VCE Chemistry Instrumental Analysis Revision I PDF

Website: contoureducation.com.au | Phone: 1800 888 300 Email: [email protected] VCE Chemistry ¾ Instrumental Analysis Revision I [4.10] Su...

Website: contoureducation.com.au | Phone: 1800 888 300 Email: [email protected] VCE Chemistry ¾ Instrumental Analysis Revision I [4.10] Summary Solutions Outline: Qualitative Tests Pg 2-09 Chromatography Pg 18-27 Purity Analysis & Concentration Pg 10-17 Volumetric Analysis Pg 28-36 1  [4.10] Website: contoureducation.com.au Phone: 1800 888 300 Email: [email protected] Section A: Qualitative Tests Sub-Section: Qualitative Physical Tests Exploration: Hexane Consider hexane (C6 H14 ): Stronger Intermolecular Bonding: [Dispersion Forces]/[Dipole-Dipole]/[Hydrogen Bonding] When hexane and water are mixed, again, two layers will initially form: Can hexane and water undergo the same type of intermolecular bonds? [Yes]/[No] Will they dissolve? [Yes]/[No] [4.10] 2  [4.10] Website: contoureducation.com.au Phone: 1800 888 300 Email: [email protected] Key Takeaways Polar and polar molecules will dissolve (as they both undergo dipole-dipole/hydrogen bonding with each other). Non-polar and non-polar molecules will dissolve (as they both undergo dispersion forces only with each other). Polar and non-polar molecules will not dissolve (as one undergoes dispersion forces only while the other undergoes stronger dipole-dipole/hydrogen bonding with itself. As such, we see that like dissolves with like. NOTE: Carboxylic Acids → ____________, acidic basic Amines → __________! ALSO NOTE: Amides are very weak bases, and are only slightly basic. You do not have to know why! Exploration: Indicators The pH ranges of different indicators can be found on page 𝟓 of the databook. Colour Change from Lower to Name 𝐩𝐇 Range Higher 𝐩𝐇 in Range Thymol blue (𝟏𝐬𝐭 change) 1.2-2.8 Red → Yellow Methyl orange 3.1-4.4 Red → Yellow Bromophenol blue 3.0-4.6 Yellow → Blue Methyl red 4.4-6.2 Red → Yellow Bromothymol blue 6.0-7.6 Yellow → Blue Phenol red 6.8-8.4 Yellow → Red Thymol blue (𝟐𝐧𝐝 change) 8.0-9.6 Yellow → Blue Phenolphthalein 8.3-10.0 Colourless → Pink [4.10] 3  [4.10] Website: contoureducation.com.au Phone: 1800 888 300 Email: [email protected] Question 1 Walkthrough. Determine the colour of phenolphthalein if it is placed in: a. A basic environment with a pH of 11. b. A neutral environment with a pH of 7. Pink Colourless Question 2 Walkthrough. Four compounds, ethanamine, octan-1-ol, ethanoic acid, and hexanoic acid were present in separate vessels. The following information about them is given, whereby they attempted to be mixed with water with bromophenol blue indicator added. Identify each of them. Vessel 𝑨 𝑩 𝑪 𝑫 Solubility in water Soluble Insoluble Soluble Insoluble Colour of water Yellow Yellow Blue Blue Substance methanoic acid hexanoic acid ethanamine Octan-1-ol Key Takeaways Physical Tests: Tests For Test Conducted Molar Mass/Size of Molecule State of Matter Solubility Add water pH Add indicator [4.10] 4  [4.10] Website: contoureducation.com.au Phone: 1800 888 300 Email: [email protected] Sub-Section: Qualitative Chemical Tests Degree of Unsaturation Definition: The amount of C = C bonds present in a molecule. Bromine/Iodine Test alkene Purpose: Bromine/iodine test indicates if an __________________ is present. When bromine is added to an: Alkene, it [reacts]/[does not react] and turns [brown]/[colourless] Alkane, it [reacts]/[does not react] and turns [brown]/[colourless] Space for Personal Notes [4.10] 5  [4.10] Website: contoureducation.com.au Phone: 1800 888 300 Email: [email protected] Iodine Number Definition: The number of grams of iodine that reacts with __________ 100 𝑔 of a chemical substance. It is typically used to determine the degree of unsaturation in large molecules such as fats. Formula: 𝐈𝐨𝐝𝐢𝐧𝐞 𝐍𝐮𝐦𝐛𝐞𝐫 = 𝒎(𝐌𝐨𝐥𝐞𝐜𝐮𝐥𝐞 𝐚𝐟𝐭𝐞𝐫 𝐫𝐞𝐚𝐜𝐭𝐢𝐨𝐧) − 𝒎(𝐌𝐨𝐥𝐞𝐜𝐮𝐥𝐞 𝐛𝐞𝐟𝐨𝐫𝐞 𝐫𝐞𝐚𝐜𝐭𝐢𝐨𝐧) Question 3 A sample of 100 𝑔 of fat was reacted with iodine, whereby the final mass of the sample was 978 𝑔. Given that the molar mass of the compound was 202 𝑔/𝑚𝑜𝑙, find the degree of unsaturation. _____________________________________________________________________________________ _____________________________________________________________________________________ _____________________________________________________________________________________ _____________________________________________________________________________________ _____________________________________________________________________________________ _____________________________________________________________________________________ _____________________________________________________________________________________ Space for Personal Notes [4.10] 6  [4.10] Website: contoureducation.com.au Phone: 1800 888 300 Email: [email protected] REMINDER: Don’t forget! Type of Reaction Reaction Full Oxidation of Primary Alcohol Oxidation of Secondary Alcohol Oxidation of Tertiary Alcohol Reaction 𝐌𝐧𝐎𝟒 − (𝐚𝐪) → 𝐌𝐧𝟐+ (𝐚𝐪) Colour Purple Pale pink/colourless Reaction 𝐂𝐫𝟐 𝐎𝟕 𝟐− (𝐚𝐪) → 𝟐𝐂𝐫 𝟑+ (𝐚𝐪) Colour Orange Green Space for Personal Notes [4.10] 7  [4.10] Website: contoureducation.com.au Phone: 1800 888 300 Email: [email protected] Exploration: Esterification Reactions Reaction: 𝐇𝟐 𝐒𝐎𝟒 (𝐥) 𝒄𝒂𝒓𝒃𝒐𝒙𝒚𝒍𝒊𝒄 𝒂𝒄𝒊𝒅 + 𝒂𝒍𝒄𝒐𝒉𝒐𝒍 → 𝒆𝒔𝒕𝒆𝒓 + 𝒘𝒂𝒕𝒆𝒓 When adding a carboxylic acid, we test for an ______________. alcohol carboxylic acid When adding an alcohol, we test for a ______________________________. Indication of esterification reaction occurred: Esters smell fruity ____________________. Exploration: Carbonate Test Consider ethanoic acid (𝐂𝐇𝟑 𝐂𝐎𝐎𝐇): Reaction with sodium carbonate: 𝟐𝐂𝐇𝟑 𝐂𝐎𝐎𝐇(𝐚𝐪) + 𝐍𝐚𝟐 𝐂𝐎𝟑 (𝐬) → (𝐂𝐇𝟑 𝐂𝐎𝐎)𝟐 𝐍𝐚(𝐚𝐪) + 𝐂𝐎𝟐 (𝐠) + 𝐇𝟐 𝐎(𝐥) Reaction with sodium hydrogen carbonate: 𝐂𝐇𝟑 𝐂𝐎𝐎𝐇(𝐚𝐪) + 𝐍𝐚𝐇𝐂𝐎𝟑 (𝐬) → 𝐂𝐇𝟑 𝐂𝐎𝐎𝐍𝐚(𝐚𝐪) + 𝐂𝐎𝟐 (𝐠) + 𝐇𝟐 𝐎(𝐥) Observations: Bubbles forming (CO2 (g)). If we see: Bubbles Form No Observations Substance [is an acid]/[is not an acid] Substance [is an acid]/[is not an acid] Space for Personal Notes [4.10] 8  [4.10] Website: contoureducation.com.au Phone: 1800 888 300 Email: [email protected] Key Takeaways Chemical Tests: Tests For Tests Conducted 𝐂 = 𝐂 Bonds Bromine/Iodine Test Add H + (aq)/MnO4 − (aq) or H + (aq)/ Cr2 O72− (aq) (reacts if primary or secondary alcohol) Hydroxyl Group Carboxylic Acid Add ________________________________ (forms smelly ester) (reacts for primary/secondary/tertiary alcohol) Alcohol Add _______________ (forms smelly ester) Carboxyl Group Carbonate Add _______________ (forms bubbles) Add H + (aq)/MnO4 − (aq) or Aldehyde H + (aq)/ Cr2 O72− (aq) Space for Personal Notes [4.10] 9  [4.10] Website: contoureducation.com.au Phone: 1800 888 300 Email: [email protected] Section B: Purity Analysis & Concentration Sub-Section: Purity Analysis Key Takeaways When the substance is pure, all parts of the molecule melt at roughly the same temperature. When the substance is impure, different parts of the molecule melt at different temperatures. [4.10] 10  [4.10] Website: contoureducation.com.au Phone: 1800 888 300 Email: [email protected] Question 1 Walkthrough. A substance is known to be primarily octan-1-amine, octan-1-ol or octanoic acid. The melting points of them are indicated in the table below: Substance Melting Point Octan-1-ol −16℃ Octan-1-amine −4℃ Octanoic Acid 16℃ The substance is then subjected to melting point determination, where the substance is found to have a melting point range of 13℃ to 16℃. What does this tell you about the molecule? The molecule is likely to be impure octanoic acid. _____________________________________________________________________________________ Key Takeaways Melting point determination can do both compound identification and purity analysis! Compound Identification: Match the melting point of the substance to the melting point of the literature value. Purity Analysis: The range of the melting point indicates the purity of the substance. Space for Personal Notes [4.10] 11  [4.10] Website: contoureducation.com.au Phone: 1800 888 300 Email: [email protected] Exploration: Simple Distillation boiling point Substances are separated based on __________________________. target temperature The ____________________________________ is between the substance’s boiling point. If we consider a mixture of ethanol and water initially at SLC (25℃), what happens as the temperature is gradually increased? Setup: distillate Ethanol forms the ___________________. [4.10] 12  [4.10] Website: contoureducation.com.au Phone: 1800 888 300 Email: [email protected] Sub-Section: Concentration Concentration (𝒎𝒐𝒍/volume) Definition: The amount of a substance present per unit of volume. SI Units Formula 𝒏 𝒎𝒐𝒍/𝑳 𝒄= 𝑽 Example: 𝟐. 𝟎𝟎 𝒎𝒐𝒍/𝑳 or 𝟐. 𝟎𝟎 𝑴. 2.00 𝑚𝑜𝑙 Meaning: In every one litre of the sample, there are ___________________ of HCl. Concentration (mass/volume) Definition: The mass of a substance present per unit of volume. SI Units Formula 𝒎 𝒈/𝑳 𝒄= 𝑽 Space for Personal Notes [4.10] 13  [4.10] Website: contoureducation.com.au Phone: 1800 888 300 Email: [email protected] TIP: Look at the units to figure out the correct formula for concentration! TIP: Think about which unit is greater! Converting: Larger Unit → Smaller Unit (e.g. 𝑘𝑔 → 𝑔) A smaller unit is worth [more]/[less], and thus there will be [more]/[less] of it for the same amount! × 103 Example: from 𝑘𝑔 → 𝑔, it is _________________. Converting: Smaller Unit → Larger Unit (e.g. 𝑔 → 𝑘𝑔) A larger unit is worth [more]/[less], and thus there will be [more]/[less] of it for the same amount! × 10−3 Example: from 𝑔 → 𝑘𝑔, it is _________________. Question 2 Walkthrough. (2 marks) The concentration for a sample of sodium chloride is known to be 1.75 𝐺𝑔/𝑑𝐿 with a volume of 50 𝑛𝐿. Find the number of moles of sodium chloride. _____________________________________________________________________________________ _____________________________________________________________________________________ _____________________________________________________________________________________ _____________________________________________________________________________________ _____________________________________________________________________________________ Space for Personal Notes [4.10] 14  [4.10] Website: contoureducation.com.au Phone: 1800 888 300 Email: [email protected] TIP: When working with % concentration, write it as a fraction and put 100 as the denominator. Examples: 𝟒𝟎% (𝒎/𝒎) 𝟑𝟎% (𝒎/𝒗) 𝟏𝟏. 𝟗% (𝒗/𝒗) 0.30 𝑔/𝑚𝐿 0.119 𝐿/𝐿 Question 3 Walkthrough. a. A 7.20 𝐿 solution which contains 19.50% (𝑤/𝑣) of NaOH is used. Find the mass of NaOH present. _____________________________________________________________________________________ _____________________________________________________________________________________ _____________________________________________________________________________________ b. A 16.50 𝐿 solution that contains 28.0% (𝑣/𝑣) ethanol is investigated. Find the volume of ethanol present. _____________________________________________________________________________________ _____________________________________________________________________________________ Space for Personal Notes [4.10] 15  [4.10] Website: contoureducation.com.au Phone: 1800 888 300 Email: [email protected] Sub-Section: Dilution Factors Key Takeaways There are two types of actions: Diluting a Sample Taking out a Sample Concentration: Concentration: [Changes]/[Stays Same] [Changes]/[Stays Same] Amount: Amount: [Changes]/[Stays Same] [Changes]/[Stays Same] [Concentration]/[Amount] changes by factor: [Concentration]/[Amount] changes by factor: 20 20 100 100 [4.10] 16  [4.10] Website: contoureducation.com.au Phone: 1800 888 300 Email: [email protected] Question 4 Walkthrough. A 10.0 𝑚𝐿 sample of beetle juice is placed into a volumetric flask and diluted to 200.00 𝑚𝐿. A 20.00 𝑚𝐿 sample is then taken out, and diluted again to 150.00 𝑚𝐿. A 30.00 𝑚𝐿 aliquot is then taken out, whereby its concentration is determined via titration to be 0.0370 𝑀. Find the concentration of beetle juice in the original 10.0 𝑚𝐿 sample. _____________________________________________________________________________________ _____________________________________________________________________________________ 150 0.037 × = 0.278 𝑀 _____________________________________________________________________________________ 20 200 0.378 𝑀 × = 5.55 𝑀 _____________________________________________________________________________________ 10 _____________________________________________________________________________________ Space for Personal Notes [4.10] 17  [4.10] Website: contoureducation.com.au Phone: 1800 888 300 Email: [email protected] Section C: Chromatography Sub-Section: Principles of Chromatography Adsorption or Adsorb Definition: The process whereby a liquid/gas/solute are attached or stuck on to a solid material (the stationary phase). sticking Think of it as ____________________________ on to a solid surface. Substances adsorb to the [stationary]/[mobile] phase. Desorption or Desorb release Definition: The __________________________ of an adsorbed substance from a solid surface. Can be thought of as the opposite of adsorption. Substances desorb to the [stationary]/[mobile] phase. Strength of Adsorption and Desorption Consider a polar mobile phase (water) and a non-polar stationary phase (river-bed): [4.10] 18  [4.10] Website: contoureducation.com.au Phone: 1800 888 300 Email: [email protected] How will the above objects interact? Object 𝑨 (Non-Polar) Object 𝑩 (Polar) Attracted to [polar]/[non-polar] Attracted to [polar]/[non-polar] [stationary]/[mobile] phase. [stationary]/[mobile] phase. [Quicker]/[Slower] rate of travel. [Quicker]/[Slower] rate of travel. [Longer]/[Shorter] retention time. [Longer]/[Shorter] retention time. Key Takeaways If an object is strongly attracted (adsorbed) to the stationary phase, it will act more stationary and will have a slower rate of travel. If an object is strongly attracted (desorbed) to the mobile phase, it will act more mobile and will have a quicker rate of travel. High Performance Liquid Chromatography (HPLC) High performance liquid chromatography (HPLC) is like column chromatography. high pressures Major Difference: Uses ___________________________________ making process faster. Setup: [4.10] 19  [4.10] Website: contoureducation.com.au Phone: 1800 888 300 Email: [email protected] Question 4 Walkthrough. A non-polar solvent is used in a HPLC column to separate three components in a mixture. The components are labelled on the chromatogram as 𝐴, 𝐵 and 𝐶 and component 𝐴 has the shortest retention time. From the chromatogram we can conclude that: A. A faster flow rate of solvent is needed to better separate components 𝐵 and 𝐶. B. Component 𝐶 is likely to be the least polar of the three components. C. Component 𝐴 is likely to be the least polar of the three components. D. Component 𝐴 adsorbs strongly on the stationary phase. Qualitative Analysis with Retention Time Qualitative Analysis Quantitative Analysis What is in the sample? How much of sample is there? Measured by: Retention Time Measured by: Peak Area [4.10] 20  [4.10] Website: contoureducation.com.au Phone: 1800 888 300 Email: [email protected] Question 5 (3 marks) Walkthrough. Blood alcohol level is the concentration of alcohol (ethanol) in blood. The breathalyser test is a form of high performance liquid chromatography employed to find the blood alcohol level. Below is a gas chromatogram of a sample of blood taken from a driver. a. Given that the retention time of ethanol is known to be 4.5 minutes when run under identical conditions and the same column, using this apparatus. Is ethanol present in this blood sample? (1 mark) Ethanol is present in this blood sample. _____________________________________________________________________________________ b. An additional test was run for the presence of THC in the driver’s blood levels, and it is also known that THC has a retention time of 1.5 minutes. Explain whether there is THC present in the blood sample. (2 marks) _____________________________________________________________________________________ There is no THC present in the blood sample as there is no observed peak at 1.5 minutes. The peak _____________________________________________________________________________________ needs to be exactly on 1.5 minutes to be considered being from THC. _____________________________________________________________________________________ Space for Personal Notes [4.10] 21  [4.10] Website: contoureducation.com.au Phone: 1800 888 300 Email: [email protected] Sub-Section: Other Factors Which Affect the Retention Time Exploration: Operational Conditions There are also other factors which will uniformly affect the retention time of all substances passing through the chromatogram. Length of Chromatograph: Shorter Length of Column Longer Length of Column [Longer]/[Shorter] Retention Time [Longer]/[Shorter] Retention Time Viscosity: Water as Solvent Honey as Solvent [Longer]/[Shorter] Retention Time [Longer]/[Shorter] Retention Time [4.10] 22  [4.10] Website: contoureducation.com.au Phone: 1800 888 300 Email: [email protected] Tightness of Stationary Phase: Less Tightly Packed Column More Tightly Packed Column [Longer]/[Shorter] Retention Time [Longer]/[Shorter] Retention Time Pressure: Low Pressure High Pressure [Longer]/[Shorter] Retention Time [Longer]/[Shorter] Retention Time Temperature: Low Temperature High Temperature [Longer]/[Shorter] Retention Time [Longer]/[Shorter] Retention Time [4.10] 23  [4.10] Website: contoureducation.com.au Phone: 1800 888 300 Email: [email protected] Sub-Section: Quantitative Analysis Key Takeaways Greater Peak Area Lesser Peak Area [Greater]/[Lesser] Concentration of Substance [Greater]/[Lesser] Concentration of Substance Standard Definition: A substance with precisely known concentration for use in quantitative analysis. Space for Personal Notes [4.10] 24  [4.10] Website: contoureducation.com.au Phone: 1800 888 300 Email: [email protected] Exploration: Constructing a Calibration Curve from Standards The concentration of ethanol is investigated in an unknown sample. Five standards are first passed through the HPLC column individually, with concentrations of 0.200 𝑀, 0.400 𝑀, 0.600 𝑀, 0.800 𝑀 and 1.00 𝑀. The peak area is measured. Graphs: [4.10] 25  [4.10] Website: contoureducation.com.au Phone: 1800 888 300 Email: [email protected] Results: Standard Number Concentration Measured Peak Area 𝟏 0.200 𝑀 𝟏𝟎𝟎𝟎 𝟐 0.400 𝑀 𝟐𝟎𝟎𝟎 𝟑 0.600 𝑀 𝟑𝟎𝟎𝟎 𝟒 0.800 𝑀 𝟒𝟎𝟎𝟎 𝟓 1.00 𝑀 𝟓𝟎𝟎𝟎 Calibration Curve: The sample with unknown amount of ethanol is passed through. Graph: [4.10] 26  [4.10] Website: contoureducation.com.au Phone: 1800 888 300 Email: [email protected] 3500 Result: Unknown sample has a peak area of __________________. Plot on Calibration Curve: 0.700 𝑀 Concentration of unknown sample of ethanol is: _______________. Question 6 (1 mark) Walkthrough. A series of pure heroin samples of different concentrations are passed through the same HPLC as that used to test the patient's blood. The results obtained are shown in the graph below. The patient's blood sample generated a heroin peak with an area of 13 𝑚𝑚2. Calculate the concentration of heroin in the patient's blood sample in 𝑚𝑔 𝑚𝐿−1. _____________________________________________________________________________________ _____________________________________________________________________________________ [4.10] 27  [4.10] Website: contoureducation.com.au Phone: 1800 888 300 Email: [email protected] Section D: Redox Titration Sub-Section: Volumetric Analysis Exploration: Volumetric analysis calculation steps Equation: 𝟓𝐂𝐇𝟑 𝐎𝐇(𝐚𝐪) + 𝟒𝐌𝐧𝐎𝟒 − (𝐚𝐪) + 𝟏𝟐𝐇 + (𝐚𝐪) → 𝟓𝐇𝐂𝐎𝐎𝐇(𝐚𝐪) + 𝟒𝐌𝐧𝟐+ (𝐚𝐪) + 𝟏𝟏𝐇𝟐 𝐎(𝐥) If provided a 150 𝑚𝐿 solution of 0.880 𝑀 permanganate, how can we find the concentration of methanol present in a 50.0 𝑚𝐿 solution? Steps: 1. Write out the balanced chemical equation (if not already provided). 𝑛 = 𝑐𝑉 2. Find the amount (in moles) of the known substance using _____________. stoichiometric 3. Find the amount (in moles) of the unknown substance using __________________________ ratios. 𝑛 𝑐= 4. Find the concentration of an unknown substance using ____________. 𝑉 Active recall: What is the concentration formula for the following concentrations? 𝟐. 𝟓𝟎 𝑴 𝟐. 𝟓𝟎 𝒈/𝑳 Formula: Formula: 𝑛 𝑚 𝑐= 𝑐= 𝑉 𝑉 Space for Personal Notes [4.10] 28  [4.10] Website: contoureducation.com.au Phone: 1800 888 300 Email: [email protected] Question 7 (3 marks) Walkthrough. The following reaction occurs between dichromate and iron (II) ions: Cr2 O72− (aq) + 6Fe2+ (aq) + 14H +(aq) → 2Cr 3+ (aq) + 6Fe3+ (aq) + 7H2 O(l) Noah uses a sample containing 12.0% (𝑤/𝑣) dichromate and reacts a solution of 90.0 𝑚𝐿 of iron (II) with the dichromate. Given that a 32.0 𝑚𝐿 sample of the dichromate was used, find the concentration of iron (II) in 𝑀. _____________________________________________________________________________________ 𝑚(Cr2 O72− ) = 12/100 × 32 × 10−3 = 3.84 × 10−3 g _____________________________________________________________________________________ 𝑛(Cr2 O72− ) = 3.84 × 10−3 /(2 × 52 + 7 × 16) = 1.78 × 10−5 𝑚𝑜𝑙 𝑛(Fe2+ ) = 6 × 1.78 × 10−5 𝑚𝑜𝑙 = 1.1 × 10−4 𝑚𝑜𝑙 _____________________________________________________________________________________ [Fe2+ ] = (1.1 × 10−4 )/(90 × 10−3 ) = 1.22 × 10−3 𝑀 _____________________________________________________________________________________ Space for Personal Notes [4.10] 29  [4.10] Website: contoureducation.com.au Phone: 1800 888 300 Email: [email protected] Sub-Section: Volumetric Analysis Instrumentation Exploration: Volumetric analysis instrumentation Volumetric analysis instrumentation: Apparatus name: Volumetric Apparatus name: Pipette Apparatus name: Burette flask Make accurate volumes of Transfer volumes. Dispense volumes solutions. Fill up to the mark. Takes out samples/aliquot. [4.10] 30  [4.10] Website: contoureducation.com.au Phone: 1800 888 300 Email: [email protected] Standard Definition: A substance with a precisely known concentration for use in quantitative analysis. Exploration: Calculating the concentration of primary standards Steps: 𝒎 1. Calculate the moles of the primary standard using 𝒏 = 𝑴. 𝒓 𝒏 2. Calculate the concentration by using 𝒄 = 𝒗. Aliquot Definition: A portion of a larger whole, especially a sample taken for chemical analysis or other treatment. This is delivered via a pipette. Titration setup [4.10] 31  [4.10] Website: contoureducation.com.au Phone: 1800 888 300 Email: [email protected] Key Takeaways Titre – means that this substance is in the burette above. Aliquot – means that this substance is in the conical flask below. TIP: The key to volumetric analysis questions is drawing out a diagram and the setup! Reminders There are two types of actions: Diluting a sample Taking out a sample Concentration: Concentration: [Changes]/[Stays same] [Changes]/[Stays same] Amount: Amount: [Changes]/[Stays same] [Changes]/[Stays same] [Concentration]/[Amount] changes by factor: [Concentration]/[Amount] changes by factor: 20 20 100 100 [4.10] 32  [4.10] Website: contoureducation.com.au Phone: 1800 888 300 Email: [email protected] Concordant titres A titration is repeated until 3 concordant titres are obtained. 0.10 𝑚𝐿 Concordant titre range: ____________________________ Range is from lowest to highest concordant titre. concordant The mean titre is the average of the ____________________ titres only! Question 8 Walkthrough. (6 marks) Will has been drinking too much alcohol, and Gracie decides to find the concentration of ethanol (CH3 CH2 OH) in a small sample of his blood. To do so, she injects a syringe into his arm and extracts 6.00 𝑚𝐿 of his blood. She then delivers this 6.00 𝑚𝐿 into a conical flask and titrates it against a secondary standard, potassium dichromate (K 2 Cr2 O7 ) with a concentration of 0.00162 𝑀. The overall equation for the reaction which takes place is shown below. 3CH3 CHOH(aq) + 2Cr2 O72− (aq) + 16H + (aq) → 3CH3 COOH(aq) + 4Cr 3+ (aq) + 11H2 O(l) a. She titrates the blood sample against the potassium dichromate five times and obtains the following titres: Titre Volume 1 17.26 2 17.29 3 17.40 4 17.31 5 17.15 Find the mean titre. (1 mark) 3 concordant titres: 17.26, 17.29, 17.31 _____________________________________________________________________________________ 17.26,17.29,17.31 Mean = 3 = 17.29 𝑚𝐿 b. Hence or otherwise, find the moles of potassium dichromate used. (1 mark) _____________________________________________________________________________________ 𝑛(K 2 Cr2 O7 ) = 𝑐𝑉 = 1.62 × 10−3 × 17.29 × 10−3 = 2.800 × 10−5 𝑚𝑜𝑙 _____________________________________________________________________________________ [4.10] 33  [4.10] Website: contoureducation.com.au Phone: 1800 888 300 Email: [email protected] c. Find the concentration of ethanol in Will’s body, giving your concentration in 𝑚𝑜𝑙/100 𝑚𝐿. (3 marks) _____________________________________________________________________________________ 3 3 _____________________________________________________________________________________ 𝑛(𝑒𝑡ℎ𝑎𝑛𝑜𝑙) = 𝑛(K 2 Cr2 O7 ) = × 2.800 × 10−5 = 4.2 × 10−5 𝑚𝑜𝑙 2 2 𝑛 4.201 × 10−5 _____________________________________________________________________________________ 𝑐(𝑒𝑡ℎ𝑎𝑛𝑜𝑙) = = = 7.0011 × 10−3 𝑚𝑜𝑙 𝐿−1 𝑉 0.006𝐿 7.0011 × 10−3 𝑚𝑜𝑙 10 _____________________________________________________________________________________ = ÷ 1000 𝑚𝐿 10 7.0011 × 10−4 𝑚𝑜𝑙 _____________________________________________________________________________________ = 100 𝑚𝐿 _____________________________________________________________________________________ d. Assuming that the density of blood is 1.000 𝑔/𝑚𝐿, find Will’s BAC (in % 𝑤/𝑤), and hence determine if he is allowed to drive, given that the legal limit for alcohol intake is 0.05% 𝑤/𝑤. (1 mark) _____________________________________________________________________________________ 𝑚(𝑒𝑡ℎ𝑎𝑛𝑜𝑙) = 𝑛 × 𝑀 = 7.0011 × 10−4 × (24 + 6 + 16) = 0.0322 𝑔 _____________________________________________________________________________________ 0.0322𝑔 𝐶(𝑒𝑡ℎ𝑎𝑛𝑜𝑙 𝑏𝑦 𝑚𝑎𝑠𝑠) = = 0.0322 % 𝑤/𝑤 100𝑔 _____________________________________________________________________________________ As 0.0322 < 0.05, he is allowed to drive. _____________________________________________________________________________________ Space for Personal Notes [4.10] 34  [4.10] Website: contoureducation.com.au Phone: 1800 888 300 Email: [email protected] Sub-Section: Rinsing of Equipment Key Takeaways When rinsing glassware, they should all be first rinsed with distilled water to get rid of other substances, before they are rinsed with the following: Container Substance to rinse with Burette [substance it holds]/[distilled water] Conical flask [substance it holds]/[distilled water] Pipette [substance it holds]/[distilled water] Volumetric flask [substance it holds]/[distilled water] Exploration: Incorrect rinsing First, consider the following scenarios: A single drop neutralises: Only a large volume neutralises: [4.10] 35  [4.10] Website: contoureducation.com.au Phone: 1800 888 300 Email: [email protected] Question 9 Walkthrough. A redox titration is undertaken between potassium dichromate and ethanol. An aliquot of ethanol is placed into the conical flask, whereby titration occurs. However, Meg accidentally rinses the burette with deionised water. a. Explain the effect this has on the calculated concentration of ethanol. _____________________________________________________________________________________ Titrant is diluted, more titre is required, and thus, the concentration of ethanol is _____________________________________________________________________________________ overestimated. _____________________________________________________________________________________ Meg then in the second run, rinses the conical flask with potassium dichromate. b. Explain the effect this has on the calculated concentration of ethanol. _____________________________________________________________________________________ Some ethanol in the conical flask is reacted away, decreasing amount of ethanol in conical flask. This decreases volume of titre delivered, thus _____________________________________________________________________________________ underestimating the concentration of ethanol. Space for Personal Notes [4.10] 36