U4 Aos 1.docx

STRUCTURE AND NOMENCLATURE OF ORGANIC COMPOUNDS The carbon atom with reference to valence number, bond strength, stability of carbon bonds with other elements and the formation of isomers (structural and stereoisomers) to explain carbon compound diversity, including identification of chiral centres in optical isomers of simple organic compounds and distinction between cis- and trans- isomers in simple geometric isomers.                                  

The valence number of carbon is four. All of the four valence electrons in a carbon atom are available to bond with other atoms. A carbon atom can form up to four covalent bonds with up to four atoms of carbon and non-metal atoms A carbon atoms can form single, double or triple bonds with other atoms. A molecule that contains only single carbon-carbon bonds is a saturated molecule. A molecule that contains one or more double and/or triple carbon-carbon bonds is an unsaturated molecule. Hydrocarbons are made up of carbon and hydrogen atoms. The covalent bonds between carbon and other atoms each have a bond energy. Bond energy is the amount of energy required to break the covalent bond. The higher the bond energy, the stronger the covalent bond. A carbon atom would be most stable bonded another carbon atom compared to other atoms bonded with atoms of the same element. A hetero group/atom is a group/atom in a hydrocarbon that is not a hydrogen or carbon atom. Isomers are molecules of the same molecular formula but different arrangement of its atoms. Isomers have different chemical and physical properties and thus behave differently. The two main types of isomers are structural and stereoisomers. The types of structural isomers are chain and positional isomers. The difference between chain and positional isomers is that positional isomers occur for organic molecules containing functional groups whereas chain isomers do not. Positional isomers consist of a carbon chain and a functional group which can be attached to different locations on the carbon chain. Chain isomers of alkanes can contain more than one alkyl group. More than one alkyl group can be attached to the same carbon atom of a chain isomer. Stereoisomers are isomers in which two molecules connect in the same order but have different arrangements in space. Stereoisomers differ to each other in terms of their arrangements three-dimensionally. The types of stereoisomers are optical and geometric isomers. In optical isomers there is a different placement of groups around an atom in a molecule. Geometric isomers have the same molecular and structural formula, but are arranged differently in space. Alkanes do not have geometric isomers since single bonds can rotate. Carbon double bonds cannot rotate. Therefore geometric isomers exist for hydrocarbons with C=C bonds are involved. A cis isomer is the geometric isomer where the two hetero groups are on the same side of the double bond. A trans isomer is the geometric isomer where the two hetero groups are on opposite sides of the double bond. Cis and trans isomers are different molecules with slightly different chemical and physical properties. Optical isomers are arranged in the same way but not the same way spatially..

Optical isomers have the same molecular and semistructural formula. Optical isomers have a different placement around a carbon atom

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The feature specific to optical isomerism is the effect of optical isomers have on light passes through them. Polarised light is light that is only oscillating in one particular direction. The plane of polarised light is related to the path traced by the wave as if is travelling and oscillating.

Chiral substances rotate polarised light. Substances that have two optical isomers make these isomers enantiomers. Each enantiomer rotates the plane of polarised light in opposite directions. Molecules that have at least one carbon atom that is bonded to four different substituents. Carbon atoms bonded to four different substituents are called chiral centres. Optical isomers or enantiomers are mirror images of one another. Optical isomers have the same physical and chemical properties. A chiral molecule is one that has optical isomers. An achiral molecule is one that does not have optical isomers.

Structures including molecular, structural and semi-structural formulas of alkanes (including cyclohexane), alkenes, alkynes, benzene, haloalkanes, primary amines, primary amides, alcohols (primary, secondary, tertiary), aldehydes, ketones, carboxylic acids and non-branched esters.          

Molecular formulas indicate the number and type of atoms in each element present in a molecule. Molecular formulas do not show how the atoms are physically arranged in a molecule. Structural formulas depict the spatial location of atoms relative to each other in a molecule. Structural formulas also give the number and location of covalent bonds. In structural formulas lone pairs are not included. Since electron pairs repel from one another the lines are to be drawn 109.5 degrees apart. Structural formulas consist of tetrahedral structure and straight chain molecules. Semistructural formulas indicate the connections in the structure of a compound without presenting its three-dimensional arrangement. Semistructural formulas are also known as condensed formulas.

IUPAC systematic naming of organic compounds up to C8 with no more than two functional groups for a molecule, limited to non-cyclic hydrocarbons, haloalkanes, primary amines, alcohols (primary, secondary, tertiary), carboxylic acids and non-branched esters.         

Alkanes are hydrocarbons where there are only single bonds between carbon atoms. Alkanes are saturated. Alkenes are hydrocarbons where there are at least one double bond between two carbon atoms and no triple bonds. Alkenes are unsaturated. Alkynes are hydrocarbons where there is at least one triple bond between two carbon atoms. Alkynes are unsaturated. Alkanols are aka alcohols. Alcohols are organic molecules with a hydroxyl group in the molecule. There are three types of alkanols.

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Primary alcohols have a hydroxyl group on the terminal carbon atom. Secondary alcohols have hydroxyl groupon a carbon atom in the middle, which is bonded to two other carbon atoms. Tertiary alcohols have a hydroxy group attached to a carbon in the middle, that is attached to three other carbon atoms. Carboxylic acids is a set of hydrocarbon derivatives containing the moiety -COOH at the end if the hydrocarbon. The suffix of a carboxylic acid is -oic acid. Esters are derivative of carboxylic acids. Esters have a -COOC- moiety. The moiety of an ester is always within the molecule, unlike carboxylic acids. Haloalkanes are organic compounds like alkanes that have a single bonds with halogen atoms. The halogen of an alkane replaces at least one of the H atoms in the molecules. Primary amines are molecules with an -NH2 group in the molecule. The -Nh2 moiety is known as the amino functional group. For the amine to be considered a primary amine, the nitrogen atoms must have two hydrogen atoms bonded to it. Ketones are hydrocarbons with a C=O moiety. The C=O moiety is a carbonyl group. The carbonyl group must be in the middle of the moiety for the hydrocarbon to be considered as a ketone. Aldehydes are groups with a carbonyl group at the terminal carbon. Primary amides are hydrocarbons with a -CONH2 moeity at a terminal carbon. The -CONH2 moiety is known as an amide functional group. Benzene is a cyclic molecule with formula C6H6. Benzene is an alkene yet does not have similar properties or undergoes the same reactions. Each carbon bond in a benzene can be seen to have 1.5 bonds with one another. The electron that forms the second bond between two carbon atoms roam around entire molecule.

CATEGORIES, PROPERTIES AND REACTIONS OF ORGANIC COMPOUNDS An explanation of trends in physical properties (boiling point, viscosity) and flashpoint with reference to structure and bonding.      

For alkane, alkene and alkyne molecules, the boiling points increase as the size of the molecules increase. These molecules are non-polar and thus compose of dispersion forces only, which increases as the length of the carbon chain increases. Straight-chain alkanes fit together closely and thus have higher boiling points than their branched-chain isomers. The strength of dispersion forces between molecules depends on the size and shape of the molecules. Haloalkanes contain polar bonds. Haloalkanes contain dipole-dipole forces and thus have higher boiling points than alkanes.

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Since hydrogen bonds are the strongest of the intermolecular forces, molecules that form these bonds have higher boiling points. These molecules with hydrogen bonds include alcohols, carboxylic acid, amines and amides. Aldehydes, ketones and esters contain a carbon-oxygen double bond. This bond is a polar bond as there is a high difference between the electronegativities of oxygen and carbon. These molecules are held together by dipole-dipole attractions. As the length of a hydrocarbon chain of an aldehyde, ketone or ester increases, so does the boiling point. In ascending order of boiling points, alkanes, alkenes and alkynes, haloalkanes, and aldehydes, ketones and esters. Small alcohols and amines are soluble in water because hydrogen bonds can be formed with polar groups and water. Lone pairs of oxygen in an alcohol can attract to the partially positive hydrogen in water molecules. Lone pairs of nitrogen in an amine can attract to the partially positive hydrogen in water molecules. Small amides and carboxylic acids can dissolve in water too. As the carbon chain length of an organic molecules increases, its solubility in water decreases. The longer hydrocarbon chains in an organic molecules disrupt the hydrogen bonds between water molecules. Hydrocarbon tail cannot form hydrogen bonds with water. Only dispersion forces can occur between the hydrocarbon chain and water molecules. Dispersion forces are weaker than hydrogen bonds. As organic molecules get larger in size, their solubility in organic solvents increase. Organic solvents are non-polar. Organic molecules with a very long hydrocarbon chain are non-polar. Like dissolves like. Primary alcohols have higher boiling points than secondary alcohols and tertiary alcohol. Secondary alcohols have higher boiling points than tertiary alcohols. The longer the hydrocarbon chain, the more soluble the organic molecule in an organic solvent. Aldehydes, ketones and esters all contain a C=O bond. Aldehydes, ketones and esters are held together by dipole-dipole attractions. Aldehydes, ketones and esters cannot form hydrogen bonds with water. The C=O bond is polar. Esters, aldehydes and ketones contain a permanent dipole. Small esters, ketones and aldehydes can be soluble in water. They cannot form hydrogen bonds with its own molecules. Hydrogen bonds can form between lone pair electrons of an O atom and the H atom of a water molecules. As hydrocarbon chain length increases, solubility in water decreases. As the non-polar hydrocarbon chain length increases, solubility in organic solvents increase. Esters, aldehydes and ketones can form dipole-dipole interactions with nearby molecules. These compounds have higher boiling points than similar sized alkanes. The similar sized alkanes are non-polar and can only form dispersion forces, which are weaker than dipole-dipole bonds. Esters, ketones and aldehydes have lower boiling points than acids, amide, amines and alcohols. Alcohols can form hydrogen bonds. Hydrogen bonds are stronger than dipole-dipole bonds. As the hydrocarbon chain length of an ester, ketone or amide increases, so does it boiling point, like with alcohols..

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Increasing the length of the hydrocarbon increases the strength of the dispersion forces Viscosity is the resistance of a liquid to pouring or flowing, The viscosity of a liquid depends on the interactions between molecules. Viscosity increases as the forces of attraction between the molecules increase. As the chain length of a hydrocarbon increases, so does the strength of dispersion forces between molecules and thus, viscosity. The flashpoint of a liquid is the lowest temperature at which it will produce a vapour that will ignite to produce a flame when an ignition source is applied. The units for flashpoint is degrees Celsius. If the dispersion forces of a molecule is weak, it has a lower flashpoint than a molecule with stronger dispersion forces Molecules with weaker intermolecular forces have lower flashpoints. Molecules with lower flashpoints are more flammable since they can ignite at lower temperature. The flashpoint of an organic liquid increases as the carbon chain length of molecules of the compound increases.

Organic reactions, including appropriate equations and reagents, for the oxidation of primary and secondary alcohols, substitution reactions of haloalkanes, addition reactions of alkenes, hydrolysis reactions of esters, the condensation reaction between an amine and a carboxylic acid, and the esterification reaction between an alcohol and a carboxylic acid.                       

Substitution reactions occur when an atom/functional group is substituted by another atom/functional group as a result of a chemical reaction. Alkanes usually relatively stable compounds but can undergo combustion reactions. The more stable a particle, the more unreactive it is. Alkanes undergo substitution reactions with halogens in the presence of ultraviolet light to produce haloalkanes. Haloalkanes undergo substitution reactions with NaOH to produce alcohols. Haloalkanes undergo substitution reactions with ammonia to produce amines. Alkanes and alkenes undergo combustion with oxygen. Alkenes are unsaturated hydrocarbons. Alkenes undergo addition reactions to produce saturated compounds- alkanes. Alkenes undergo addition reactions with hydrogen and a metal catalyst to produce alkanes. Alkenes undergo addition reactions with halogens to produce dihaloalkanes. Alkenes undergo addition reactions with hydrogen halides to produce haloalkanes. Alkenes undergo addition reactions with water and phosphoric acid catalyst to produce alcohols. Combustion and substitution reactions involving alcohols are redox reactions. Alcohols can be oxidised by other strong oxidising agents. Primary alcohols are first oxidised to aldehydes. Primary alcohols first oxidesd to aldehydes may be further oxidised to carboxylic acids. Secondary alcohols are oxidised to ketones. Tertiary alcohols are resistant to oxidation by these oxidising agents. Higher temperatures and longer reaction times favour the production of carboxylic acid over aldehydes. Carboxylic acids can react with alcohols in the presence of an acid catalyst to produce esters. Carboxylic acids react with ammonia and amines to form amides. Addition polymerisation is the joining of monomers.

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Alkenes undergo polymerisation reactions to form polymers containing only carbon-carbon single bonds. Double bond in alkene breaks to form single covalent bonds with neighbouring monomer in addition polymerisation. The empirical formula of each monomer is the formula of the polymer. Organic molecules with triple carbon bonds can also take part in addition reactions. Esterification is the process of forming esters. Esterification requires a catalyst and a little bit of heat. Esterification occurs between primary alcohols and carboxylic acids. Esterification is a reversible reaction whereby the ester can react with water to form alcohol and carboxylic acid in a process called hydrolysis. Hydrolysis is also known as a hydrolytic reaction. Hydrolysis requires a catalyst. Esters are not the only compounds that undergo hydrolysis. Esters can be hydrolysed by acids or alkalis to produce alcohols and carboxylic acids or their sodium salts, respectively.

The pathways used to synthesise primary haloalkanes, primary alcohols, primary amines, carboxylic acids and esters, including calculations of atom economy and percentage yield of single-step or overall pathway reactions. o o o o o o o o o o

A reaction pathway summarises the reactions required to produce a product from simple starting materials. A reaction pathway is a sequence of more than one reaction used to convert a reactant into a product. Reaction pathways indicate the reaction conditions and reagents required for each steps. The theoretical yield of a chemical reaction is the mass of the product that would be formed if the limiting reactant reacted completely. To calculate the percentage yield, divide the actual yield by the theoretical yield that would be obtained, if all the limiting reactant reacted completely and multiply by 100. When a reaction proceeds by a number of steps, the overall percentage is reduced at each step. More reactions causes more energy to be lost. The overall yield of the product of a multistep reaction is found by multiplying the percentage yield of each step together and expressing as a percentage by multiply by 100. The atom economy for a chemical reaction is a measure of how many atoms in the reactants end up in the desired product for the reaction. Atom economy can be calculated by dividing the molar mass of the desired product by the molar mass of all reactants, then multiplying by 100.

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ANALYSIS OF ORGANIC COMPOUNDS The principles and applications of mass spectroscopy (excluding features of instrumentation and operation) and interpretation of qualitative and quantitative data, including identification of molecular ion peak, determination of molecular mass and identification of simple fragments .        

Mass spectroscopy analyses samples of solids, liquids and gases. This technique is sensitive and quantitative. Mass spectroscopy can detect concentrations in ppb to parts per trillion. In a mass spectrometer ions are formed in the ionisation chamber where the sample is exposed to high voltages. The ions are separated in a magnetic field on the basis of their mass-to-charge ration (m/z) The number of ions with different m/z values are measured by a detector and the data is recorded as a mass spectrum. A mass spectrometer does not use electromagnetic radiation. This device gives very precise information about the mass of cations formed in it.

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A mass spectrum has the relative intensity (%) on the vertical axis. A mass spectrum has mass-to-charge ratio (m/z) on the horizontal axis. The small peaks in a mass spectrum are caused by molecular ions. The m/z value is equal to the relative molecular mass. A molecular ion is formed when the entire molecule loses an electron and becomes positively charged. Peaks smaller than ones caused by the molecular ion, represent fragment ions Fragment ions form when high-energy electrons in the ionisation chamber cause bonds and consequently molecules to break. The most intense peak is known as the base peak. The base peak is produced by the most abundant fragment ion. The base peak has a relative intensity of 100%. The relative intensities of a peak on the ass spectrum depends on the: - energy of the ionising electrons, - ease with which fragments can be formed. - stability of the fragment ions formed. The mass-to-charge ratio determines the formula of the fragment ion. Fragments can form by the breaking of almost any bond in the molecular ion .

The principles and applications of infrared spectroscopy (IR) (excluding features of instrumentation and operation) and interpretation of qualitative and quantitative data including use of characteristic absorption bands to identify bonds.                      

Infrared (IR) spectroscopy is a powerful analytical tool that can be applied to the analysis of many organic and inorganic compounds. This form of spectroscopy can analyse solids, liquids and gases. It gives information about the functional groups present in an organic molecule. IR light has a lower energy and longer wavelength than visible and ultraviolet light. This energy is not enough to promote electrons to high electrons but can change the vibration of the bonds in the molecules. This form of spectroscopy exploits the ability of molecules to bend and stretch. This technique is powerful as nearly all molecules absorb infrared radiation. When molecules absorb infrared radiation, the stretching/bending vibrations of the bonds become more energetic. The mass of atoms attached to a bond affects the frequency of the IR radiation that the molecules absorbs, The frequency of electromagnetic radiation in IR spectroscopy is expressed in wavenumbers (cm-1) The wavenumber is the number of waves per centimetres The wavenumber is inversely proportional to wavelength. A bond that vibrates at a higher frequency absorbs IR radiation with a higher wavenumber and greater energy than a bonds that vibrates at a lower frequency. The horizontal axis of an IR spectrum shows the wavenumber. The vertical axis of an IR spectrum shows the percentage transmittance. The percentage transmittance is on a scale from 0 to 100. The baseline of the spectrum is where all the light is passed through the sample. The baseline is at 100%. When the spectrum lowers in transmittance, the molecules has absorbed IR radiation. Absorption bands are where the molecules absorb IR radiation and appear as lower peaks in the spectrum. The following image depicts how absorption bands are described. Absorption bands with wavenumbers above 1400 cm-1 are used to identify functional groups.

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The region below 1400cm-1 is called the fingerprint region as absorption bands of this frequency are unique to each compound. Each type of bond absorbs IR radiation over a typical range of wavenumbers. This form of spectroscopy is useful to distinguish between very similar compounds. Different types of covalent bonds absorb IR radiation within a characteristic range of frequencies/wavenumbers, allowing the functional groups in an organic compound to be identified. IR spectroscopy can also be used in quantitative analysis.

The principles (including spin energy levels) and applications of proton and carbon-13 nuclear magnetic resonance spectroscopy (NMR) (excluding features of instrumentation and operation); analysis of carbon-13 NMR spectra and use of chemical shifts to determine number and nature of different carbon environments in a simple organic compound; and analysis of high resolution proton NMR spectra to determine the structure of a simple organic compound using chemical shifts, areas under peak and peak splitting patterns (excluding coupling constants) and application of the n+1 rule.                    

Nuclear magnetic resonance (NMR) spectroscopy is a powerful technique used to determine the structure of complex molecules. It analyses materials using the interaction of the nucleus of particular isotopes with an external magnetic field and electromagnetic radiation. NMR spectroscopy uses electromagnetic radiation in the radio frequency range to obtain information about the structure of molecules. The nucleus of atoms must have a property known as the nuclear spin to interact with radio waves. A nucleus with the nuclear spin has an odd number of protons or odd number of neutrons or both. This property causes the nucleus to behave like a small bar magnet. In an external magnetic field, these nuclei line up in the same direction as the field or lines up in the opposite direction. If the nucleus is of lower energy, it lines up in the same direction as the magnetic field. If the nucleus is of higher energy, it lines up in the opposite direction. A magnet or nucleus is in an unstable arrangement if it lines up against an external field. Inside the NMR spectrometer, nuclei are usually of lower energy. A radio transmitter provides the energy to flip the nuclei into a high-energy state. Over time, nuclei spin back to a lower- energy spin and release a pulse of energy. This difference in energy between the spin states depends on the type of nucleus and the chemical environment surrounding the nucleus. A chemical environment is made up of atoms and electrons that surround a specific atom. The pulse of energy is measured and represented in graphical form on a NMR spectrum. Carbon-13 NMR spectroscopy is useful in investigating the carbon atoms inside molecules. An NMR spectrum provides information about the number and type of hydrogen and carbon nuclei in an organic compound. Atoms in the same chemical environment absorb the same energy and produce a single signal in the NMR spectrum. Inside a molecule atoms have the same chemical environment if they are attached in the same way to the same atoms.

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More planes of symmetry leads to fewer unique chemical environments. In different spectrometers nuclei emit different frequencies of radio energy. To ensure results can be obtained accurately, chemists compare signals the nuclei emit to the signal a reference compound, usually TMS. The difference in energy needed to change spin state in a sample is compared to the energy needed to change spin states in TMS. This energy difference is known as the chemical shift. Chemical shift is a characteristic of an atom's environment. Chemical shift is measured in parts per million, ppm. Chemical shift is denoted by the lower case delta sign. The chemical shift of TMS is 0. Remember a hydrogen atom is equivalent to a proton. Proton NMR spectroscopy gives information about the structure of any molecule containing hydrogen atoms. Proton NMR spectroscopy involves finding the number of chemically distinct hydrogen environments in a molecule. The number of signals in a proton NMR spectrum = the number of different hydrogen environments. Hydrogen atoms in the same chemical environments are equivalent. Equivalent atoms have the same chemical shift. Equivalent atoms form one signal. A NMR spectrum shows the chemical shift on the horizontal axis. The size of the signals is measured by the area under the curve of each signal. The peak area of each signal is proportional to the number of hydrogen atoms in the environment it corresponds with. The relative peak areas are indicated above each peak. The relative peak areas correspond to the number of hydrogens in each environment. Each signal is not a single peak but a series of fine peaks. In high-resolution NMR spectra, some signals are seen to split into line patterns. An environment is neighbouring if it is up to three bonds away from the hydrogen atoms in question. The number of neighbours plus one give the number of lines in the line pattern. Eg. A compound have three neighbours and thus a four-line pattern. The splitting pattern informs which environments are caused by hydrogen atoms close to one another in a molecule, A one-line single is a singlet, a two-line signal is a doublet, then triplet then quartet. Signal splitting acts over small distances only. The interacting hydrogens must be bonded to adjacent atoms for signal splitting to occur.

Determination of the structures of simple organic compounds using a combination of mass spectrometry (MS), infrared spectroscopy (IR) and proton and carbon-13 nuclear magnetic resonance spectroscopy (NMR) (limited to data analysis).   

The structure of organic compounds are often determined using all three techniques. The infrared spectrum of a compound provides evidence about functional groups present in a molecule. The proton and carbon NMR spectra provide detailed information that can be used to determine the connectivity of atoms and overall structure of a molecule.

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Steps to deduce the structure and name of a compound may be: 1. use the mass spectrum to identify the molecular ion and relative molecular molar mass. 2. use the IR spectrum to identify the functional groups present in the compound. 3. use the proton NMR spectrum to identify different hydrogen environments. 4. use the carbon NMR spectrum to identify the different carbon environments. 5. use the data from the spectra to deduce the structure of the compound. 6. name the compound.

The principles of chromatography including use of high performance liquid chromatography (HPLC) and construction and use of a calibration curve to determine the concentration of an organic compound in a solution.         

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Chromatography is a technique to separate and analyse complex mixtures of organic compounds. It is a technique for separating the components of a mixture. All method of chromatography have a stationary phase and a mobile phase. The mobile phase (a liquid or gas) carries the components. The stationary phase (a solid or a liquid) is what the mobile phases carries over. The stationary phase has an adsorbent surface. Adsorption describes the adhesion/stickiness of molecules or substances to the surface of a solid or liquid. As the components move over the stationary phase they undergo a continual process of adsorption, followed by desorption and dissolving into the mobile phase. The rate of movement of each component depends on how: -strongly the component adsorbs onto the stationary phase. -readily the component dissolves in the mobile phase. High-performance liquid chromatography (HPLC) is also known as High-pressure liquid chromatography. HPLC is a very sensitive technique used to: -separate the components in a mixture, -identify the components -measure their concentrations. HPLC is a modern technique based on column chromatography. In HPLC, the mobile phase is a liquid under pressure. The retention time is used to identify the components in a mixture. The retention time is the time taken for a component to pass through the column. HPLC can be used for both qualitative and quantitative analysis. A chromatogram helps in qualitative analysis, to identify the chemicals present in a substance. A chromatogram represents the time on the horizontal axis. A chromatogram represents the absorbance on the vertical axis. The same compound will give the same retention time if all the conditions remain the same. Each component forms one peak on the chromatogram. The size of the peaks is due to how much light the components absorb. The smaller, more polar components are soluble. These components are eluted more quickly and thus have a shorter retention time. Quantitative analysis determines the concentration of an individual component in a mixture. To determine the concentration, compare the component's peak area to the peak areas of samples of the same chemical at known concentrations. A solution with an accurately known concentration is called the standard solution.

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A calibration curve can be drawn by plotting the peaks areas against the concentrations of the standard solutions, to find the unknown concentration.

Determination of the concentration of an organic compound by volumetric analysis, including the principles of direct acid-base and redox titrations (excluding back titrations).   

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Volumetric analysis determines the amount or concentration of a substance dissolved in solution. Standard solutions must be prepared from primary standards to use in the form of analysis. A primary standard should: - be readily obtainable in pure form. - have a known chemical formula - easily storable - unreactive with the atmosphere - high molar mass - cheap To prepare standard solutions dissolve and accurately measured mass of a primary standard in water to make an accurately measured volume. To prepare a standard solution you can also perform a titration with another standard solution in order to determine its exact concentration. The steps taken are shown in the following image: To undertake a titration of highly concentrated stock solutions, the solutions must first be diluted. A stock solution is a large volume of a common chemical. The form of analysis involves reacting a measured volume of a standard solution with a measured volume of a solution of unknown concentration. The solutions react completely in the mole ratio that the balanced equation indicates. The process described above is known as performing a titration. Once the concentration of a solution is determined, it can be called a standard solution or said to be standardised. The steps for an acid-base titration are: 1. measure a known volume of one of the solutions using a pipette. 2. transfer this volume of solution to a conical flask. 3. add a few drops of an appropriate acid-base indicator so that a colour change will signal the point at which the titration should stop. 4. dispense the other solution from a burette slowly into the conical flask until the indicator changes colour permanently. 5. measure the volume of the titre. 6. calculate to find required unknown. As a liquid is delivered from a burette into a conical flask, the pH of the solution in the conical flask changes. The graphical depiction of a change in pH is illustrated by a pH/titration curve. The pH curve shows the pH on the vertical axis. The pH curve shows the volume of the acid on the horizontal axis. The equivalence point occurs when the gradient of the pH curve is the steepest. The equivalence point occurs when two chemicals have reacted in the mole ratio indicated by the balanced chemical equation.

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The end point is the point during a titration when the indicator changes colour. Must select indicator that changes colour at the steep section of the pH range. This is so that the end point and equivalence point occur at the same time. Concordant titres vary within narrowly specified limits. Three concordant titres Are usually obtained during a titration. Glassware is often rinsed before a volumetric analysis is conducted. Rinsing glassware removes traces of any chemicals and increases accuracy and precision of the results. The burette and pipette should be finally rinsed with the acid or based the glassware will transfer before use. The volumetric flask and conical flask should only be rinsed with deionised water. The conical flask can be rinsed with tap water. Volumetric analysis involving redox reactions can be used to determine the composition of a range of substances. Some organic compounds undergo redox reactions, which can form the basis of volumetric analysis, Primary alcohols can react with strong oxidising agents. Some redox titrations involve a change in colour without the addition of an indicator.