*Your ConquerHSC Notes are consistently being revised throughout the 2019 Year to ensure quality
Week 14 Content
Overview of Week 14 Inquiry Question – How are the ions present in the environment identified and measured?
Learning Objective #1 – Analyse the need for monitoring the environment
Learning Objective #2 – Conduct qualitative investigations – using flame tests, precipitation and complexation reactions as appropriate – to test for the presence in aqueous solution of the following ions:
- Cations: Barium, Calcium, Magnesium, Lead (II), Silver, Copper (II), Iron(II), Iron (III)
- Anions: Chloride, Bromide, Iodide, Hydroxide, Acetate, Carbonate, Sulfate, Phosphate
Learning Objective #3 – Conduct investigations and/or process data involving:
- Gravimetric analysis
- Precipitation Titrations
- Ultraviolet-visible (UV-Vis) Spectrophotometry
- Atomic absorption spectroscopy (AAS)
Overview of Week 14 Inquiry Question
Learning Objective #1 - Analyse the need for monitoring the environment
The main environmental focus will be on water quality and relating water quality’s effect on the environment as outlined in Module 8’s content focus in the NEW HSC Chemistry Syllabus.
Anyhow, let’s get start learning the material. The concern here why is it necessary to monitor the water quality?
Water pollution can basically be defined as the alteration in the water quality that negatively affects living organisms or render the water being unsuitable for desired uses.
While it rare to encounter water with zero impurities as many substances in nature as water-soluble such as ions. That being said, it does not mean that humans should pollute water.
It is important to monitor water quality by determining the amount and type of pollutants that is affecting the water quality so that the health of living organisms and the condition of surrounding environment can be maintained at a desirable state.
For instance, the use of fertilisers and pesticides (such as insecticides) in agriculture to grow crops may pollute local waterways. This is because water from farm irrigation or rain can dissolve and carry fertilisers & pesticides to into the waterway. Pesticides are toxic to fish as it can reduce the fish’s ability to regulate its internal temperature which can affect enzymes’ ability to catalyse necessary metabolic processes like cellular respiration to sustain life.
Hydrocarbon released into the sea due to oil spillages could be washed near and onto beaches. The hydrocarbons that enter the human body (e.g. the people in the water at beaches) could cause severe respiratory irritation. The long term effects of hydrocarbon exposure is also currently not fully understood.
Sulfur dioxide that is produced from factories involving in petroleum refining, metal smelting, paper manufacturing, etc can cause acidic rain.
SO2(g) + 1/2 O2(g) -> SO3(g)
SO3(g) + H2O(l) -> H2SO4(aq)
The sulfuric acid can then dissolve in waterways when it rains which lowers the pH of the water, resulting the death of living organisms. For example, fish larvae could die if the pH of the water is 5.5 or lower. This would threaten the biodiversity of aquatic organisms in the country which can result in economic implications as we have explored in Week 9’s notes such as declining export revenue, the price of fish increasing and thus reduces the affordability of seafood.
Furthermore, heavy metal pollution is toxic as they can prevent substrates from binding to enzymes’ active site and thus prevents normal metabolic processes from occurring in living organisms.
Also, most heavy metal bioaccumulates which means that it gets passed along in the food chain can accumulate to lethal quantities.
The infamous Minamata disaster in 1950s whereby a chemical plant in Japan released chemical waste containing large quantities mercury (a heavy metal) led to a serious disease, called the Minamata disease. The organisms that were exposed to the large quantities of mercury were affected in many ways. Some of these included narrow vision, numbness, slurring of speech and, in severe conditions, it resulted in paralysis, loss of consciousness, convulsions, fever and death.
As mercury is capable of bioaccumulating like many heavy metals, it means that the mercury is accumulate in the food chain so that consumption of surviving organisms with appreciable amounts of mercury could result in death. Many of the fish that resided in the polluted water had high levels of mercury which were eaten by the unsuspected local population, which resulted in the dangerous health conditions we have mentioned above.
Learning Objective #2 - Conduct qualitative investigations - using flame tests, precipitation and complexation reactions as appropriate.
Flowchart to test for unknown cation in a sample solution
Relevant complexation reactions in the above flowchart to test for unknown cation:
AgCl(s) + 2NH3(aq) <-> Ag(NH3)2+(aq) + Cl–(aq)
- NOTE: When the concentration of ammonia is high (6M is used), two ammonia molecule can both share its lone pair of electrons to bond with the silver ion, forming a water-soluble complex ion.
- [H3N: -> Ag+ <- :NH3]
- As a result, the silver chloride precipitate dissolves which can be used as a confirmatory test for the silver ions.
Copper hydroxide (blue precipitate) can be dissolved by adding excess ammonia by producing a water-soluble complex ion:
- Cu(OH)2(s) + 4NH3(aq) -> [Cu(NH3)4]2+(aq) + 2OH–(aq)
The resulting solution has a deep blue colour which can be used as a confirmatory test for the presence of Cu2+ ions in unknown sample solution.
Step #1: Dip the loop of the platinum wire into concentrated HCl then heat it to clean the wire in blue flame, freeing it from any impurities.
- NOTE: Sodium ion is common in most environments (such as coming from sweat on your hands). When sodium ion becomes a neutral sodium atom after regaining electron from excited state, it produces a bright orange-yellow flame that may obscure the colour of the metal ion in the metal chloride that you want to test.
- NOTE: Usually a blue (oxidising) flame is used and you stop heating the wire when no more coloured flame (that is not blue) is produced.
Step #2: Dip the wire into the HCl again and then into powdered metal chloride salt so that the salt is on the wire.
- NOTE: Chlorides of metal are used as salts as chloride ion is quite volatile in oxidising flame.
Step #3: Heat the wire in the blue flame and observe the flame’s colour.
Results of flame test:
- Ba2+ = Apple Green Colour
- Ca2+ = Reddish – Brown (Orange – Red is accepted too)
- Cu2+ = Blue – Green
- Na+= Orange – Yellow
- Pb2+ = Colourless
- Ag+ = Colourless
- Mg2+ = Colourless
- Fe2+ = Colourless
- Fe3+ = Colourless
Flowchart to test for unknown anion in a sample solution
Relevant complexation reactions in the above flowchart to test for unknown anion:
Complexation Reaction Test for Phosphate Anion
PO43- (aq) + 3NH4+(aq) + 12MoO42-(aq) + 24H+(aq) -> (NH4)3[PO4(MoO3)12](s) + 12H2O(l)
NOTE: (NH4)2MoO4(aq) dissociated into NH4+ and MoO42- ions.
NOTE: (NH4)3[PO4(MoO3)12] has a yellow colour, confirming the presence of phosphate anion.
Complexation Reaction Test for Acetate Anion
A new sample is used and neutral, aqueous ferric chloric is added as the complex ion is not formed under acidic conditions.
3Fe3+(aq) + 6CH3COO–(aq) + H2O(l) -> [Fe3(OH)2(CH3COO)6]+ (aq) + 2H+(aq)
NOTE: Fe3+ ions came from: FeCl3(aq) -> Fe3+(aq) + Cl–(aq); note that Cl– ion is a spectator ion.
NOTE: [Fe3(OH)2(CH3COO)6]+ is a complex ion with a crimson red colour.
After adding warm and boiling it, a brown-red precipitate forms.
[Fe3(OH)2(CH3COO)6]+ (aq) + 4H2O(l) -> 3Fe(OH)2CH3COO(s) + 3CH3COOH(aq) + H+(aq)
NOTE: Fe(OH)2CH3COO(s) precipitate has a brown-red colour.
Learning Objective #3 - Conduct investigations and/or process data involving:
- Gravimetric analysis
- Precipitation Titrations
Similar to how titration is a type of quantitative analysis that we have explored in Module 6, gravimetric analysis is also an quantitative analysis.
Typically, the form of gravimetric analysis that we are interested here is a precipitation procedure that is used to measure the weight of a substance (atom or compound) that is chemically related to the analyte that we are interested in.
Thus, through gravimetric analysis, we are able to obtain information about the analyte such as the percentage compositions of all the atom that makes up the analyte.
Example Question for Gravimetric Analysis
Question: Suppose that you wish to determine the percentage of iron in an iron ore sample.
- In order to do this, you were instructed to dissolve 1.24 grams of iron ore sample in hydrochloric acid.
- After this, water and ammonia was added to the solution to form hydrated iron oxide precipitate, Fe2O3•H2O.
- After filtering, washing and igniting the precipitate, you obtained 0.63 grams of pure ferric oxide solid.
- Calculate the percentage of iron in the sample of analyte.
Solution to Question
Moles of hydrated iron oxide = m/MM = 0.63g / (160g/mol) = 0.0040 moles
Moles of Fe3+ = 2 x moles of hydrated iron oxide due to 2:1 mole ratio = 0.0079 moles
Mass of Fe3+ = moles of Fe3+ x MM(Fe3+) = 0.0079 x 55.85 = 0.44 grams
% of Fe3+ = (0.44 / 1.24) x 100% = 35%
Advantages & Problems associated with gravimetric analysis
The benefit is that the results obtained from gravimetric analysis provide us with standardised values of the substance which are used instrument calibration. One example is the use of buffer solutions for calibration as we have explored in Module 6 to test the pH of a range of acids and bases.
Due to the need of standardisation and calibration by instruments, performing gravimetric analysis may be more accurate than using instruments to analyse the percentage composition of elements in an analyte.
That being said, gravimetric analysis is more time consuming than using instruments. There are also many problems that we need to remove and minimise when performing gravimetric analysis which we will explore soon.
Generic Gravimetric Analysis Procedure
Step 1: Weigh the sample of analyte which you want to measure.
Step 2: Dissolve the sample in water or an alternative, appropriate solvent.
Step 3: Transfer reason amount of precipitant in excess into the solution containing the sample. 
Step 4: Flocculate the precipitate using an appropriate method (e.g. heating, slowly mixing the solution, etc). 
Step 5: Filter the precipitate using an appropriate method.
Step 6: Wash the precipitate to remove impurities. 
Step 7: Dry the precipitate to a constant weight (and sometimes ignite the precipitate) then weigh the precipitate. 
Step 8: Repeat step 7 to ensure that the weight is constant.
NOTE: In HSC Chemistry, you will know what the appropriate solvent and appropriate method for coagulate and filtration will be. We will discuss about this now.
 The precipitant that you select to add to and react with the sample to produce a precipitate should be chosen on the basis that the resulting precipitate is very sparingly soluble (typically the Ksp should not exceed 10-6).
- You should avoid adding very large amounts of excess precipitant as it may increase the solubility of any precipitates that may be formed.
 The appropriate filtration method is decided based on cost-to-benefits as well as the nature of the precipitate. Suction filtration can be used to increase the speed compared to normal filtration that only relies on gravity.
 There may be some impurities that strongly adhere to the precipitate and thus will not be removed through washing.
- Before using water to wash the precipitate, it must be certain that the water will not dissolve amounts of precipitate that would be considered appreciable or significant in our quantitative analysis calculations. If yes, then an alternative wash liquid must be used such as a water-alcohol solution to reduce solubility of precipitate.
- You should always use a wash liquid in which the impurities have much higher solubility than the precipitate itself.
Precipitation titration is different to the normal titration that we explored in Module 6 in that a precipitate is produced when the titrant reacts with the analyte.
In order for a precipitation titration to be successfully performed whereby useful results can be extracted (e.g. unknown concentration of element in the analyte), many factors must be satisfied.
- These include that the precipitate must be quickly formed and be reproducible across multiple (of the same) precipitation titration reaction. 
- The produced precipitate also need to have low solubility (i.e. solubility of precipitate less than 5%) so that it appreciate amount of it will not re-dissolved in the salt solution after titration. 
- Furthermore, an end point must be capable of being identified. 
 The fact that many precipitation reactions are slow, it limits the range of reagents that we can use for the precipitation reaction and thus limits the method’s use in practice. The precipitation reaction should be quick so that it limits the amount of impurities that may be present in the resulting precipitate as a result of coprecipitation or during post-precipitation.
 The precipitate must be sufficiently sparingly soluble such that it drives the precipitation as close to completion as possible in order to have distinctive change in the ions’ concentration that form the precipitate at equivalence point (or endpoint in an experimental POV).
 There may be problems identifying when the end point has been reached if the precipitate formed is highly coloured, blending in with the indicator’s colour.
- Furthermore, the formation of a precipitate may yield a cloudy solution that obscures the endpoint.
- Due to the lack of indicators to detect the endpoint in precipitation titration, this process is not very commonly used as a quantitative analysis technique.
Precipitation Titration Curve
If you recall from Module 6, we have examined the titration curve of titrating a strong acid with a strong base. The precipitation titration curve is has a similar shape as that. We will explore this using an example below.
EXAMPLE: Precipitation Titration involving titrating Potassium Chloride with Silver Nitrate
When silver nitrate is as a precipitating agent, the precipitation titration is called an argentometric titration.
A silver nitrate solution is a common precipitating agent as the precipitate formed satisfies the criteria of precipitation titration that we have discussed earlier.
We can express this precipitation titration’s in our example involving potassium chloride and a standardised silver nitrate solution* in the form a chemical equation:
Ag+(aq) + Cl–(aq) <-> AgCl(s)
*Silver nitrate can be standardised by reacting with a standardised NaCl solution with three drops of potassium chromate solution as an indicator. We will explore the details about this in Volhard’s method which is a method of argentometric titration which we will examine shortly.
Notice that our reaction involves the formation of silver chloride, we have written it in a way such that the silver chloride precipitate is written on the right hand side of the equation.
That is, the above formula is opposite to the Ksp expression, also known as Kf.
If we know that the Ksp for the silver chloride is 1.8×10-10, then Kf = 1/Ksp = 1/1.8×10^-10 = 5.9 x 109.
- Notice that the Kf value is very large which is essential for the precipitation titration as it facilitates in driving the precipitation reaction to closer to completion.
- You should keep this in mind that, to verify whether or not the precipitant is appropriate (and thus valid) for a certain precipitation, you should calculate the Kf value.
- This is because we want to determine the concentration of chloride ions in the resulting silver chloride precipitate by titrating potassium chloride against a standardised solution of silver nitrate.
- However, it is important to remember that some of the precipitate will dissociate into its ions at equivalence point to establish equilibrium as per AgCl(s) <-> Ag+(aq) + Cl–(aq), although this will be small due to the sparingly insoluble nature of the precipitate. The value
- [Ag+] = [Cl–] since the precipitation reaction is at a 1:1 ratio.
- Also, 1.8 x 10-10 = [Ag+] [Cl–].
- Suppose that the [Cl]equilibrium = x. Therefore, 1.8 x 10-10 = x2.
- So, the x or the [Cl–]equilibrium = Squareroot (1.8 x 10-10) = 1.3 x 10-5.
- pCl (equivalence point) = -log10[Cl–] = -log10[1.3 x 10-5] = 4.87
- Therefore, the analyte that could be tested NOT limited to those containing only the chloride ion. However, we will be using chloride ion as our example in all three for be consistent and show differences between the different method’s procedures.
Mohr's Method in Argentometric titration
In Mohr’s method, a soluble chromate salt such as potassium chromate, whereby the chromate ions gives a yellow colour in solution, is used as an indicator to allow silver ions to react with chromate ions to form a reddish-brown silver chromate precipitate. This allows the observation that endpoint has been reached.
- To reach the endpoint, excess silver ion is added in the form of standardised silver nitrate so that the silver ions react with the chromate ions to form the reddish-brown silver chromate precipitate.
This is possible because the molar solubility of silver chromate is higher than the molar solubility of silver chloride. So, silver chloride will precipitate before silver chromate which allows us to determine that the endpoint has been reached when the yellow coloured solution produces a reddish-brown precipitate.
- NOTE: This method can only be performed if the analyte containing the chloride ion that we want to measure the concentration of yields a solution that has a pH of approximately 8 or else the precipitation of silver chromate will be affected. Thus, endpoint and equivalence will be inaccurate and, thus, affect our calculation of the chloride concentration in the analyte sample.
- NOTE: We cannot determine analytes containing iodides using Mohr’s method as it will form a yellow colour precipitate with silver ions which obscures the endpoint. Typically, the Mohr’s method is only used for analytes containing bromides and chloride ions.
Volhard's Method in Argentometric titration
Compared to the Mohr’s method which involves a direct titration, the Volhard method uses back titration. Hence, like all back titration, two titrants are used.
First, an aqueous chloride solution (e.g NaCl) of known concentration and volume is titrated against the silver nitrate to standardise the silver nitrate. Similar to the Mohr Method, the a soluble chromate salt such as potassium chromate is used as an indicator to determine when the endpoint is reached.
In the second step, the analyte containing the chloride ions we wish to determine concentration for is titrated against excess quantities of excess standardised silver nitrate to form silver chloride with excess quantities silver nitrate.
- Ferric Ammonium Sulfate (green colour) is used as an indicator here where excess silver nitrate will react with iron alum to form silver sulfate precipitate where the green colour will become colourless as silver sulfate forms.
Thirdly, step three involves the solution that contains the excess silver nitrate solution from step 2 to be titrated against a standardised, soluble thiocyanate salt such as KSCN or NH4SCN to produce AgSCN precipitate whereby a ferric ammonium sulfate solution is used as an indicator to make endpoint observable.
- The endpoint of the titration occurring in step 3 can be determined whereby all silver nitrate would have reacted with KSCN or NH4SCN (depending on which you choose to use) and thus excess thiocyanate ion (from KSCN or NH4SCN) will react with Fe3+ in the indicator to form iron (III) thiocyanate ion which has a crimson colour in solution which marks the endpoint
- From this result of the amount of SCN- used, we can determine the concentration of the excess silver nitrate that was remaining after the titration between the analyte of unknown concentration in step 2. Thus, we can determine the unknown concentration of the analyte introduced in Step 2 using mole ratio.
NOTE: If the indicator is more stable than the silver ions, then the indicator will precipitate the chloride ions immediately in step 2 which we don’t want for purpose of back titration.
NOTE: Volhard’s Method should be carried out in acidic medium as a basic solution may result in the Fe3+ forming iron (III) hydroxide rather than the red iron thiocyanate ion formed in step 3 which means that endpoint will not be seen.
NOTE: Chloroform can be added in step 2 in the conical flask containing the analyte so that it lowers the solubility of the silver chloride that is formed.
NOTE: This method can be used to determine concentration of a variety of analytes as it can determine many halogen anions including iodide as well as other classes of anion such as sulfides, carbonate, thiocyanates, cyanides and chromates.
Fajan's Method in Argentometric titration
Basically in Fajan’s Method, silver nitrate can be standardised using a sodium chloride standard solution, similar to Step 1 in the Volhard’s method.
The analyte of unknown concentration containing the chloride ions can then be titrated against the now-standardised silver nitrate to form silver chloride precipitate where fluorescein is used as a yellow-green indicator.
When excess silver nitrate is added, the excess silver ions will be form a layer surrounding (adsorb on the surface of) the silver chloride precipitate whereby the indicator will react with the excess silver ions that is absorbed on the silver chloride precipitate the pink colour due to the formation of fluoresceinate (negatively charged) which is pink in colour that binds with the inner excess Ag+ layer that is absorbed onto the AgCl precipitate.
So, the endpoint will change from yellow-green to pink, marking the precipitation of silver chloride.
NOTE: This method can be used to determine chlorides, bromides and iodides.
Learning Objective #4 - Conduct investigations and/or process data to determine the concentration of coloured species and/or metal ions in aqueous solution, including but not limited to, the use of:
- Ultraviolet-visible (UV-Vis) Spectrophotometry
- Atomic absorption spectroscopy (AAS)
Before we explore colourimetry, let’s examine a few related terms.
The term ‘photometry’ refers to the measurement of the light intensity whereas the term ‘spectrophotometry’ is used to describe the measurement of the light intensity of a specific wavelength.
- So, spectrophotometry is a more specific branch of photometry.
Spectroscopy is an even broader term that encompasses both photometry and spectrophotometry as it involves the study of absorption and emission of energy or radiation (e.g. light energy) by a substance to examine the relationship between energy and matter.
Every unique molecule have their (usually unique) absorption and emission spectrums. Here, we are dealing with the absorption spectrum.
Therefore, we fire (visible) light radiation at a specific wavelength to the sample solution to determine the unknown concentration of the molecule of concern. We will discuss how we can fire a light radiation at a specific wavelength shortly.
In essence, colourimetry is a type of spectrophotometry whereby a colorimeter is used to examine how a specific wavelength/frequency (colour) of visible light that is absorbed by a molecule (or atom) of concern in the test sample to infer the molecule’s (or atom’s) concentration.
This is because the absorbance value of the test sample can be compared against standardised solutions with known concentration & absorbance of the molecule/sample of concern using a calibration curve.
- The lower the intensity the light exiting, the higher the absorbance value and the higher the concentration of the molecule or atom that is present in the test sample.
A colourimeter is an instrument that uses an interchangeable/replaceable colour filter to isolate wavelengths of light that the unknown molecule most readily absorbs so that the intensity of the wavelengths absorbed can be measured using a detector and used to determine the unknown concentration of a molecule of concern that is present of the test sample.
If no filter is used, the accuracy with more lowered as the wavelengths of light in which the molecule must readily absorbs is less intense and may not produce the total/true maximum absorbance value.
What coloured filter should I use?
The workings of a Colourimeter
Although it is possible to perform colourimetry using the human eye by comparing colour intensities, it is often difficulty to quantify the intensity that is observed and relate it to concentration.
Hence, a colourimeter can be used to determine the intensity of the specific wavelength that is absorbed by the test sample (difference b/w light before and after passing through the sample) which will help us determine the concentration using a calibration curve whereby we have constructed using standardised solutions.
It is important to note that colourimetry only deals with light with different wavelengths in the visible light region of the electromagnetic spectrum. This is the reason why colourimetry can be using the naked eye (although difficult). This is because the eye cannot see light with wavelengths in the ultraviolet and infrared region of the EM spectrum.
- We will explore other types of spectroscopy that involves other kinds of electromagnetic radiation that are outside the visible light region in next week.
As you would have learn in junior science, the visible light region consists of wavelengths of light ranging from 380 – 780nm in which humans can see.
- Sometimes, you may see 400-700nm as the visible light region but this will vary depending on the light source and sensitivity of the eye.
- Radiation of different wavelengths in the visible light region of the EM spectrum appear as different colours when seen using the human eye.
- Sunlight, which the human eye sees as white light, is composed of a mixture of light at different wavelengths.
It is worth mentioning that light can be interpreted in two forms as it has both wave and particle properties.
As a wave form, the wavelength of light is inversely proportional to its frequency.
Frequency = Speed of Light / Wavelength (λ).
Light also have particle properties in which it exist as discrete units (particles) known as photons.
Each photon carries energy equal to the plank’s constant (h) multiplied by the frequency of light (f).
E = Plank’s constant x Frequency = h x f
As light is transmitted through the cuvette containing the sample with the molecule of concern, some of the light may be absorbed by the the cuvette’s wall and the solvent.
Typically distilled water is used to dissolve the test sample containing the molecule or atom of interest.
Therefore, a reference cuvette consisting of an identical cuvette and same solvent is measured for its absorbance to set it as a reference point for colourimeter before testing the sample containing the molecule/atom of concern.
Ultraviolet-visible (UV-Vis) Spectrophotmetry
Compared to a colourimeter, a UV-Vis spectrophotometer uses a diffraction grating or (glass/quartz) prism to split visible light and ultraviolet radiation into different wavelengths.
As a result of this, radiation with wavelength ranging from ultraviolet to the visible light region can be measured for its intensity, i.e. 200 – 780nm.
Through the use of UV-Vis spectroscopy (using ultraviolet and visible light radiation), we can examine the extent of conjugation of the unknown molecule of concern that is present in the sample solution.
Therefore, UV-Vis allows us to determine the structure of a molecule which is often used alongside other spectroscopy methods such as IR (using infrared radiation) and NMR (using radiowaves radiation). These spectroscopy techniques allows chemist to gain a complete understanding of the overall structure and molecular formula of the unknown molecule.
Conjugation essentially refers to a structure of alternating double and single bonds.
In UV-Vis spectroscopy, the incident radiation’s wavelength applied to the sample starts from the 780nm to the 200nm (i.e. high to low).
Recall that the lower of the radiation’s wavelength, the higher the radiation’s energy.
So, if 200nm is used at first, the molecule’s electrons may absorb the wavelength instantly and transition from a lower energy electron orbital to an electron orbital of higher energy, (more on this later).
Vice versa, if the radiation’s wavelength starts from the 800nm and is eventually reduced to the higher energy wavelengths, it is possible to pinpoint the highest specific wavelength (λmax) in which the molecule’s electrons will successfully absorb and be moved to an excited state.
When radiation passes through the sample solution, electron can absorb the radiation and be move to a an electron orbital of higher energy. This transition can be done in four ways and, in each of the ways, electrons are moved from the HOMO (highest occupied molecular orbital) to the LUMO (lowest occupied molecular orbital).
Transition Type 1: An electron may move from a sigma HOMO orbital to LUMO sigma* orbital. This transition requires a certain amount of energy which corresponds to a certain wavelength of light (E=hf).
- Compared to the remaining three other types of transitions, the excitation of an electron from sigma orbital to sigma* orbital requires highest energy and corresponds to a wavelength of light required of approximately ~ 150nm.
- However, UV radiation has energy with wavelength between 200nm to 400nm.
- Therefore, this transition of from electrons in the HOMO to the LUMO is not possible in normal UV-Vis spectroscopy conditions.
- To measure this electron transition, vacuum UV radiation must be used.
- If electron makes this type of orbital transition, our molecule of concern would be hydrocarbons molecules with no conjugation (e.g. saturated alkanes).
Transition Type #2: Another type of electron transition is from HOMO ‘non-bonding (n)’ orbital to LUMO sigma* orbital. The electron would need to absorb an energy with wavelength approximately equal to ~175nm.
Generally, the molecule of concern here will have saturated molecules where the bond consists of two different atoms other than C-H bonds such as – e.g. halogens bonded to carbon (C-Cl, C-Br), alcohols (e.g. C – O -H), aldehydes, amines (C – N), water, ketones. Again, since this is outside the range of normal UV-vis spectroscopy conditions, vacuum UV radiation needs to be used.
- This is why understand UV-Vis spectroscopy that operates using normal UV radiation wavelengths have water used as a solvent to make the sample solution containing for the dissolved molecule that you wish to test for concentration.
Transition Type #3: The third type of electron orbital transition/excitation is from a HOMO Pi orbital to LUMO Pi* orbital. To excite an electron from Pi orbital to Pi* orbital, the energy required is more than 200nm (range depends on degree of conjugation).
- Generally, when this electron excitation occurs, the compound typically contains double or triple bond (unsaturated) hydrocarbons with conjugation OR aromatic (contains a ring structure) compounds.
Transition Type #4: The last type of transition involves electron moving from a HOMO ‘non-bonding (n)’ orbital to LUMO Pi* orbital. Here, the energy required to excite electrons to cause such orbital transition is from 200nm – 800nm (range depends on conjugation of compound).
- Generally, these are carbonyl compounds (C=O), Cyanides (C≡N) contained in the sample solution.
NOTE: For a molecule to have colour, the electron make a transition from the HOMO to the LUMO at a wavelength corresponding to the visible light region.
In general, the trend is that the more stable the molecule, the higher wavelength it is able to absorb.
For example: Increasing conjugation makes compound more stable (since resonance structures can be formed) which means that higher wavelength (or less energy) radiation is required to promote electron from LUMO to HOMO.
Also, higher degree of substitution of central atoms would also increase stability as the negatively charged electrons are spread across a large area.
- For instance, substitution can involve replacing a hydrogen atom with a methyl group.
Atomic absorption spectroscopy (AAS)
Through atomic absorption spectroscopy, it is possible to analyse the concentration of a metal element (in the form of a metal ion) that may be present in the sample that contains many other elements.
- NOTE: AAS machines is very sensitive as they can measure up to parts-per-billion (ppb).
- For such reasons only small volume such as few millimetres of sample solution is required to precisely determine the concentration of the metal ion present in the sample solution.
- Compared to other quantitative analytical methods such as gravimetric analysis, AAS does not require as large volume solutions that containing more moles of the element.
At the start of the AAS, after dissolving our sample in a solvent (usually distilled water), the sample solution is passed through an atomiser to heat the solution. By doing this, the solution can be converted into a vapour state whereby all molecules in the vapourised sample will be atomised (turning all molecules in sample into individual atoms).
It is worth noting that some of the atoms that were atomised have electrons that are in the unstable, excited state whereby some of the atoms have their electrons in their ground state.
In AAS, we will be dealing with the atoms with electrons in the ground state after atomisation. Comparatively, in atomic emission spectroscopy (AES), the atoms that are in the excited state after vapourisation are analysed.
After atomisation, a monochromatic beam of radiation with a specific wavelength is passed through the vapourised atoms in the flame so that only the electrons of the specific atom of our interest will absorb the radiation and the electrons will move from the ground state to the excited state.
It is important to note that: Electrons in atoms are located in discrete energy levels and they absorb some intensity of the electromagnetic radiation at a specific wavelength (the wavelength of radiation or energy which is absorbed is different for electrons in different element). The electron that absorbs the specific wavelength of energy can be moved to a higher, unstable energy level (excited state). When an electron is unexcited, it is located in its ground energy level state and an electron said to be ‘excited’ if it is moved to a higher, unstable energy level than when it’s in the ground state.
The decrease in the intensity of the radiation, due to the absorption of energy (radiation) by electrons, will be directly proportional to the concentration of the specific atom of our interest that is present in our sample. This is the principle of AAS which utilises that knowledge that electron exists in discrete energy level.
NOTE: The basis of AES (atomic emission spectroscopy) is similar to AAS. However, in AES, it deals with the fact that the electrons of different elements emit a specific wavelength of energy when the excited electrons returns to its ground state.
Structure and operation of Atomic Absorption Spectroscopy Machine
1. Hollow-cathode lamp – A hollow-cathode lamp containing the specific metal to be analysed is present in the cathode. For instance, if we wish to measure the concentration of calcium in a water sample, we would use a calcium cathode lamp. We do not know the concentration of calcium in the water sample (or maybe there isn’t any calcium ions) so we can use the AAS machine or atomic absorption spectrometer to determine the presence and concentration of the metal ion that we wish to measure.
- As a current (electricity) travels through the cathode lamp, the electrons in the specific metal in the lamp are excited. When the excited electrons return to their ground state, they produces a specific wavelength of light (corresponding to the element’s emission spectrum) that can be absorbed by vapourised element (in the form of metal ions due to heat of flame in step 2) that may be present in the sample solution.
2. Flame – The sample solution that is contained inside a nebuliser is aspirated or sprayed in the flame. The flame, with a temperature of approximately a thousand degrees celsius, is produced using a slot-type burner to enhance the absorbance of the specific-wavelength electromagnetic radiation by the vapourised metal ions of concern as the radiation passes through the flame.
- NOTE: Sometimes a graphite furnace is used to produce triple the temperature to enable analysis of sample solutions of extremely small volume (thus potentially smaller quantity of metal ion of concern).
- NOTE: The flame can scatter the light into different wavelengths therefore we need to use a monochromator to direct only the radiation with the specific wavelength absorbed by the element of concern to the detector.
3. Monochromator – If the sample solution contains the metal ion of concern, it will absorb the radiation of specific wavelength produced by the lamp as it passes through the flame. The remaining light of the specific wavelength, alongside any wavelengths of light that is scattered by the flame, is passed through the flame and through the monochromator. The monochromator contains mirrors, a diffraction grating and a slit. The mirrors and diffraction grating directs the light through the monochromator so that all of it reaches the slit whereby the slit only allows light of the specific wavelength (monochromatic light) corresponding to the metal ion to pass through the monochromator and into the photomultiplier.
4. Photomultiplier (Detector) – this is a detector instrument that measures the intensity of the specific-wavelength of light and convert it into an absorbance reading that is carried via electrical signal.
5. Display screen – Here, a display screen receives the electrical signal and conveys the absorbance value (which is related to the concentration of the metal ion of concern present in the sample solution).
Prior to testing the absorbance reading of the sample solution, you would have used diluted, standard solutions containing the metal of known concentration and obtained absorbance value for each of them using the AAS machine to create a calibration curve. After that, the sample solution containing unknown concentration of the metal can be tested for its absorbance value using the AAS and the concentration of the metal of concern can be obtained using the calibration curve.
NOTE: Within the standard solutions, a blank solution containing zero concentration of the metal is tested for absorbance (should be zero) and be used calibrated the machine as a reference point.
Explain the principle of flame test whereby two named metal atoms emit an unique colour.
Explain why not all metal atoms are suitable for flame test?