The Excited State of β-naphtol
A laboratory experiment in chemical physics
The Excited State of β-naphtol
Introduction
In this laboratory we will use the absorption and fluorescence spectroscopy to study the
proteolysis of β−naphthol (a week aromatic acid) in its ground and excited state. In the
theoretical part of the lab we will derive an expression for the fluorescence intensity as a
function of pH for the acid and base form of the b-Naphthol using the reaction kinetics of the
proteolysis reaction. The pH dependence of the fluorescence will then be used to determine the
pKa value for the excited state. The theoretical and experimental principles give in this lab
This lab instruction does not give comprehensive instruction on how to perform the lab; instead
it gives the background for self study for both the theoretical and experimental part of the
course.
Theory
A central question in this lab is how the molecular properties changes after photoexcitation.
Photoexcitation means that an electron in the molecule has moved into a higher orbital which is
often an antibonding orbital. An electron in an antibonding orbital will change the bonding
properties in the molecule such as bonds distances and angles, a proton for example might
become weakly bounding in a molecule and it is therefore reasonable to assume that the acidbase equilibrium will be different in the excited state ac compared to the ground state.
A molecule in the excited state will also have a different electron distribution than the ground
state and therefore we can assume the dipole moment will be different in both magnitude and
direction in the excited state. This will influence both the absorption and fluorescence spectra.
A molecule in a solvent will be in a”cage “of solvent molecules, the solvent molecules will
orient around the molecules to achieve the lowest possible energy for the system. What will
happen to the molecule is excited? How fast can the solvent molecules reorient to around the
new dipole of the excited state.
Franck Condon principle tells us that it is reasonable to assume that the excitation of an
electron is so fast, that nuclear changes have no time to happen. So will the solvent cage look
the same around the excited molecule immediately after absorption of light? Will the cage that
surrounded the ground state to be optimal also for the excited state? If not, what can you expect
to happen, and how fast will the change occur compared to the lifetime of the excited state?
These factors will be important to consider when we look at different methods for studying the
acid-base equilibrium β-naphthol in both the ground and excited states.
In order to study the excited state properties of b-naphtol the sample will be excited by
appropriate?+ wavelength. Under steady state condition, the following reaction scheme can be
written.
k3
B*+H+
*
A
k’4 = k4 [H+]
k1
IB
IA
k2
B+H+
A
Ka
A,B,A*, B* Are β-naphtol acid and base forms in the ground and excited state.
k1, k2
k3, k4
Ka
IA, IB
Are 1:a reaction rate for deactivation of A* and B* by fluorescence, internal
conversion and quenching.(1/k = τ= fluorescence lifetime)
are reaction rates for the protolysis equilibrium for the excited state
Is the acid constant for β-naphtol
are excitation rate; Ix= ωεX [X], where ω is a instrumental constant, εX and
[X] are the absorption coefficient and concentration of X.
(I)-Determination of pKa of the absorption spectra
pKa can be determined if we can measure the ratio [B] / [A] at different pH. One
way is to set a high and a low pH, determine the absorption coefficient for the base
and acid form at two wavelengths and then measure the absorbance at intermediate
pH values.
Here we use a simpler method using a single wavelength. Under the assumption that
the absorbance of the pure acid and base form (A0A and A0B) at this particular
wavelength can be measured separately by setting the low and high pH, we can we
write the ratio:
[ B]
[ A]
=
(ε B − ε A )[ B]
(ε B −ε A )[ A]
[ ]
=
AB − ε A[ B]
AB − ε A(Ctot −[ A])
=
=
ε B[ A]− AA
εB (Ctot −[ B ])− AA
AB + AA − ε A Ao
Atot − Ao A
= o
A B − Atot
ε B B o − AB −AA
[]
And since Atot = AA + AB and Ctot = [A°] = [B°], and summing that the total content
of beta-naphthol is a constant. [A°] and [B°] are the concentration of pure acid and
pure base form. Observe that Atot will change with pH, but A°A and A°B are
constants at a given wavelength.
Question:
Think of an experiment to determine the pKa and makes an estimate based on the
absorption spectra (see attached figure). At what wavelength would you measure?
How do you change the pH while keeping the β-naphthol concentration constant?
(II) - Determination of pKa* from the absorption spectra
Under chemical equilibrium, the change in the Gibbs free energy under standard
condition can be written as:
ΔG ° = - kT lnk
where k is the Boltzmann constant. This means that both K and K* can be expressed
using the difference in (free-) energy between protolysed and non-protolysed form.
Information on the energy differences can be obtained from absorption spectra. The
energy level scheme below can be used:
(1)
∆EA = hνA and ∆EB = hνB, where ν is the absorption frequency corresponding to the
transition between the lowest vibration levels in the ground and excited state (00transition). ΔH °> (ΔH °)* is explain the ”red shift” in the absorption spectra in the
base form.
Question: It is expected that ΔH ° and (ΔH °) * is positive? Motivate!
You should now be able to write an expression of difference pKa * - pKa as a
function of νA and νB
Hint: Apply Hess's law on the energy diagram. Then use: ΔG° = ΔH°-TΔS°, ΔG° =kT lnK, lgK=inK/ln10 and the approximation ΔS ° ≈ (ΔS °)*.
Question: is this a reasonable approximation? What will be the largest contribution
to the entropy of the reaction?
One problem is that we do not know which wavelength in the absorption spectrum
that correspond to the 0-0-transition. The spectral resolution is not sufficient to
enable us to distinguish individual peaks. Note that the maxima only indicate the
most likely transition. A good approximation is that λ00≈ (λabs,max + λem,max) / 2, i.e.
the mean value for the of absorption and emissions maxima.
Question: Explain with a simple figure why this approximation is reasonable. What
factors affect the red shift between the emission and absorption spectra .
Write an approximate expression of pKa*, make the change from frequency to
wavelength. Use the given absorption spectra to provide an estimate of pKa *.
(III) ß-naphthol fluorescence
We begin by formulating general expressions for how the fluorescence intensity
from the excited acid and base form changes with changing pH. If we introduce
expressions for how the concentration of A * and B * changes as a function of time
as shown in the reaction scheme and assuming a steady sate condition,
−
d[ A* ]
=k3[ A*]+k1[ A*] −k' 4[ B* ]− IA =0
dt
(2)
d[ B*]
=k' 4[ B*]+ k2[ B*] −k3[ A*]−I B =0
dt
The concentration of excited state can be expressed as:
−
(3)
k2
IA + IA + IB
'
k
A∗ = 4
k1 k 2 k 2 k 3
+ ' + k1
k 4'
k4
[ ]
(4)
Factor out IA from the numerator and k1 from the denominator we obtain
I ⎞
⎛ k2
⎜ ' +1+ B ⎟
IA ⎟
I
k
A∗ = A ⎜ 4
⎟
k1 ⎜ k 2 k 2 k 3
1
+
+
⎟
⎜ '
'
⎠
⎝ k 4 k1 k 4
[ ]
(5)
Multiply both numerator and denominator by k4'/ k2 gives
⎛
k 4' k 4' I B ⎞
⎜1+
⎟
+
I
k
k
I
⎜
A ⎟
2
2
A∗ = A ⎜
'
⎟
k1
k3 k4
⎜⎜ 1 + +
+ 1 ⎟⎟
k
k
1
2
⎝
⎠
[ ]
(6)
If we use the fact that IA= ω εA [A], IB= ω εB[B], k4'=k4 10-pH, [B]/[A]=10pH-pKa,
and that [A]=Ctot-[B], where Ctot is the total concentration of ß-naphthol, we obtain
a very useful expression for the fluorescence of the acid form :
(
⎛ k4 ⋅10 −pH ε B pH −pKa
DA
⎜ 1+ k 2 1+ ε A ⋅10
FA* =
pH −pKa ⎜
−pH
k k ⋅10
1+10
⎜
1+ 3 + 4
k1
k2
⎝
)
⎞
⎟
⎟⎟
⎠
DA= (PA* ω εA Ctot) / k1 is a constant if Ctot is a constant. We have also used the fact
that FA* = PA*·[Α∗], where PA* is the probability if fluorescence from the exited acid
(7)
form of the molecule.
An expression for the base form fluorescence (FB*) can be derived ia an
approximately similar method
(
⎛ k ε pK −pH
pH −pK a 1+ k3 1+ εA ⋅10 a
D ⋅10
⎜ 1
B
FB* = B pH− pK
−pH
a ⎜
1+10
⎜ 1+ k3 + k 4 ⋅10
k1
k2
⎝
)
⎞
⎟
⎟⎟
⎠
where DB= (PB* ω εB Ctot) / k2.
Question: perform this derivation. Hint: Multiply numerator and denominator by k3
and factor out k1 from the numerator and k1/k2 the denominator)
By examining the properties of expressions for FA* and FB* as a functions of pH
ranges (pH experimentally, a reasonable of pH is in the range 0 to 13), and at various
excitation wavelengths, we can arrive at a method for determining pKa and pKa*
respectively.
(IV) - Determination of pKa of fluorescence spectra
In this section we shall study in detail the functions FA* =f(pH) and FB* =f(pH).
First, we see that one can determine the pKa also from the fluorescence
measurements, i.e. the excited molecules can provide information on the
characteristics of the ground state. Can we in the absorption spectrum find a
wavelength where only the base absorbs? That is εA= 0, εB > 0?What happened to
the expression for FB* ?
Task:
Determine the value of FB* when pH = pKa. Note that we already know
how DB can be determined.
It is necessary to estimate the value of k4·10-pH / k2: k4 ≈ 1010 M-1 s-1. If
we assume that all collisions lead to reaction, i.e a diffusion controlled
process The diffusion constant in water is of size 1010 M-1s-1.
k2 ≈ 108 s-1 , 1/k2 = τ =fluorescence lifetime. This can be easily
determined for example with single photon counting. For a simple
aromatic compounds τ is typically a few nanoseconds.
We therefore have k4·10-pH/ k2 ≈ 102-pH<< 1 för pH>4. For a weak acid such as
β-naphthol if is certainly that pKa> 4.
Suggest an experiment to determine the pKa! Suggest an appropriate excitation
wavelength and at which you would measure the emission. Outlining how FB*
(8)
should look like as a function of pH?
(V) - Determination of pKa * from fluorescence spectra
It is a bit trickier to determine pKa *. The easiest way is to use the expression for
FA* and select the appropriate values for εA, εB . The method is based on the fact
that we can show that at pH values higher than pKa * but lower than the pKa; the
fluorescence intensity from both the acid base form are independent of pH, i.e it will
form a plateau.
Consider the expression for FA*. If 5 <pH <pKa -2, we have 0pH-pKa << 1 and
k4·10-pH/ k2 << 1.
A good approximation in this interval is therefore:
FA* =
DA
k
1+ 3
k1
⇔
FA*
=
DA
1
k ≡ Rp
1+ 3
k1
(9)
Rp is defined as the function's plateau value. We see that in principle the ratio k3/k1
can be determined.
Task: Show that the FA * has a plateau even at very low pH values. k3 / k1 can
be estimated using the approximate value you worked out for pKa*.
Note: k2 ≈ k1. Sketch the form of the function, i.e. FA* = f(pH) for pH 0-13.
If pH<<pKa we know that 10pH-pKa << 1.Then we can for reasonable values εB/εA
make the simplification:
⎛ k4 ⋅10−pH ⎞
1+
k2
⎜
⎟
FA* =DA
⎜⎜ k 3 k4 ⋅10 −pH ⎟⎟
⎝ 1+ k1 + k 2 ⎠
(10)
Immediately we see that for the pH value when k3/k1 = k4·10-pH /k2 , the expression
become
k
1+ 3
k1
FA*
=
≡R
D A 1+2⋅ k3 a
k1
And we find the relationship
(11)
Ra =
1
2−Rp
(12)
From an experimental determination of the plateau value Rp, Ra can be calculated.
From the experiential curve when the FA* / DA = Ra , we can obtain the pH value
that satisfies the equality k3/k1 = k4·10-pH /k2
-pH
which is equivalent to
/k 2. Vi have already shown that Ka*= k3 / k4,, therefore
k3/k4 = k1·10
Ka* = k1·10-pH /k2. If the resulting pH is inserted into the expression
pKa* = pH - lg (k1/k2) can pKa* be determined if we know k1 and k2. The
fluorescence lifetimes are not so difficult to measure experimentally using for
example time resolved single photon counting, for this lab however we can use the
literature values: k1= 1.38·108 s-1 and k2 = 1.06·108 s-1.
Task: We can similarly use FB* = f(pH) to determine the pKa*. Show that in this
case.
Ra = Rp / (1+ Rp) if εA/εB=1 (is there such a wavelength?), Sketch FB*!
Experiment:
It remains now to design experiments in which the pKa and pKa * can be determined. We have
access to the absorption and fluorescens spectrophotometers and a pH meter! Make a summary
of the experiments to be performed, the wavelengths of excitation and emission to be used in
the various experiments.
Miscellaneous:
- Beta-naphthol (2-hydroxy) is a potential carcinogen. Gloves are recommended.
- Keep in mind that beta-naphthol is a light-sensitive substance.
References:
(1) - Laws W.R. ; Brand L. Journal of Physical Chemistry, Vol. 83, page 759th
(2) - van Stam J.; Löfroth J.-E. Journal of Chemical Education, Vol. 63, page 181
(3) - Lakowicz, Principles of Fluorescence Spectroscopy,
Absorption spectra of β-naphtol at different pH values:
Curve (a), pH=13.3; (b), pH=10.0; (c), pH=9.6; (d), pH=9.0; (e), pH=0.0
Fluorescence spectra of β-Naphtol at different pH values
Curve (a), pH=13.6; (b), pH=9.2; (c), pH=8.3; (d), pH=7.8; (e), pH=2.8; (f), pH=1.0