...acid it can be titrated with a base. Titration is the process of adding a known amount of a solution of known concentration to a known amount of solution of unknown concentration. The more accurately the concentration of the solution of known concentration is known, the more accurately the concentration of the unknown solution can be determined. Some chemicals can be purchased in a pure form and remain pure over a long period or time. Other chemicals are easily contaminated by the absorption of carbon dioxide or water from the air. Sodium hydroxide absorbs moisture from the air and often appears wet. Thus if a solution of sodium hydroxide is prepared by weighing the sodium hydroxide, the concentration of the solution may not be precisely the intended concentration. Potassium hydrogen phthalate on the other hand, has a lesser tendency to absorb water from the air and when dried will remain dry for a reasonable period of time. Potassium hydrogen phthalate may be purchased in pure form at reasonable cost. Potassium hydrogen phthalate is a primary standard. This means that carefully prepared solutions of known concentration of potassium hydrogen phthalate may be used to determine, by titration, the concentration of another solution such as sodium hydroxide. The equation for the reaction of potassium hydrogen phthalate with sodium hydroxide is: KCO2C6H4CO2H + NaOH ( KCO2C6H4CO2Na + H2O The equivalence point of a titration occurs when chemically equivalent amounts of...
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...(Na2CO3) and sodium hydroxide (NaOH) in a mixture Objective To determine the respective weight per cent of sodium carbonate and sodium hydroxide in a mixture by acid-base titration. Result and calculation Part A Titration 1 Titration number 1 2 3 Initial volume of burette( cm3) 5.10 2.70 9.70 Final volume of burette (cm3) 34.40 31.80 39.20 Total volume of HCl used (cm3) 29.30 29.10 29.50 Average volume of HCl required for titration =(29.30+29.10+29.50)/3 cm3 = 29.30 cm3 Titration 2 Titration number 1 2 3 Initial volume of burette( cm3) 4.50 14.00 2.70 Final volume of burette (cm3) 25.00 21.70 22.80 Total volume of HCl used (cm3) 20.50 20.30 20.10 Average volume of HCl required for titration =(20.50+20.30+20.10)/3 cm3 = 20.30 cm3 Part B Titration number Rough 1 2 3 Initial volume of burette( cm3) 4.9 4.80 3.60 2.20 Final volume of burette (cm3) 28.3 28.90 27.70 26.20 Total volume of HCl used until phenolphthalein decolourised (cm3) , x 23.4 24.10 24.10 24.00 Initial volume of burette after adding methyl orange indicator ( cm3) 28.3 28.90 27.70 26.20 Final volume of burette (cm3) 34.1 33.40 32.10 30.70 Total volume of HCl used until phenolphthalein decolourised (cm3) , y 5.8 4.50 4.40 4.50 Average volume of HCl required to react with Na2CO3 (2y) =2(4.50+4.40+4.50)/3 cm3 = 2(4.4667) cm3 =8.9333 cm3 Average volume of HCl required to react with NaOH (x-y) =(24.10+24.10+24...
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...acid it can be titrated with a base. Titration is the process of adding a known amount of a solution of known concentration to a known amount of solution of unknown concentration. The more accurately the concentration of the solution of known concentration is known, the more accurately the concentration of the unknown solution can be determined. Some chemicals can be purchased in a pure form and remain pure over a long period or time. Other chemicals are easily contaminated by the absorption of carbon dioxide or water from the air. Sodium hydroxide absorbs moisture from the air and often appears wet. Thus if a solution of sodium hydroxide is prepared by weighing the sodium hydroxide, the concentration of the solution may not be precisely the intended concentration. Potassium hydrogen phthalate on the other hand, has a lesser tendency to absorb water from the air and when dried will remain dry for a reasonable period of time. Potassium hydrogen phthalate may be purchased in pure form at reasonable cost. Potassium hydrogen phthalate is a primary standard. This means that carefully prepared solutions of known concentration of potassium hydrogen phthalate may be used to determine, by titration, the concentration of another solution such as sodium hydroxide. The equation for the reaction of potassium hydrogen phthalate with sodium hydroxide is: KCO2C6H4CO2H + NaOH ( KCO2C6H4CO2Na + H2O The equivalence point of a titration occurs when chemically equivalent amounts of...
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...allowed the molar concentration of a strong acid to be assessed in a standardized solution. The Average molar concentration of NaOH is approximately 0.279mol/L. Vinegar analysis was conducted with the help of three trials to determine the percent by mass of acetic acid in vinegar instead of a standardized solution. The unknown vinegar being utilized was “Shaggy”. The average percent by mass of CH3COOH “Shaggy” is 1.41%. According to “Laboratory Manual for Principles of General Chemistry”, volumetric analysis is a chemical analysis that is performed primarily with the aid of volumetric glassware. For this procedure, a known quantity or carefully measured amount of one substance reacts with a to-be-determined amount of another substance with the reaction occurring in an aqueous solution. This is conducted by a titration procedure. A burette dispenses a liquid called the titrant to a receiving flask containing the analyte (Beran, 133). The reaction is known to be completed when amounts corresponding to the mole ratio of the balanced equation is reached – this is referred to as the stoichiometric point. In the lab, standardization of a sodium hydroxide solution was performed, thus creating the secondary standard solution (Beran, 135). Solid sodium hydroxide is very hygroscopic, meaning it is capable to absorb water vapor readily (Beran, 134). In part A of our experiment KHP is utilized as the primary acid standard for determining the molar concentration of a sodium hydroxide...
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...notebook for an outline of the general procedure. The unknown acid number was 6553, and the concentration of NaOH used in the experiment was .09912 M. Also, three drops of phenolphthalein indicator were added to the initial titration and the titration curve. EXPERIMENTAL DATA: Initial Titration: * Volume of NaOH added at the endpoint was 29.8 mL Titration Curve: * Volume of NaOH added at the endpoint was 29.0 mL CALCULATED RESULTS: Acid concentration from first titration was .118M Ka from initial pH was 1.08x10^-5 Acid concentration from titration curve was .115M Titration | Volume of NaOH (mL) | pH | (base)/(acid) | pKa | Ka | 1/4 | 7.25 | 4.1 | 1/3 | 4.577 | 2.65x10^-5 | 1/2 | 14.5 | 4.6 | 1 | 4.6 | 2.55x10^-5 | 3/4 | 21.8 | 5.19 | 3 | 4.713 | 1.94x10^-5 | Average: | | | | 4.663 | 2.18x10^-15 | DISCUSSION: The purpose of the experiment was to titrate a weak acid of unknown concentration with a strong base, NaOH, and then utilizing an initial titration and titration curve to determine that acid concentration and Ka. After performing the initial titration of the acid concentration, we calculated it to be 0.118 M, with a Ka of 1.08x10^-5. On the other hand, when we performed the titration curve, it calculated an acid concentration of 0.115 M and a Ka of 2.18x10^-5. The results I obtained seemed reasonable since the acid concentration we acquired from both the initial titration and titration curve yielded a 0.003 difference...
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...experiment was to prepare an aqueous solution of NaOH, and to determine the concentration of the secondary standard NaOH solution by titrating it with a solute of primary standard, KHP. Another objective of this experiment was to learn how to plot a 2nd derivative graph using LoggerPro, and determining the concentration of the secondary standard from the 2nd derivative data. The primary standard was KHP and the secondary standard was NaOH. The difference between the 2 types of standards is that the primary standard is a has powerful reactants and isn't sensitive to the the environment and the secondary standard is something that is determined will react with a highly pure primary standard that can be standardized. It was necessary to standardize the NaOH solution because NaOH absorbs moisture from the air, making the compound not 100% NaOH, so to obtain precise concentrations, the NaOH needed to be standardized with KHP. The molarity of the NaOH solution was determined by dividing the average number of moles of NaOH (determined by taking the moles of KHP and using the 1:1 ratio) and diving it by the average volume (number of L) at the indicator point. The difference between the equivalence point and the indicator point is that the indicator point is where the indicator changes color, and the equivalent point is the place in a titration where the amount of indicator added neutralizes the solution. Yes the calculated molarities of NaOH for the indicator method and 2nd derivative...
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...compounds, and their reactions are very important. Perhaps the most important reaction is the one in which an acid and base are combined, resulting in the formation of water (in aqueous solution) and a salt; this reaction is called neutralization. A buffer solution is a solution that contains both an acid and a salt containing the conjugate base anion in sufficient concentrations so as to maintain a relatively constant pH when either acid or base is added. In this experiment you will prepare a buffer solution and observe its behavior when mixed both with an acid and a base. You will also compare the behavior with that of solutions containing only the acid. Theory In his theory of ionization in the 1880’s, Svante Arrhenius defined acids are substances which form H+ and bases as substances which form OH- in water. He further defined a salt as a substance other than an acid or base which forms ions in aqueous solution. Such substances are thus capable of producing an electric current and are called electrolytes. The amount of electricity produced is directly proportional to the concentration of ions in solution. With regard to electrolytes we have learned previously that strong acids and strong bases ionize completely, and are therefore strong electrolytes because they produce a large electric current. Soluble salts are the other type of strong electrolytes. We also learned that weak acids and weak bases ionize only partially in solution, producing...
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...Purpose of the work: The purpose of this experiment was to experimentally determine the concentration of the unknown solution as well as what molecule the unknown solution was. In order to do that, we found Ka1 and Ka2 of unknown acid and Ka for the salt of the acid. Finally, we compared data to reveal the unknown solution identity and concentration. Experimental measurement and data analysis: The concentration of NaOH we used for this experiment was approximately 0.0804M. For titration of the H3PO4, we had two equivalence points and two half-equivalence points because H3PO4 has three ionizable H+ ions, however it only loses two because by the time it reaches HPO4 the last H+ is not recognized by the water. Therefore, since it can lose two...
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...Titrations Practice Worksheet Find the requested quantities in the following problems: 1) If it takes 54 mL of 0.1 M NaOH to neutralize 125 mL of an HCl solution, what is the concentration of the HCl? 2) If it takes 25 mL of 0.05 M HCl to neutralize 345 mL of NaOH solution, what is the concentration of the NaOH solution? 3) If it takes 50 mL of 0.5 M KOH solution to completely neutralize 125 mL of sulfuric acid solution (H2SO4), what is the concentration of the H2SO4 solution? 4) Can I titrate a solution of unknown concentration with another solution of unknown concentration and still get a meaningful answer? Explain your answer in a few sentences. 5) Explain the difference between an endpoint and equivalence point in a titration. Solutions to the Titrations Practice Worksheet For questions 1 and 2, the units for your final answer should be “M”, or “molar”, because you’re trying to find the molarity of the acid or base solution. To solve these problems, use M1V1 = M2V2 . 1) 0.043 M HCl 2) 0.0036 M NaOH For problem 3, you need to divide your final answer by two, because H2SO4 is a diprotic acid, meaning that there are two acidic hydrogens that need to be neutralized during the titration. As a result, it takes twice as much base to neutralize it, making the concentration of the acid appear twice as large as it really is. 3) 0.1 M H2SO4 4) You cannot do a titration without knowing the molarity of at least...
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...POINT LISAS CAMPUS Esperanza Road, Brechin Castle, Couva www.utt.edu.tt LAB 1 Decomposition reaction Aim: Determination of the number of moles of water molecules of crystallization present in hydrated Magnesium Sulphate (MgSO4.xH2O) Apparatus: Mass balance, test tube, test tube holder, heat-proof mat and bunsen burner. Reagents: Hydrated sodium carbonate. Theory: Chemical decomposition, analysis or breakdown is the separation of a chemical compound into elements or simple compounds. A more specific type of decomposition is thermal decomposition or thermolysis, which is caused by heat. ABA+B, the reaction is endothermic, since heat is required to break the chemical bonds. Most decomposition reaction require energy either in the form of heat, light or electricity. Absorption of energy causes the breaking of the bonds present in the reacting substance which decomposes to give the product. When a hydrated salt is heated it decomposes into a pure form of the salt and water. MgSO4.xH2O MgSO4 + H2O Procedure: Refer to Handout Results: A. Mass of test tube/g = 21.77 B. Mass of the tube and salt/g = 24.0 A table showing the mass of the test tube and salt after 3 consecutive heating: Heating | Mass of the test tube and salt/g | 1st | 23.96 | 2nd | 23.81 | 3rd | 23.81 | Calculations: G. Mass of anhydrous magnesium sulphate/g = F - A = 23.81 – 21.77= 2.04 H. Mass of water of crystallization evaporated/g...
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...Industries. Your first big assignment is to investigate the strength of several commercial antacids for the Food and Drug Administration (FDA). They have sent five antacids to be tested with a back-titration that works as follows: • • • First, each antacid tablet is mixed with 40 mL of 0.1 M HCl—this acidic solution is the same stuff that is in stomach acid, and one antacid pill is nowhere near enough to neutralize all 40 mL of the acid. So, to see how much extra help each antacid pill needs to neutralize 40 mL of 0.1 M HCL, you add 0.05 M NaOH drop-by-drop to back-titrate the solution until the pH is neutral. What this means is that, the stronger the antacid tablet, the less NaOH it will take to help bring the acid to neutral. (In other words, the stronger antacid tablets counteract more of the original HCl, leaving the solution closer to neutral before the NaOH is added.) Here are your results: Maalox Mass of one dose antacid mL NaOH used in backtitration 20.0 g Tums 21.0 g Mylanta 18.0 g CVS brand 18.3 g Rennies 17.5 g 24.1 mL 22.4 mL 20.0 mL 19.9 mL 24.4 mL 1. Which is the strongest antacid, on a single-dose basis? Which is the weakest? Explain and show your calculations. 2. Which are the strongest and weakest, on a by-weight (mass) basis? 3. When people do back titrations, they usually watch the solution for a color change when the solution becomes neutral. What might you have used in the above experiment to get this color change to happen in the solution...
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...of monobasic acid which is 3.5% dissociated in N/20 solution at 20o C. (6.34x10-5) 2. Calculate the hydrogen ion concentration in 10 liters of 0.1 N solution of an acid having dissociation constant 4.0x10-10 . (6.32x10-6g ion liter-1 ) 3. Nicotinic acid (Ka =1.4x10-5) is represented by the formula HNic. Calculate its percent dissociation in a solution which contains 0.10 mol of nicotinic acid per 2.0 liter of solution. (1.66%) 4. A certain weak acid has Ka=1.0x10-4. Calculate the equilibrium constant for its reaction with strong base. (1010) 5. 0.16 g of N2H4 are dissolved in water and the total volume made upto 500 ml. Calculate the percentage of N2H4 that has reacted with water in this solution. The Kb for N2H4 is 4.0x10-6M. (2%) 6. An aqueous solution of aniline of concentration 0.24M is prepared. What concentration of sodium hydroxide is needed in this solution so that anilinium ion concentration remains 1x10-3M? (Ka for C6H5NH3+ =2.4x10-5M) (10-2M) 7. An aqueous solution contains 10% ammonia by mass and has a density of 0.99g cm-3. Calculate hydroxyl and hydrogen ion concentration in this solution.(Ka for NH4=5.0x10-10M) (9.28x10-13 mol L-1) CALCULATION OF pH 8. Calculate pH value of (a) .00001M HCl solution, and (b) 0.04M HNO3...
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...CALCULATION EXERCICES To the LABORATORY EXPERIMENTS IN MEDICAL CHEMISTRY Edited by: Zoltán Matus Compiled by: Péter Jakus László Márk Anikó Takátsy Pécs, 2007 Table of content: Introduction 3 1. Stoichiometry I. Balancing equations 4 2. Stoichiometry II. Calculation exercises 9 3. The gaseous state 13 4. Concentrations of solutions 17 5. Calculations connected to titrimetry 28 6. Electrolytic dissociation 33 7. Dilute solutions 38 8. Hydrogen ion concentration, pH, buffers 44 9. Heterogeneous equilibria. Crystallisation, solubility product, partition coefficient 55 10. Thermochemistry 64 11. Electrochemistry 67 2 INTRODUCTION The chapter is devoted to helping the students practice the most important topics of General Chemistry. The order of the sections follows the schedule of the lectures and seminars, and their volume indicates the importance of the topic. Each section begins with a few solved problems. They represent the minimum requirement at the exam. The worked-out solutions are not the only ones. For an easier self-checking, the numerical results of the unsolved calculation exercises are given in parentheses after each question. Sources: 1.) Laboratory experiments in medical chemistry, ed. György Oszbach, Pécs, 1998. 2.) Villányi Attila: Ötösöm lesz kémiából, (6. ed.) Mőszaki Könyvkiadó, Budapest., 1999 3.) Charles E. Mortimer...
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...CHEM1903 – Chemistry 1A (SSP) Michael West (305159240) 1. Experiment 2.2 – Titrimetric determination of the molecular mass of an organic acid Method An unknown organic acid was supplied in solid form. The acid was known to be diprotic and had the reference number 19. Using an analytical balance, 1.5397 g of the acid were weighed out, and made up with deionised water into 250 mL of solution. 25 mL of the acid solution was added to a conical flask with phenolphthalein indicator and titrated against standardized 0.0983 M NaOH solution. Three titrations were performed and the results averaged. The molar mass of the acid was then calculated and compared to a list of given possibilities. Results and Calculations The three titres were 26.30 mL, 26.50 mL and 26.30 mL. The mean titre volume was hence 26.37 mL. The number of moles of NaOH was thus moles. Because the acid was diprotic, reaction stoichiometry dictates that there was one mole of acid for every two moles of NaOH. Accordingly, in 25 mL of the acid solution, there were moles of acid. The molar mass of the acid is then g⋅mol-1. This matches most closely with succinic acid, for which the given molar mass was 118.1 g⋅mol-1. Although this represents a 0.6% discrepancy, the error is small enough to identify the acid as succinic acid with a high degree of certainty, given the possibilities listed. 2. Experiment 2.3 – Determination of the carbon dioxide and hydrogencarbonate contents of soda water by indirect titration Method...
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...Chemistry Acids and Bases I. Acids A. Definition Covalent molecular compound Ionizes in water Forms H+(aq) as the only cation B. Common acids Dil./ conc. HCl Very dil. HNO3 Dil. H2SO4 C. Basicity Number of H+(aq) produced per acid molecule Monobasic HCl, HNO3, CH3COOH Dibasic H2SO4, H2SO3 Tribasic H3PO4 Stepwise ionization for polybasic acids H2SO4 (aq) H+(aq) + HSO4-(aq) HSO4-(aq) H+(aq) + SO42-(aq) D. Physical properties Exist in all three states in pure form Solid (citric acid) Liquid (sulphuric acid H2SO4, carbonic acid H2CO3) Gas (hydrogen chloride HCl) Electrical conductivity Electrical conductor Electrolytes Conduct electricity in aqueous/ molten states Chemically decomposed upon conduction of electricity Mobile ions present: H+(aq) and other anions Conductivity of dibasic acid > monobasic acid More H+(aq) produced by dibasic acid of the same molarity Action on damp blue litmus paper Bluered Action on universal indicator/ damp pH paper Greenred/ turns red 1 © 2013 Victor Fong Chemistry Acids and Bases E. Chemical properties Species responsible for acidic properties: H+(aq) Reaction with metals Acid + metal salt + hydrogen Mg + 2HCl MgCl2 + H2 Only reacts with metals above Cu in MRS/ ECS Observations Metal dissolves to form a colourless/ coloured solution Colourless...
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