Sultan Qaboos University
Collage of Science
Department of Chemistry
Physical Chemistry Laboratory (CHEM3335)
Determination
of pka
Name: Al Hamdani.
ID: .
Date: 16/32010.
1
Objective:
This experiment has two theory objectives:
1. To determine the molar absorptive of a dye and estimate an unknown
concentration of the dye in solution.
2. To estimate the pKa of an indicator dye from spectrophotometric data.
Introduction:
Typically acid-base indicators are themselves weak acids or bases whose acid
and base forms have different colors in solution. As the result of the reaction
with excess titrant, we convert one form to the other causing a color change
that indicated the endpoint of a titration. If we represent the indicator's acid
form as HIn and its basic form as In-, then the following equilibrium describes
the chemical reaction that occurs as the [H+] is changed. If HIn and In- have
different colors, then the solution's color will change as a function of [H+]
depending on which of the compounds is present in the greater amount.
H ln
aq

H
  
lnaq
aq
The acid dissociation equilibrium constant (Ka) for the indicator that describes
this reaction is given by Eqn. 1, in terms of the concentrations of the hydrogen
ion, In- and HIn. Because we are working in aqueous solution, it is
convenient to rearrange Eqn. 1 to Eqn. 2 by taking the negative logarithm of
both sides, and then recognizing the definitions of pKa and pH, rewriting
Eqn. 2 as Eqn. 3, which is simply another version of the HendersonHasselbach equation. Note that Eqn. 3 predicts that the indicator's pKa
2
corresponds to the pH of an indicator solution when the logarithmic term
equals zero (i. e., when [In-] equals [HIn]).
[H  ][In  ]
Ka 
[HIn] …………………………………………………………1
 [In  ] 

 log K a   log [H  ]  log 
 [HIn]  ………………………………………….2
 [In  ] 

pK a  pH  log 
 [HIn]  ……………………………………….………….3
A convenient way to determine the equilibrium constant of a reaction
involving colored species and H+ is to use absorbance spectroscopy. If we
monitor a wavelength at which either one of the two species strongly absorbs
we will see the absorbance as a function of pH change as that species'
concentration in solution changes. From the equilibrium between HIn and In, given above, and considering Le Chatelier's principle, we can see that when
the [H+] is large (low pH ), the equilibrium will shift completely to the left
and the indicator will be completely in the HIn form
 
 In  
  pK a
pH  log 
 HIn  
………….………………………………………4
To get a more precise measure of the pKa, Eqn. 3 is rearranged to give Eqn.
4. This equation gives a straight line when the solution's pH is graphed as a
function of log([In-]/[HIn]). The slope of this line should be +1 and the yintercept, where log([In-]/[HIn]) is zero (i. e. [In-] = [HIn]), is the pKa,
3
To use Eqn. 4 to determine the pKa of the indicator, it is necessary to know
the pH of solutions that have different ratios of the two indicator species HIn
and In-. Since the pH of the solution will determine the amount of the total
indicator that will be in each form, it can be difficult to control exactly how
much of the indicator exists as In- and HIn. It is possible, however, to use
absorbance to obtain the ratio [In-]/[HIn].
A 1  ε ( HIn,λ1)  b  [HIn ]
…………….……………………………………5
is the cell pathlength. Since the amount of the indicator that is in the form of
HIn depends on the pH, [HIn] can be difficult, or impossible, to determine.
What is known, however, is the total concentration (CT) of the indicator in
both forms since a known amount of the indicator was added to the solution
at the beginning. At any pH the indicator's CT is given by Eqn. 6. In
solutions where the pH is sufficiently low, all of the indicator is in the acidic
form, and consequently CT = [HIn]. Substituting CT = [HIn] into Eqn. 6
gives Eqn. 7.
C T  [HIn ]  [In - ] …………………………….………………………6
A 1 (lowest pH)  ε ( HIn,λ1)  b  C T
…………………………………………7
we can write an equation analogous to Eqn. 6 (Eqn. 8) where we have simply
replaced the molar absorptivity of HIn with that of In- and the concentration
of [HIn] with [In-]. In basic solution, CT = [In-], which leads to Eqn. 9.
A  2  ε ( In-,λ2)  b  [In - ]
………………………………..………………….8
4
A  2 (highest pH)  ε ( In-,λ2)  b  C T
………………….……………………….9
So, in this experiment I have to determine the pKa indicator by the
relationship between pH and log([In-]/[HIn]) for an indicator. The pKa of
the indicator corresponds to the intersection of the line with the pH axis.
Experimental:
Apparatus and chemicals:
0.10M NaOH –fennels- burette- beakers- pipettes- conical flask -calibrated
PH meter - 0.10HCL- KH2P04, K2HP04- bromothymolblue solutionscanning spectrophotometer, see fig 1.
Figure1: The Perkin Elmer Lambda 11™ Single Beam UV/Vis Spectrophotometer. When the
cavetti containing the reference sample is placed into
the compartment, its taped side should be perpendicular to the light source
5
Condition:
I have done this experiment at a constant pressure (atmospheric pressure) and
room temperature (25ºC).
Procedure:
1- prepare 6 buffer solution by mixing fallowing amount of (KH2P04,
K2HP04) :
KH2PO4
K2HPO4
flask
(mL)
(mL)
1
44
6
2
36
14
3
31
19
4
25
25
5
8
42
6
3
47
Table1: the amount of KH2PO4 and
K2HPO4 on each buffer
2- pipette (25ml) each buffer solution to another conical flash and add
exactly 1.0ml of indicator.
3- use the remaining solution in the original flask to determine the PH value
of each buffer by using calibrated pH meter
4-We prepared strong acid solution by mix (25ml) HCL with 1.0 ml indicator
,also we prepared strong base by mix 25 ml NaoH with 1.0 ml indicator. These
are the solution needed to find  ln and  H ln .
6
5-We used 1cm cell to measure the
visible absorption spectrum of each
buffer solution , we recorded the
data ( wavelength, isobestic & the
absorbance of each solution from
computer)
Figure 2: the solutions that we prepared.
6-Finally we used these data to
calculate the pKa.
Results:
# of
Name of
buffer
buffer
abs(432nm)
abs(616)nm
1
acid
0.19742
-2.91E-03
2
Sample1
0.20488
2.64E-02
3
Sample2
0.20821
7.47E-02
4
Sample3
0.17954
1.01E-01
5
Sample4
0.1708
1.44E-01
6
Sample5
0.11787
3.72E-01
7
Sample6
7.78E-02
4.87E-01
8
base
5.29E-.2
6.43E-01
Table2: the result that we got it from
scanning spectrophotometer
7
0.7
0.6
0.5
acid
base
0.4
ads
s1
isobistic
point
0.3
s2
s3
s4
0.2
s5
s6
0.1
0
0
100
200
300
400
500
600
700
800
900
-0.1
Graph1: the graph of ads ve wavelength
Wavelength
that we got it from scanning
spectrophotometer
# of
buffer
sample
1
sample
2
sample
3
sample
4
sample
5
sample
6
HIn(mol/L)
In(mol/L)
f(x)
0.0000995
0.0000005
0.0475162
9.88E-05
0.0000012
0.1365652
0.0000984
0.0000016
0.1917159
0.0000977
0.0000023
0.2944088
1.3231585
0.8646600
0.7173419
0.5310492
0.0000981
0.0000019
1.3834317
0.0000987 0.0000013 3.1404487
1.248749
Intercept
slope
uncertainty in
log f(x)
0.073374
R2
0.986378
in intercept
Table3: the result that we got it from our
experiment.
Notice:
f ( x) 
abs (616)n(buffer)  abs (616)nm(acid )
abs (616)nm(base)  abs (616)nm(buffer)
8
pH
-2.34
6
-1.91
6.39
-1.78
6.55
-1.63
6.75
0.1409577
-1.72
7.65
0.4969917
-1.90
8.24
7.512387
Uncertainty
slope
Log(In)/(Hin
0.056533
pH vs log(In-/HIn)
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y = 1.2487x + 7.5124
R2 = 0.9864
8
7
pH
6
5
pH
4
Linear (pH)
3
2
1
-1.5000000
-1.0000000
-0.5000000
0
0.0000000
log(In-/HIn)
Graph2: PH vs log(In-/HIn)
Calculations:
Calculate the concentration of the indicator:
C0=.0.0001 mol/cm3
For strong acid:
 HIn 
A
 2.91 10 3
1
1

  2.91cm M
bc (1.00cm)(0.0001M )
For strong base:
A
bc
6.43  10 1
1
1

 6.43  10  4 cm M
(1.00cm)(0.0001M )
 In  
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0.5000000
1.0000000
For sample 1:
( A )  ( ln )(C )
[ H ln]  b
 H ln   ln
(2.64  10 2 )
(1.00cm)

 (6.43  10  4 cm  1 M  1) (0.0001M )
(29.1 M 1)  (6.43  10  cm  1 M  1)
4

 0.0000995 M
  
c  HIn   In 
In   c  HIn 



 0.0001  (0.0000995)  0.5  10 6 M
Log [(ln-)/ (Hln]) = Log [(0.5×10-6)/( 0.0000995)] = -2.30
pH= pKa + log[(In-)/(HIn)]
pKa= pH – log[(In-)/(HIn)]
=6.00-(-2.30) =8.30
**Other way to get pH>>is:
f ( x) 
abs (616)n(buffer)  abs (616)nm(acid )
abs (616)nm(base)  abs (616)nm(buffer)
So f(x) for sample 1 is:
f ( x) 
(2.64  10 2 )  (2.91  10 3 )
 0.0475
(6.43  10 1 )  (2.64  10  2 )
log f ( x)  log( 0.0475)  1.32
pKa  pH  log f ( x)
 6.00  (1.32)  7.32
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Discussion:
In the experiment the pKa of bromothymol blue (3',3"- dibromothymolsulfonephthalein) is determined by the two methods which have been
discussed. At pH less than 6, the indicator is yellow and at pH greater than
7.6, the indicator is blue. At an intermediate pH, the blue and yellow combine
to yield a green solution.
In our graph we obtained the pKa = Y- intercept =7.512387 and the actual
value of pKa is 7.00. So our result is close to the actual value.
Error analysis:
Slop is 1.2±0.07
Intercept is 7.51±0.06
I think this error because of :
- Maybe we did not transfer accurate volume.
- Also maybe some apparatus were not clean enough.
- pH reading is not accurately depending to pH meter
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