Volumetric Analysis PDF

**[Volumetric Analysis: Acid-Base]** Volumetric Analysis is a method of quantitative analysis which involves reacting volumes of solutions with each other. **[Expressing the Concentrations of Solutions]** \*\*Definition: A solution is a completely perfect mixture of a solute and a solvent. In a s...

**[Volumetric Analysis: Acid-Base]** Volumetric Analysis is a method of quantitative analysis which involves reacting volumes of solutions with each other. **[Expressing the Concentrations of Solutions]** \*\*Definition: A solution is a completely perfect mixture of a solute and a solvent. In a solution that particles of the solute are uniformly distributed throughout the solvent, i.e., the mixture is homogenous.\*\* \*\*Definition: The concentration of a solution is the amount of solute that is dissolved in a given volume of solution.\*\* There are a number of ways of expressing the concentration of a solution. [Percentage of Solute] \- Weight per weight (w/w) -- This means the grams of solute per 100 g of solution, or g/100g.\ - 10% NaCl w/w means that there are 10 g of sodium chloride per 100 g of solution. 10 g of the salt is dissolved in water and more water is added until the total weight of the solution is 100 grams. -Weight per volume (w/v) -- This means the grams of solute per 100 cm3 of solution, or g/100 cm3.\ - 10% w/v NaCl means that there are 10 g of sodium chloride per 100 cm^3^of solution. Volume per volume (v/v) -- This means the volume of solute per 100 cm3 of solution, or cm3 /100 cm3\ - 10% v/v ethanol has 10 cm^3^ of ethanol per 100 cm^3^. A solution of salt is marked 0.85% w/v. What mass of sodium chloride is needed to make up 250 cm^3^ of this solution? 0.85% w/v means that there are 0.85 g NaCl per 100 cm^3^ solution. Therefore, there are 0.85 x 2.5 = 2.125 g NaCl per 250 cm^3^ solution. Answer: 2.125 g NaCl **[Parts per Million (ppm)]** This is used for very dilute solutions e.g., water analysis. 1 mg per litre = 1 mg per 1000 grams of water\ = 1 mg per 1000 x 1000 mg\ = 1 mg per million mg\ = 1 ppm\ 1 ppm = 1 mg/L **[Moles of Solute Per Litre of Solution (Molarity)]** \*\*Definition: The molarity of a solution is the number of moles of solute per litre of solution.\*\* A 1 M (one molar) solution is one that contains one mole of the solute is dissolves in one litre of solution. **[Converting Moles per Litre to Grams per Litre]** How many grams of NaCl per litre are present in a solution marked 0.25 M NaCl? Rel. Molecular Mass of NaCl = 23 + 35.5 = 58.5 A 1 M NaCl solution contains 58.5 grams per litre Therefore, a 0.25 M NaCl solution contains 58.5 x 0.25 = 14.625 g/L Answer: 14.625 grams NaCl per litre **[Converting Grams per Litre to Moles per Litre (Molarity)]** What is the molarity of a solution that contains 3.68 g of NaOH per litre of solution? Rel. Molecular Mass of NaOH = 23 + 16 + 1 = 40 40 g in 1 L solution 1 M NaOH solution 1 g in 1 L solution [\$\\frac{1}{40}\$]{.math.inline} M NaOH solution 3.68 g in 1 L solution [\$\\frac{3.68}{40}\$]{.math.inline} M NaOH solution = 0.092 M NaOH solution Answer: 0.092 M NaOH solution **[Calculating the Number of Moles Given the Volume and the Molarity of a Solution]** Number of moles = [\$\\frac{\\text{volume\\ x\\ molarity}}{1000}\$]{.math.inline} How many moles of NaOH are present in 25 cm^3^ of 0.55 M NaOH? 1 litre of 0.55 M NaOH contains 0.55 moles NaOH 1000 cm^3^ of 0.55 M NaOH contains 0.55 moles NaOH 1 cm^3^ of 0.55 M NaOH contains [\$\\frac{0.55}{1000}\$]{.math.inline} moles NaOH 25 cm^3^ of 0.55 M NaOH contains [\$\\frac{25\\ x\\ 0.55}{1000}\$]{.math.inline} moles NaOH = 0.014 Answer: 0.014 moles NaOH **[Reaction between a Solution and a Solid]** What mass of Magnesium will react with 50 cm^3^ of 0.5 M H~2~SO~4~. The balanced equation for the reaction is\ Mg + H~2~SO~4~ MgSO~4~ + H~2~ No. of moles of acid = [\$\\frac{\\text{volume\\ x\\ molarity}}{1000}\$]{.math.inline} = [\$\\frac{50\\ x\\ 0.5}{1000}\$]{.math.inline}\ = 0.025 moles One mole of sulfuric acid reacts with one mole of magnesium. Therefore moles of magnesium which react = 0.025 moles 0.025 x 24 g = 0.6 g\ Answer: 0.6 g magnesium **[Dilution of Solutions]** No. of moles of solute in diluted solution = No. of moles of solute in conc. Solution [\$\\frac{\\text{Vol}\_{\\text{dil}}\\text{\\ x\\ }M\_{\\text{dil}}}{1000}\$]{.math.inline} = [\$\\frac{\\text{Vol}\_{\\text{conc\\ \\ x}M\_{\\text{Conc}}\\ }}{1000}\$]{.math.inline} What volume of 12 M HCl solution is required to make up 500 cm^3^ of 3 M HCl solution? [\$\\frac{\\text{Vol}\_{\\text{dil}}\\text{\\ x\\ }M\_{\\text{dil}}}{1000}\$]{.math.inline} = [\$\\frac{\\text{Vol}\_{\\text{conc\\ \\ x}M\_{\\text{Conc}}\\ }}{1000}\$]{.math.inline} 250 x M~dil~ = 15 x 2 M~dil~ = [\$\\frac{15\\ x\\ 2}{250}\\ \$]{.math.inline}= 0.12 M The concentration of the diluted solution is 0.12 M **[Standard Solutions]** \*\*Definition: A standard solution is a solution whose concentration is accurately known.\*\* \*\*Definition: A primary solution is a substance that can be obtained in a stable, pure, and soluble solid form so that it can be weighed out and dissolved in water to give a solution of accurately known concentration.\*\* A primary standard is a substance that meets the following criteria;\ - Must be available in a highly pure state\ - Must be stable in air\ - Must dissolve easily in water\ - Should have a fairly high relative molecular mass\ - Should undergo complete and rapid reaction\ - Should not be hydrated **[Mandatory Experiment: To prepare a standard solution of sodium carbonate]** 1. Using a balance, measure accurately 2.65 g of pure anhydrous sodium carbonate on a clock glass. 2. Slowly transfer the sodium carbonate with stirring, to about 50 cm^3^ of deionised water in a clean 250 cm^3^ beaker. To ensure that all the sodium carbonate is transferred, use a wash bottle to rinse the clock glass with deionised water, and add the rinsings to the beaker. 3. Continue stirring the mixture with a stirring rod until the sodium carbonate has fully dissolved. 4. Using a wash bottle, wash off the solution on the stirring rod with deionised water into the beaker. 5. Pour the solution through a clean funnel into the 250 cm^3^ volumetric flask. 6. Using a wash bottle, rinse out the beaker several times with deionised water, and add the rinsings to the solution in the flask 7. Rinse the funnel with deionised water, allowing the water to run into the flask. 8. Fill the flask to within about 1 cm of the calibration mark, and then add the water dropwise, using a dropping pipette, until the bottom of the meniscus just rests on the calibration mark. 9. Stopper the flask and invert it several times to ensure a homogeneous (evenly mixed) solution. Label the flask. **[Volumetric Analysis: Titrations]** \*\*Definition: A titration is a laboratory procedure where a measured volume of one solution is added to a known volume of another solution until the reaction is complete.\*\* **[Apparatus used in Volumetric Analysis]** [Graduated Cylinder\ ]- Not particularly accurate and should only be used for measuring out approximate volumes of liquids\ - Should always be rinsed with deionised water before use A measuring cylinder with red numbers Description automatically generated(1) [Volumetric Flask\ ]- Designed to contain a definite volume of solution\ - There is an engraved graduation mark on the narrow neck of the flask\ - The capacity and the temperature at which the stated volume is most accurately measures, are usually marked on the flask\ - Should always be rinsed with deionised water before use\ - The last few cm^3^ of water must be added carefully so that the bottom of the meniscus rests on the calibration mark. A dropper is used for this purpose.\ - After making the solution up to the mark, the contents must be well mixed by inverting the stoppered flask about 20 times. This ensures a homogenous mixture.\ (2) [The Pipette\ ]- Used to deliver an exact volume of solution\ - Filled by drawing the solution into it by suction. Liquid is drawn into the pipette above the graduation mark that is etched on the surface. This is done with a pipette filler\ - The liquid is then allowed to run into a beaker until the bottom of the meniscus is level with the graduation mark\ - The tip of the pipette is touched against the side of a glass beaker to remove the drop adhering to the tip\ - It is then allowed to discharge slowly into a conical flask, keeping the tip of the pipette in contact with the side of the flask\ - When the discharge is complete, touch the tip against the inside of the flask, and wait a few seconds before removing the pipette from the flask\ - Should be rinsed with deionised water before use and then rinsed with the solution it is to contain to remove any water from the pipette\ A close-up of a measuring device Description automatically generated (3) [Conical Flask\ ]- Designed so it can be swirled around without the liquid splashing out of it\ - The shape allows any droplets adhering to the side of it to be easily washed down into the flask\ - Should always be washed with deionised water before use\ - Should not be washed out with the solution it is to contain\ (4) **[Mandatory Experiment: To use the standard solution of sodium carbonate to standardise a given hydrochloric acid solution]** 1. Pour approximately 50 cm^3^ of the hydrochloric acid solution into a clean dry beaker. 2. Rinse the burette with deionised water, and clamp it vertically in the retort stand. 3. Using the funnel, add a little of the hydrochloric acid solution to the burette; rinse the burette with this solution and discard it. 4. Fill the burette with hydrochloric acid solution above the zero mark. Remove the funnel. Using the tap at the base of the burette, allow the acid to flow into a beaker until the level of liquid is at the zero mark. Ensure that there are no air bubbles in the nozzle of the burette. 5. Hold a piece of white card behind the burette in order to see the level of liquid more clearly. Record the level of liquid in the burette in your notebook. Remember to record the lower level of the meniscus at eye level. 6. Rinse the pipette with some deionised water and then with some of the sodium carbonate solution. Rinse the conical flask with deionised water only. 7. Using the clean pipette and a pipette filler, transfer 25 cm^3^ of the sodium carbonate solution into the clean conical flask. Add 2-3 drops of methyl orange indicator. Note the colour of the solution. A white tile underneath the conical flask will make the colour more obvious. 8. Carry out a rough titration by adding hydrochloric acid solution from the burette in approximately 1 cm^3^ lots to the conical flask, swirling the flask constantly, until the colour of the solution in the conical flask changes. During the titration, any solution adhering to the sides of the conical flask during the titration should be washed down with deionised water. 9. Note the burette reading and calculate how much acid was used. This is only a trial titration and gives the approximate value of the end point. 10. Wash out the conical flask with water. Pipette 25 cm^3^ of sodium carbonate solution into the flask and add 2-3 drops of methyl orange indicator. 11. Repeat the titration for a more accurate reading, by adding in hydrochloric acid solution from the burette rapidly to within 2 cm^3^ of the end point. Swirl the conical flask vigorously, and, while continuing to swirl, add the hydrochloric acid solution dropwise from the burette. When a definite permanent colour change occurs, note the reading. 12. Repeat the titration (steps 10 and 11) until two readings agree within 0.1cm^3^ 13. Calculate the concentration of the hydrochloric acid solution. 1. Rinse the burette, pipette and conical flask respectively with deionised water. 2. Rinse the burette with hydrochloric acid solution, and rinse the pipette with sodium hydroxide solution. 3. Using the pipette, transfer 25 cm^3^ of the unknown sodium hydroxide solution into the clean conical flask. Add 5 drops of methyl orange indicator. 4. Fill the burette to the 0 cm^3^ mark with hydrochloric acid solution. 5. Carry out one rough and two accurate titrations.\ - The Methyl Orange changes from yellow to red/pink marking the end point 6. Calculate the concentration of the sodium hydroxide solution. **To obtain a sample of salt** 1. To 25 cm^3^ of the sodium hydroxide solution in a beaker, add just enough hydrochloric acid to exactly neutralise it. The indicator should not be added. 2. Gently heat the solution until all the water has evaporated to dryness. A sample of sodium chloride will remain in the beaker.\ - White crystals are formed in the evaporating dish **[Mandatory Experiment: To determine the concentration of ethanoic acid in vinegar]** 1. Place 25 cm^3^ of vinegar in a 250 cm^3^ volumetric flask and dilute with water to the calibration mark, mixing well. 2. Stopper the flask and invert several times to ensure a homogeneous solution. Label the flask. 3. Rinse the burette, pipette and conical flask respectively with deionised water. 4. Rinse the burette with diluted vinegar solution, and rinse the pipette with sodium hydroxide solution. 5. Place 25 cm^3^ of 0.1 M sodium hydroxide solution in the conical flask, and add 3 drops of phenolphthalein indicator. 6. 7. Fill the burette to the 0 cm^3^ mark with diluted vinegar. 8. Carry out one rough and three accurate titrations. 9. Calculate the concentration of ethanoic acid in the diluted vinegar solution. 10. Calculate the percentage (w/v) of ethanoic acid in the vinegar.\ - At the end point, the phenolphthalein changes from pink to colourless **[Mandatory experiment: To determine the amount of water of crystallisation in hydrated sodium carbonate]** 1. Weigh accurately approximately 1.5 g of hydrated sodium carbonate into a beaker. 2. Add about 50 cm^3^ of deionised water and stir to dissolve the sample. 3. Transfer all of the solution into a 250 cm^3^ volumetric flask. Rinse the beaker with deionised water and add the washings to the volumetric flask. 4. Make up the volumetric flask to the mark, using a dropper to add the final amounts of deionised water. Stopper the flask and invert several times. 5. Rinse the burette, pipette and conical flask respectively with deionised water. 6. Rinse the burette with hydrochloric acid solution, and rinse the pipette with sodium carbonate solution. 7. Titrate the sodium carbonate solution against 0.1 M hydrochloric acid, using 25 cm^3^ of sodium carbonate solution in the conical flask and methyl orange as indicator. 8. Repeat the titrations until the readings agree to within 0.1 cm^3^. 9. Calculate the concentration of the sodium carbonate solution. 10. Determine the formula of hydrated sodium carbonate\ - At the ned point, methyl orange indicator changes from yellow to red/pink **[Exam Questions]** [2014 -- Hl -- Section A -- Question 1] A batch of washing soda crystals (hydrated sodium carbonate, Na~2~CO~3~. xH~2~O) had lost some of its water of crystallisation by a process called efflorescence. A chemist was required to determine the percentage water of crystallisation in the crystals and the value of x, the average number of water molecules in the formula. A sample of the crystals was accurately weighed and found to have a mass of 2.50 g. The sample was dissolved in deionised water and made up to 250 cm^3^ of solution. A number of 25.0 cm^3^ portions of this solution were titrated with a previously [standardised] 0.10 M hydrochloric acid (HCl) solution. The mean volume of the hydrochloric acid solution required to reach the end point was 21.6 cm^3^. The balanced equation for the titration reaction is: 2HCl + Na~2~CO~3~ → 2NaCl + H~2~O + CO~2~ \(a) Explain the underlined term. Solution whose concentration is exactly known\ (b) Describe in detail how the chemist should have dissolved the weighed sample of washing soda crystals and made the solution up to exactly 250 cm^3^.\ - Wash into beaker of deionised water.\ - Stir to dissolve.\ - Pour through funnel into volumetric flask adding rinsings of beaker.\ - Bring bottom of meniscus level with mark\ (c) State one precaution that should have been taken as the end point of the titration was approached. Explain how this precaution would have contributed to the accuracy of the titration result.\ - Add drop by drop.\ - Add dropwise so that end point will be accurately detected.\ (d) Name a suitable indicator for this titration. State the colour change in the titration flask at the end point. Methyl orange turns from Yellow to Pink.\ (e) From the mean volume of the hydrochloric acid solution, calculate the concentration of sodium carbonate (Na~2~CO~3~) in the original solution in\ (i) moles per litre,\ (ii) grams per litre.\ HCl Na~2~CO~3~\ V~a~ = 21.6 V~b~ = 25\ M~a~ = 0.1 M~b~ = ?\ N~a~ = 2 n~b~ = 1\ [\$\\frac{v\_{a}\\text{\\ x\\ }M\_{a}}{N\_{a}}\$]{.math.inline} = [\$\\frac{v\_{b}\\text{\\ x\\ }M\_{b}}{N\_{b}}\$]{.math.inline}\ [\$\\frac{21.6\\ x\\ 0.1}{2}\$]{.math.inline} = [\$\\frac{25\\ x\\ ?}{1}\$]{.math.inline} M~b~ = 0.0432 moles/L\ 0.0432 × 106 = 4.58 g/L\ \ (f) Calculate the percentage water of crystallisation in the crystals and the value of x, the average number of water molecules in the formula Na~2~CO~3~.xH~2~O. Hydrated sodium carbonate = 2.50 g / 250 cm^3^\ Anhydrous sodium carbonate = 1.14 g / 250 cm^3^\ Water = 1.36 g / 250 cm^3^ [1.36]\ 2.5 × 100 = 54% water of crystallisation Value of x\ [0.0108 moles] [2.5 g]\ 250 cm^3^ = 250 cm^3^ Mass of hydrated sodium carbonate = [2.5]\ 0.018 = 231.5 Na~2~CO~3~.x H~2~O = 231.5\ 106 + 18x = 231.5\ x = 6.9 average number of water molecules [2012 -- HL -- Section A -- Question 1\ ]1. A student determined the concentration of a hydrochloric acid solution by titration with 25.0 cm^3^ portions of a 0.05 M [primary standard] solution of anhydrous sodium carbonate. The portions of sodium carbonate solution were measured into a conical flask using a 25 cm^3^ pipette. The hydrochloric acid solution was added from a burette. The mean titre was 20.8 cm^3^. The balanced equation for the titration reaction was: 2HCl + Na~2~CO~3~ 2NaCl + H~2~O + CO~2~\ (a) Explain the underlined term. Pure, stable, soluble material from which solutions of known concentration can be made.\ (b) Describe how the student should have prepared 500 cm^3^ of the 0.05 M primary standard solution from a known mass of pure anhydrous sodium carbonate, supplied on a clock glass.\ - Rinse from clock glass into beaker and dissolve.\ - Pour using funnel into 500cm^3^ volumetric flask and add rinsings of beaker.\ - Add deionised water until bottom of meniscus on mark.\ - Read at eye level.\ - Stopper and invert several times\ Calculate the exact mass of anhydrous sodium carbonate (Na~2~CO~3~) required to prepare this solution.\ [500 × 0.05]\ 1000 × 106 = 2.65g\ (c) (i) Describe how the liquid level in the burette was adjusted to the zero mark. Fill above mark and adjust level until bottom of meniscus is on the graduation mark with tap\ (ii) Why was a pipette filler used to fill the pipette with 25.0 cm^3^ of the sodium carbonate solution? Avoid solution getting into mouth for hygiene reasons.\ (d) Name a suitable indicator for this titration. State the colour change observed at the end point. Methyl Orange from orange to red\ (e) Calculate, correct to two decimal places, the concentration of the hydrochloric acid solution in (i) moles per litre, (ii) grams per litre. HCl Na~2~CO~3~\ V~a~ = 20.8 V~b~ = 25\ M~a~ = ? M~b~ = 0.05\ N~a~ = 2 n~b~ = 1\ [\$\\frac{v\_{a}\\text{\\ x\\ }M\_{a}}{N\_{a}}\$]{.math.inline} = [\$\\frac{v\_{b}\\text{\\ x\\ }M\_{b}}{N\_{b}}\$]{.math.inline}\ [\$\\frac{20.8\\ x\\ ?}{2}\$]{.math.inline} = [\$\\frac{25\\ x\\ 0.05}{1}\$]{.math.inline} M~a~ = 0.12 moles/L 0.12 × 36.5 = 4.38g/L **[References]** 1. Twinkl.ie 2. Esawindia.com 3. Glossary.periodni.com 4. Betterequipped.ie