Effect of pH changes on UV absorbance

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SUMMARY

The discussion centers on the effect of pH changes on the UV absorbance of phenol solutions. At pH 13, phenol exhibits an absorbance of 1.5 at 400 nm, while at pH 4, it shows 0.0 at 270 nm and 1.0 at 400 nm. The participants conclude that the phenoxide ion, formed at higher pH levels, absorbs at 400 nm due to increased electron donation from the lone pair of electrons, which stabilizes the ground state. The discrepancy in expected absorbance wavelengths is attributed to the influence of additional hydroxyl groups on the phenol structure, which may further shift the absorbance maximum.

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  • Understanding of UV-Vis spectroscopy principles
  • Knowledge of acid-base equilibrium and Le Chatelier's principle
  • Familiarity with the structure and properties of phenol and phenoxide
  • Basic skills in calculating pKa and molecular weight from absorbance data
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sven222
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So I have an chem assignment which I've been going through for the past week or so, but the last question has provided somewhat of a stumbling block. Here it is,

At pH 13, the absorbance of a particular phenolsolution is 1.5 at 400 nm and 0.0 at 270 nm. At pH 4, the values for the solution of the same cocentrations are 0.0 and 1.0 at these two wavelengths respectively. At pH 9, the values are 0.9 and 0.4 respectively.
(a) Explain these spectral changes
(b) Calculate the pKa of the phenol.
etc etc (few more related Q's)

So from the question, I have gathered the following,

High pH solutions absorb at higher wavelenghts. Whereas low pH solutions absorb at lower wavelengths. Given the equation C6H5OH --> C6H5O- + H+, one would assume that when the equilibrium of the reaction lies to the right, then the wavelength of the absorbance would decrease, because the concentration of H+ has increased (ie lower pH).

Now, if this is the case, then why? Does the increased H+ concentration stabilise the ground state as opposed to the excited electronic state? This would explain the decrease in wavelength (ie hypsochromic (blue) shift), which in turn is an increase in energy. In other words, more energy is required for an electronic excitiation to occur (stabilised ground state by the H+).

That sounds alright, but then I realized that phenol absorbs at around 270 nm. According to my theory, it should absorb at 400 nm (equil. would lie to the left, thus less H+, thus higher pH, thus high wavelength). And then it turns out the phenolate ion absorbs at about 280 nm, not the 400 nm that one would expect from the initial data. This begs the question, what causes the 400 nm absorption?

I think I'm almost there, but I've tangled myself up in all of this information. I'm not looking for a direct answer, but rather just a push in the right direction. If anyone could do that, it would be much appreciated :)
 
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Alright, you're pretty turned around. At a low pH the equilibrium lies to the left, not the right. You're added H+ with acid. So Le Chatelier's principle forces it to the left, you're protonating it. Where as with a high pH you're adding base, so you move the equilibrium to the right, you're deprotonating phenol.

So you've got to ask yourself what is UV spectroscopy measuring? In infrared you're measuing the vibrations of covalent bonds. In NMR you're measuring the relaxation of protons in a magnetic field. So what are you measuring in UV?

Once you've got that, then look at phenol and phenoxide. What is different in these two molecules? What has changed in context of UV?
 
Ah, you're my hero! :biggrin:

I knew there was a simple mistake I was making, that in turn ruined the rest of my workings. I must admit it has been some time since I studied equilibrium, so the french guys principle wasn't exactly slapping mei n the face. But it all makes sense now :smile:

So here's what I've come up with so far,

In highly basic solutions, more product (phenoxide) is synthesised in order to satisfy Le Chatelier's Principle. Therefore it is phenoxide that absorbs at 400 nm, as opposed to Phenol, which absorbs at 270 nm.

Now, in order to explain the jump in wavelength, I've focused on the newly formed lone pair of electrons resulting from the loss of the proton. Given that this lone pair is in conjugation with the pi electrons on the aromatic ring, then it is acting as an electron donor, which in turn increases the max wavelength of absorbance of UV radiation. Would this be correct? It seems to correlate with data obtained from various texts.

If this is the case, then why is the difference so much? According to the literature, the wavelength at which phenoxide absorbs should only be around 290 nm. 400 seems extremely high. What would this jump be a result of?
 
OK well I thought I had the answer, but some complications have arisen.

In the next part of the question, you needed to find the molecular weight of the phenol compound by calculating concentration through absorbance etc etc. So anyway, I worked out a molecular weight of 158, which is extremely close to that of 1,2,3,4,5-pentahydroxyphenol. In other words, phenol with 5 OH groups attached to the ring structure. Would this explain the abnormally high wavelength the ion absorbs at? After all, if one O- can cause a shift of about 20, then 5 should send it up even further yes? Each O- would be working in conjugation with the ring. Or so I think

Thoughts?
 

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