+ ] = 8.80 ⋅ 10-7 , pH = 6.06, [HL] = F = 0.05M, [H 2L+ ] = 9.36 ⋅ 10-6M, [L- ] =1.02 ⋅ 10-5M
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Polyfunctional Acids and Bases
Sept 24, 2009
9.1 Shortcut to the Intermediate forms of HL
Ka1 = 4.7•10-3
H2 L
Ka2 = 1.8•10-10
HL
Kb2 = 2.13•10-12
L-
Kb1 = 5.5•10-5
Using the analysis from the previous page....
it can be shown that
[H+] = { [(Ka1 • Ka2 • HL) + (Ka1 • Kw)] / (Ka1 + HL)}
or
[H + ] =
½
K a1K a2HL + K a1K w
K a1 + HL
Simpler form is based on Ka2•HL >> Kw and Ka1 << HL, in this case
€
K a1K a2HL + K a1K w
[H ] =
+
K a1 + HL
log [H+ ] =
pH =
12
1
2
1
2
(logK
a1
K a1K a2HL
=
HL
=
K a1 •K a2
)
+ logK a2 = log K a1 •K a2
(pK a1 + pK a2 )
or
pH = -log K a1 •K a2
Polyfunctional Acids and Bases
for our example pH = 6.04
Sept 24, 2009
9.1 Summary of Diprotic Acid Calculation
How to calculate pH and composition of solution from different forms of acid, H2A, HA-, A-2…
I. H2A ie, H2CO3 & H2L+ (Acid form)
1. Treat H2A as monoprotic Acids, Only the first Ka is important
2. Use Ka2 to solve for the equilibrium concentration for [A-2] since [HA-] and [H+] will be equal.
II. HA-
ie, NaHA, NaHCO3. & HL (Amphotheric form)
1. Use approximation of [HA-] = F (Formal conc) and find pH by:
2. From the result of first suggestion,
[H+ ] =
[HA-] = F, and [H+] = { [(Ka1 • Ka2 • F) + (Ka1 • Kw)] / (Ka1 + F)}
[H2A] =
€
[HA - ] [H + ]
K a1
½
K a1K a2F + K a1K w
K a1 + F
K a2 [HA - ]
and [A ] =
-2
[H + ]
€
III. A-2 i.e, CO3-2 & L- (Basic form)
1. Treat A2- as monobasic:
2.  Use K1 to solve for H2A
€
[H + ] =
Kw
[OH - ]
[H2A] =
=
kw
x
[HA - ] [H + ]
K a1
=
[HA - ] (Kw/[OH - ])
K a1
Polyfunctional Acids and Bases
13
€
= K b2
Sept 24, 2009
9.2 Diprotic Buffers
A buffer from a diprtoic (or polyprotic) acid is treated in the same way as a buffer
from a monoprotic acid. pKa1 = 2.950, pKa2 = 5.408
pH = pK a1 + log
[HA- ]
pH = pK a2 + log
[H 2A]
[A-2]
[HA- ]
Example: pH of solution with 1.00 g KHC8H4O4 (KHP) with 1.20 g Na2C8H4O4?
Is the information of Volume important?
KHP (204.22 g/mol) Na2P (210.094 g/mol)
1.00 g KHP = 4.90 mmol, 1.20 g Na 2P = 5.71mmol
pH = pKa2 + log
14
€
5.71mmol
4.90mmol
= 5.408 + 0.0664 = 5.474
Polyfunctional Acids and Bases
Sept 24, 2009
9.3 Polyprotic Acids and Bases
Phosphoric acids is a triprotic acid. Calculate the pH of a 0.100 M solution of phosphoric acid.
Name
Phosphoric
Formula
K a1
H3PO4
K a2
7.31⋅10 -3 6.2⋅10 -8
K a3
4.2⋅10 -13
Acid
H3PO4 + H2O
€
 H3 O +
Ka1 =7.3•10 -3
H2PO4- + H2O
 H3 O +
+
HPO4 2-
Ka2 = 6.2•10 -8
HPO42- + H2O
 H3 O +
+
PO4 3-
Ka3 = 4.2•10 -13
Ka1 = [H3O+] [H2PO4-] =
x2  7.3 •10
[H3PO4 ]
0.100
Ka2 = [2.7•10-2+x] [HPO42-] = x
[2.7•10-2-x ]
15
H2PO4-
+
-3
(0.100 ) = x2 
x = [H3O+] = 2.70•10
= 6.2•10-8  x = [H3O+] = 6.2•10
Ka3 = [2.7•10-2+x] [PO43-] =
[6.2•10-8-x ]
4.35•105 x = 4.2•10-13 
[H3O+ ] = 2.7•10-2 M +
6.2•10
-8
M + 9.64•10
-8
-19
M ~ 2.7•10 -2 M
Polyfunctional Acids and Bases
M
M
x = [H3O+] = 9.64•10
-19
-2
M
pH =
1.57
Sept 24, 2009
9.3 Summary of Polyprotic Systems
How to calculate pH and composition of solution for polyprotic acid,
H3A, H2A-, HA-, A-3...
1. H3A Treat H3A as monoprotic Acids. Only the first Ka is important
2.  H2A-
Treat as the intermediate form of a diprotic aicd
[H+ ] =
3. HA-2
€
16
K a1 + F
Also treated as the intermediate form of the polyprotic,
with HA2- having conjugates of H2A- and A-3
[H+ ] =
4.  A-3
K a1K a2F + K a1K w
K a2K a3F + K a2K w
K a1 + F
Treat A3- as monobasic:
K b = K b1 =
Polyfunctional Acids and Bases
Kw
K a3
Sept 24, 2009
9.4 Principle Species under acidic – basic conditions
Under the condition in which pH = pKa, then conjugates are
equal in concentrations [HA] = [A-]
HA predominant form:
.
pH
major species .
< pKa
HA
> pKa
A-
← more acidic more basic →
pka ^
HA
A-
H2A predominant form:
.
pH
pH < pKa1
17
major species .
H2A
pKa1 < pH > pKa2
HA-
pH > pKa2
A2-
← more acidic
pka1 ^
HA
Polyfunctional Acids and Bases
more basic →
HA-
A-2
pka2 ^
Sept 24, 2009
Example: What is the Principal form of benzoic acid
What is the principal form of benzoic acid at pH = 8? pka = 4.20
HBz D Bz- + H+
Ka = 6.3•10-5
HA predominant form:
.
pH
← more acidic more basic →
pka ^ = 4.2
major species .
< pKa
HA
> pKa
A-
HA
A-
The math
Ka = 6.310 •10-5 =
6.310 •10-5 =
6.310 •10-5
[1.00 •10 ]
-8
=
18
[HBz]
,
pH = 8, H+ = 1.00 •10-8
pKa = pBz- + pH+ - pHBz
4.2 - 8 = -3.8 = pBz- - pHBz
6300 [Bz- ]
10−(−3.8) = 6310 =
=
;
1
[HBz]
[Bz- ] [1.00 •10-8 ]
[HBz]
[Bz- ]
[HBz]
[Bz ] : [HBz] →
-
[Bz- ] [H+ ]
6310
6311
Total = [Bz- ] + [HBz] = 6300 + 1
> 1 ; so [Bz- ] > [HBz] by 6310
⋅ 100 = 99.98% Bz
[Bz- ] : [HBz] →
-
Polyfunctional Acids and Bases
6310
6311
⋅ 100 = 99.98% Bz-
Sept 24, 2009
Fraction Decomposition Equations
Consider the diprotic chemical:
HA D H+ + AαHA =
α
A−
=
[HA]
[A- ] + [HA]
[A− ]
[A- ] + [HA]
[HA]
=
=
F
[A− ]
F
=
=
[H+ ]
[H+ ] + K a
Ka
[H+ ] + K a
Consider the diprotic chemical:
H2A D H+ + HA
HA D H+ + AH2A form : α H2A =
HA - form : α HA- =
A -2 form : α A-2 =
19
€
[H2A]
F
[HA - ]
F
[A -2 ]
F
=
=
=
[H + ]2
[H + ]2 + [H + ] K a1 + K a1K a2
K a1 [H + ]
[H + ]2 + [H + ] K a1 + K a1K a2
K a1 K a2
[H ] + [H + ] K a1 + K a1K a2
+ 2
Polyfunctional Acids and Bases
Sept 24, 2009
Isoelectric and Isoionic pH
Isoionic point, or isoionic pH is the pH of the pure neutral polyatomic acid.
CH 3
H 3N+ C H CO 2H
CH 3
CH 3
H 3N+ C H CO−2
H 3N+ C HCO−2
!
CH 3
!
H 2N C HCO+2
+
+
H+
H+
pK 1 = 2.34
[H ] =
+
pK 2 = 9.87
K a1K a2F + K a1K w
K a1 + F
Isotonic point, or isoelectric pH is the pH at which the average charge
of the polyprotic acid is 0 or when the [H2A+] = [A-2]
[H+] = K a1 ⋅K a2
pH = -log K a1 ⋅K a2
pH =
20
1
2
(pK a1 + pK a2 )
Polyfunctional Acids and Bases
Sept 24, 2009
Sample Problem:
p6 pH of a weak acid
a) Calculate the pH of 0.300 M piperazine.
b)  Calculate the concentration of each form of piperazine in this solution.
HPH
i
D
+
H 2O !
0.300
-x
-
[e] 0.300-x
NH2+
HN
NH + H2O
HN
HPH+2
+ OH-
0
+x
0
+x
x
x
+
OH- ,
HPH2+ D
K b1 = 5.38 ⋅ 10-5 , K b2 =
Kw
K a1
=
H2PH22+ + OH1.0 ⋅ 10-14
4.65 ⋅ 10
-6
= 2.15 ⋅ 10-9
F-x
K b1 =
x2
F-x
= 5.38 ⋅ 10-5 → x = [ HPH+2] = [OH- ] = 5.38 ⋅ 10-5 (F-x)
x = [ HPH+2] = [OH- ] = 5.38 ⋅ 10-5 (0.300) = 4.017 ⋅ 10-3M, 5% Rule: 1.33% < 5% Assumption OK
a) [ HPH+2] = [OH- ] = 4.017 ⋅ 10-3M, pOH = 2.40, pH = 11.60
b) K b2 = 2.15 ⋅ 10
-9
=
[H 2PH 2 ]´ =
2+
21
[H 2PH2+
] [OH-]
2
[HPH 2]
+
2.15 ⋅ 10-9 [HPH+2]
[OH-]
→
=
2.15 ⋅ 10-9 [HPH+2]
[OH-]
=
[H 2PH2+
]
2
1
2.15 ⋅ 10-9 [4.017 ⋅ 10-3M]
[4.017 ⋅ 10 M]
-3
Polyfunctional Acids and Bases
= 2.15 ⋅ 10-9 M
Sept 24, 2009
Sample Problems:
p20 Buffer
Which of the two following compounds would you mix to make a buffer of pH a) 7.45 b) What mass of these two buffers is required to prepare 1.0L buffer with a total
phosphate concentration of 0.050M. c) What is the exact procedure to prepare a buffer that is
exactly 7.45?
pK a1 =2.148
a) H3PO4
H2PO4
!
-
pK a2 =7.198
HPO4
!
pK a3 =12.3758
2-
!
PO4
3-
Use NaH2PO4 and Na2HPO4 that will have pH of 7.198 at 50 : 50 mix
b) 7.45 = 7.198 + log
HPO4
HPO4
2-
H2PO4
-
⇒ log
HPO4
H2PO4
2-
H2PO4
-
= 100.252 = 1.786 ;
2-
2-
2-
-
[HPO4 ] +
plugging in : [HPO4 ] = 0.0205 M
= 7.45 - 7.198 = .252
-
[H2PO4 ] = 0.050
-
[H2PO4 ] = 0.01795 M
Use 4.55 g of Na2HPO4 and 2.15 g of NaH2PO4
c) Weigh out 0.050 mol Na2HPO4 and dissolve in 900mLof water.
Add HCl while monitoring the pH with pH - sensor. Stop titration
when pH reads 7.45. Dilute to 1.0 L with H2O.
22
€
Polyfunctional Acids and Bases
Sept 24, 2009
Sample Problems:
p25 Polyprotic acids
The diprotic acid H2A has pKa1 = 4.00 and pKa2 = 8.00.
i) At what pH is [H2A] = HA-] ?
ii) What pH is [HA-] = [A2-]
iii) Principle species at pH 2.00
iv) pH = 6.00
i)
At [H2A] = HA-] , pH = pKa1 =4.00
iii)  H2A
iv)  HA-
ii) At [HA-] = [A-2] , pH = pKa2 =8.00
← more acidic
pka1 ^=4.00
HA
v) pH = 10.00
more basic →
HA-
A-2
pka2 ^=8.00
v)  A-2
23
Polyfunctional Acids and Bases
Sept 24, 2009