CHM 130 Acid-Base Titration Molarity of Acetic Acid in Vinegar

Transcription

1 CHM 130 Acid-Base Titration Molarity of Acetic Acid in Vinegar INTRODUCTION One of the most important techniques for chemical analysis is titration to an equivalence point. To illustrate this procedure, let us examine a typical problem. Suppose that an investigator wishes to know the exact quantity of an acid present in a certain mixture. He can find how much by determining the quantity of a base required to just neutralize the acid (the equivalence point). The quantity of base needed can be measured by preparing a base solution of known concentration and measuring the volume of it needed for the neutralization. The volume of base is determined by adding it from a buret until an indicator * in the mixture changes colors to signify that the end point of the titration has been reached. The volume of base solution used would then be read from the buret. The end point of a titration is not quite the same thing as the equivalence point. The end point has been reached when the indicator changes color, the equivalence point has been reached when enough base has been added to exactly neutralize the acid. Ideally, an indicator should be chosen so that the end point and the equivalence point should be as close to each other as possible. *Phenolphthalein indicator is suitable for titrations of a strong base with a weak acid. It is colorless in acid and pink in base. It also changes color very close to the equivalence point. The base solution used in a titration must be standardized, in other words the concentration of the solution has been accurately determined. A widely used procedure for preparing and standardizing solutions for titrations is to prepare a solution of approximately the desired concentration and then to titrate it against an accurately weighed quantity of a compound of known purity. The compound of known purity used in this manner is called a primary standard. The primary standard that will be used in this experiment is oxalic acid dihydrate, H 2 C 2 O 4 2H 2 O. Oxalic acid dihydrate is a solid acid, so it can be weighed very accurately on a balance. This acid contains two acidic hydrogen ions (H + ) that can be released into solution to react with the hydroxide ions (OH - ) from the base. Oxalic acid dihydrate is derived from the stems of the rhubarb plant and is highly TOXIC. It should be handled with care and gloves should be used while working with it. 1

2 BACKGROUND The reaction of an acid and base to form a salt is called neutralization. Consider the reaction between nitric acid and calcium hydroxide. The balanced "molecular" equation is: 2HNO 3 (aq) + Ca(OH) 2 (aq) Ca(NO 3 ) 2 (aq) + 2 H 2 O (l) This type of equation is useful for calculations but it does not represent what is really going on! Let's examine the total ionic equation which indicates what species are actually in the solution. 2H + (aq) + 2NO 3 (aq) + Ca +2 (aq) + 2OH (aq) Ca +2 (aq) + 2NO 3 (aq) + 2H 2 O (l) Since a calcium ion and two nitrate ions appear on both sides of the total ionic equation (they are spectator ions), they may be eliminated to give the net ionic equation. Dividing this by 2 gives: 2H + (aq) + 2OH (aq) 2H 2 O (l) H + (aq) + OH (aq) H 2 O (l) This net ionic equation describes the reaction of all strong acids with all strong bases to form soluble ionic salts and water. Concentrations of solutions may be expressed in a variety of ways. One of the most useful is molarity. The molarity (M) of a solution is defined as the number of moles of solute per liter of solution, i.e., In acid-base reactions, one mole of hydrogen ions (H + ) from the acid will exactly neutralize one mole of hydroxide ions (OH - ) from the base. HOWEVER, one mole of acid does not necessarily neutralize one mole of base. In the nitric acid-calcium hydroxide example above for instance, two moles of nitric acid react with one mole of calcium hydroxide. This is because nitric acid only produces one hydrogen ion in solution, while calcium hydroxide produces two hydroxide ions in solutions. If you were to use phosphoric acid and potassium hydroxide as shown below, the ratio would be one mole of phosphoric acid to three moles of potassium hydroxide. H 3 PO 4 (aq) + 3KOH (aq) K 3 PO 4 (aq) + 3H 2 O (l) We will keep this stoichiometric ratio in mind when we come to the calculation section. 2

3 PROCEDURE A.) Standardization of a Sodium Hydroxide solution. 1. Weigh out 3 samples of oxalic acid that are about 1.0g each and record the exact mass to the nearest mg as Mass of Oxalic Acid in the data table. 2. Place the oxalic acid samples in three separate 250 ml Erlenmeyer flasks, label the flasks 1, 2, and 3 to avoid confusion later. 3. Add about 50 ml of deionized water to each flask to dissolve the oxalic acid. 4. Add 3 drops of phenolphthalein to each flask. 5. Clean and rinse a buret with distilled water. If the water beads up on the inside of the buret instead of forming an even sheet of water on the glass, reclean the buret. 6. Rinse the buret three times with small portions of the sodium hydroxide solution to be standardized. 7. Fill the buret above the zero mark and briefly open the stop-cock all the way to drive the air out of the tip and rinse the stop-cock and tip with the sodium hydroxide solution. 8. Adjust to the liquid level in the buret to between 0 and 5mL, record this value to the nearest 0.01mL as Volume NaOH Initial on the data table. If you are unsure how to read a buret accurately, please ask your instructor. 9. Place the first flask under the buret and place a piece of white paper under the flask. The white paper will make the indicator's color change more visible. 10. Titrate the acid by allowing the base to flow into the flask. By slowly rotating the stop-cock you can adjust the flow rate of the buret from a steady stream to single drops. The flask should be continuously swirled. Eventually, a pink color will appear where the base hits the acid solution and will linger for a short time while the solution is swirled then it will go clear again. This is a signal to slow down the addition of the base. As you get nearer to the end point the pink color will linger longer and longer before the solution goes clear again. 11. When it appears that only a few more drops are needed, rinse down the walls of the flask with distilled water. 12. Continue to add base only a drop at a time until a pale pink color persists throughout the whole solution for at least 30 seconds the solution does not go back to being clear. If the solution is dark pink, you have added too much base and have gone past the end point. 3

4 13. Read the volume in the buret to the nearest 0.01mL, and record this value as Volume NaOH Final on the data table. The volume of sodium hydroxide used can be determined by subtracting Volume NaOH Initial from Volume NaOH Final. ml NaOH added = Volume NaOH Final Volume NaOH Initial 14. Repeat steps 7-13 to titrate the second and third samples. 15. Pour the titrated solutions into the waste container, clean each of the flasks and rinse them with deionized water. B.) Determination of the Molarity of Acetic Acid in Vinegar. 1. Use the preset repipettor to dispense 15.00mL of vinegar into each of the three 250mL Erlenmeyer flasks. Place your flask under the tip and lift the pump very slowly and carefully to avoid creating air bubbles. Allow the pump to drop by itself, do not push on it. 2. Add 3 drops of phenolphthalein to each flask. 3. Titrate each vinegar sample following steps 7-13 in Part A. 4

5 CALCULATIONS A.) Standardization of the Sodium Hydroxide solution. The balanced equation involved is: H 2 C 2 O 4 2H 2 O (s) + 2NaOH (aq) Na 2 C 2 O 4 (aq) + 4H 2 O (l) Therefore, the calculation steps would be as follows: ( ) ( ) ( ) A = the exact mass of oxalic acid that was used B = the molar mass of oxalic acid C = the volume of NaOH used, converted to Liters. Once you have determined the Molarity of the NaOH for each of the three trials, average the values to determine an average Molarity for the NaOH. B.) Determination of the Molarity of Acetic Acid in Vinegar. The balanced equation involved is: HC 2 H 3 O 2 (aq) + NaOH (aq) NaC 2 H 3 O 2 (aq) + H 2 O (l) Therefore, the calculation steps would be as follows: ( ) ( ) D = the volume of NaOH used, converted to Liters Use the average M NaOH calculated from Part A Once you have determined the Molarity of the acetic acid for each of the three trials, average the values to determine an average Molarity for the acetic acid. 5

6 Name: Section Lab Partner DATA TABLE Molarity of Acetic Acid in Vinegar Part A: Standardization of Sodium Hydroxide Trial 1 Trial 2 Trial 3 A Mass of Oxalic Acid (g) B Molar mass of Oxalic Acid H 2 C 2 O 4 2H 2 O Volume NaOH Initial (ml) Volume NaOH Final (ml) C Volume NaOH used (ml) Molarity of NaOH (mol/l) Average Molarity of NaOH (mol/l) Show Calculations for all three trials below: 6

7 DATA TABLE Part B: Molarity of Acetic Acid in Vinegar Trial 1 Trial 2 Trial 3 Volume of Acetic Acid (ml) Volume NaOH Initial (ml) Volume NaOH Final (ml) D Volume NaOH used (ml) Molarity of HC 2 H 3 O 2 (mol/l) Average Molarity of HC 2 H 3 O 2 (mol/l) Show Calculations for all three trials below: 7

8 ADVANCDED STUDY PROBLEMS mL of a sulfuric acid solution was standardized by titrating it with M NaOH using phenolphthalein indicator ml of the NaOH was required. Calculate the molarity of the sulfuric acid. H 2 SO 4 (aq) + 2NaOH (aq) Na 2 SO 4 (aq) + 2H 2 O (l) 2. Calculate the molarity of an acetic acid solution if 32.0 ml of a M NaOH is required to neutralize 25.0 ml of the acetic acid g of a mixture of solid Oxalic Acid (molecular weight g/mole) plus an inert material was dissolved in water and titrated with M NaOH ml of the NaOH was required. Calculate the percent acid in the mixture. 4. What volume (in ml) of 11.7 M HCl is required to make1.0 liter of 0.10 M hydrochloric acid solution? 8