Acid-Base Titrations - Lab Report Example

_______________________________________ Drawer number _________ Lab Day and Time: ____________________________ _________________ Experiment 9: PRELAB EXERCISE FOR ACID-BASE TITRATIONS 1. HCl and NaOH are mixed in equal molar ratios. a. Write the balanced equation for this reaction: HCl (aq) + NaOH (aq)  NaCl (aq) + H2O b. What is this kind of reaction called? This is called Neutralization reaction, in which an acid and a base react to form a neutral compound or salt. 2. The definition of molarity is the number of mole of solute dissolved in a liter of solution. 3. a. What is the purpose of an indicator? An acid-base indicator is either a weak acid or a weak base that changes its color, over a range of pH, in an alkaline or acidic solution. The conjugate acid or conjugate base of an indicator takes a color different from that of the original substance. In titration, an indicator is added to a mixture with an unknown amount of analyte. An indicator changes its color once reacted to the titrant, signifying the equivalence point and end point of titration. b. What acid-base indicator will you be using today? In this experiment, phenolphthalein indicator will be used. c. The color of this indicator in basic solution ranges from pink to red. 4. Should you use a graduated cylinder or a pipet to measure out the vinegar? Explain. In measuring or obtaining small volume of acids or bases from the reagent bottle and delivering such into a flask, pipette is typically used. Pipet not only measures the required amount, but facilitates the transfer of acids or base from the reagent bottle into a flask as well. 5. You dilute the vinegar with about 25 mL of deionized water. Should you use a pipet or a graduated cylinder to measure out the water? Explain your answer. In measuring or obtaining the required volume of liquids, a graduated cylinder is typically used. Since water is a non-caustic substance, a graduated cylinder is just enough to safely measure its required volume. 6. Why is the 25 mL of deionized water not included in the calculations? Deionized water is tantamount to distilled water, wherein the natural ion components of a tap water are removed via ion exchange process. Deionized water is excluded in the calculation because it only serves as a reaction medium. 7. Does it matter whether the buret has water droplets in it when you add the standard NaOH? Explain your answer. Prior to titration, it is important to ensure that the buret is clean and dry as much as possible. Excessive amount of water clinging at the sides of the buret may affect the concentration of the standardized titrant. Since the titrant is prepared using water as the dissolving medium, water from the sides of the buret adds to the water component of the titrant, affecting the standard concentration. 8. If an air bubble is left in the buret tip before the titration but disappears during the titration, how will this affect the number of moles reported for the acid? Explain your answer. Air in the buret, in the forms of bubbles, displaces the titrant from its occupied space. This affects the accuracy of the titrant’s volume reading from the indicated graduations of the buret. 9. Sodium benzoate and calcium propionate, common food preservatives, can be prepared by the reactions below. Complete and balance the equations: HC6H5COO + NaOH  NaC6H5COO (aq) + H2O (aq) 2 HC2H5COO + Ca(OH)2  Ca(C2H5COO)2 (aq) + 2 H2O (aq) PRELAB EXERCISE FOR ACID-BASE TITRATIONS, continued Complete this report form using the data in bold. (Your numbers will be different from these.) Molarity of the standard NaOH solution: 0.117 M Trial 1 Trial 2 Trial 3 1. volume of vinegar pipetted 5.00 mL 5.00 mL 5.00 mL 2. initial buret reading 0.45 mL 16.15 mL 31.80 mL 3. final buret reading 16.15 mL 31.80 mL 47.60 mL CALCULATIONS 4. mL NaOH required (#3 – #2) ________ ________ ________ 5. molarity of acetic acid in vinegar samples ________ ________ ________ (see calculations p. 5) 6. average molarity of acetic acid in vinegar ________ (average of three values in #5) 7. average percent acetic acid in vinegar sample ________ (use molarity from #6; follow calculation on p. 5) Unknown number: _________ Molecular weight 144 (from instructor) Molarity of the standard NaOH solution: 0.147 M (M = moles per liter) Trial 1 Trial 2 8. Weight of the unknown (use about 0.001 mole) 0.173 g 0.216 g 9. initial buret reading 4.10 mL 2.95 mL 10. final buret reading 28.55 mL 33.60 mL CALCULATIONS 11. mL NaOH required (#10 – #9) ________ ________ 12. moles of NaOH used (#11 in L x M NaOH) ________ ________ 13. moles of acid titrated (g acid ÷ mol. wt. acid) ________ ________ 14. moles of NaOH ÷ moles acid (#12 ÷ #13) ________ ________ 15. average of 2 values in #14, rounded to nearest integer ________ this is the number of protons in the unknown acid (average of three values in #12) Solutions: 4. trial 1 16.15 mL – 0.45 mL = 15.70 mL trial 2 31.80 mL – 16.15 mL = 15.65 mL trial 3 47.60 mL – 31.80 mL = 15.80 mL 5. trial 1 M = (0.117 M) (15.70 mL) 5.00 mL = 0.367 M trial 2 M = (0.117 M) (15.65 mL) 5.00 mL = 0.390 M trial 3 M = (0.117 M) (15.80 mL) 5.00 mL = 0.370 M 6. Average molarity = 0.367 M + 0.390 M + 0.370 M 3 = 0. 376 M 7. Percent Acetic Acid 0.376 mole HC2H3O2 x __1 L__ x _60 g HC2H3O2__ L 1000 mL 1 mole HC2H3O2 = 0.02256 g HC2H3O2 x 100 mL = 2.26% 11. trial 1 28.55 mL – 4.10 mL = 24.45 mL trial 2 33.60 mL – 2.95 mL = 30.65 mL 12. trial 1 0.147 mole NaOH x __1L _ x 24.45 mL 1 L 1000 mL = 3.59 x 10 -3 mole trial 2 0.147 mole NaOH x __1L _ x 30.65 mL 1 L 1000 mL = 4.51 x 10-3 mole 13. trial 1 0.173 g ÷ 144 g/mole = 1.20 x 10-3 mole trial 2 0.216 g ÷ 144 g/mole = 1.50 x 10-3 mole 14. trial 1 3.59 x 10 -3 mole ÷ 1.20 x 10-3 mole = 2.99 trial 2 4.51 x 10-3 mole ÷ 1.50 x 10-3 mole = 3.00 15. Average 2.99 + 3.00 2 = 2.99 or 3 as rounded to the nearest whole number Questions for Experiment 9: Acid Base Titrations 1. 100.0 mL of 0.100 M HCl was required to reach the endpoint when titrated against 50.00 mL of NaOH. What is the molarity of the NaOH solution? HCl (aq) + NaOH (aq)  NaCl (aq) + H2O Mb = 0.100 M HCl x 100.0 mL HCl 50.0 mL Mb = 0.200 M NaOH 2. 26.55 mL of a 0.719 M standard solution of NaOH was required to reach the endpoint when titrated against 200.0 mL of sulfuric acid, H2SO4. a. Write the balanced equation for this reaction. Note that the molar ratio of H2SO4 to NaOH is not 1:1. H2SO4 (aq) + 2 NaOH (aq)  Na2SO4 (aq) + 2 H2O b. What is the molarity of the sulfuric acid? = Ma = 26.55 mL NaOH x 0.719 M NaOH x 1 mole H2SO4 (200.0 mL H2SO4) 2 moles NaOH Ma = 0.0477 M H2SO4 3. If 30.25 mL of 0.1544 M NaOH were required to titrate 0.278 g of an unknown acid (molecular weight 234), how many titratable hydrogens did the acid have? HxA + X NaOH  NaxA + X H2O 0.1544 mole NaOH x 30.25 mL NaOH x __1 L__ = 4.671 x 10 -3 mole NaOH L 1000 mL _0.278 g__ = 1.188 x 10 -3 mole of unknown acid 234 g/mole X = ____4.671 x 10 -3 mole NaOH____ 1.188 x 10 -3 mole of unknown acid X = 3.93 or 4 as rounded to the nearest whole number There are about four titratable hydrogens in the acid. 4. What effect on the final molarity of acetic acid in vinegar would each of these have? Explain. a. If water had been in the pipet when you pipetted the vinegar, would the calculated M of acetic acid be: lower than the actual M higher than the actual M no different from the actual M Since water is only the dissolving medium, it does not affect the actual amount of acetic acid present in the vinegar. b. If water had been in the Erlenmeyer into which you added the vinegar, would the calculated M of acetic acid be: lower than the actual M higher than the actual M no different from the actual M Since water is only the dissolving medium, it does not affect the actual amount of acetic acid present in the vinegar. c. If water had been in the buret into which you poured NaOH solution, would the calculated M of acetic acid be: lower than the actual M higher than the actual M no different from the actual M The water from the pipet adds to the volume of water required for the indicated molarity of NaOH, increasing the value of the denominator. The resulting quotient or value for molarity then is lower. A lower molarity of the standard base, with constant volume of base and acid, results to a lower computed molarity of the acid. d. You pumped vinegar into a beaker to bring to your bench. If there had been water in that beaker, would the calculated M of acetic acid be: lower than the actual M higher than the actual M no different from the actual M Since water is only the dissolving medium, it does not affect the actual amount of acetic acid present in the vinegar. e. You took NaOH solution from a carboy into a beaker to bring to your bench. If there had been water in that beaker, would the calculated M of acetic acid be: lower than the actual M higher than the actual M no different from the actual M The water from the pipet adds to the volume of water required for the indicated molarity of NaOH, increasing the value of the denominator. The resulting quotient or value for molarity then is lower. A lower molarity of the standard base, with constant volume of base and acid, results to a lower computed molarity of the acid. 5. If 10.0 mL of an acetic acid solution of unknown concentration were titrated with 28.75 mL of 0.117 M NaOH, what is the concentration (molarity) of acetic acid? HCl (aq) + NaOH (aq)  NaCl (aq) + H2O Ma = 28.75 ml NaOH x 0.117 M NaOH 10.0 mL Ma = 0. 336 M of acetic acid Read More