Name: Prem Patrick Parcon Date performed: Jan. 18, 2011 Co-worker: Kate Alyssa Caton Date submitted: Jan. 25, 2011 Experiment No. 4 PRECIPITATION TITRATIONS DETERMINATION OF THE CHLORIDE IN THE SAMPLE I. THEORETICAL FRAMEWORK Precipitation titrimetry is one of the oldest analytical techniques, dating back to the mid 1800s. The most widely used and most important precipitating reagent is silver nitrate. Precipitation titration is sometimes based on reactions that yield ionic compounds of limited solubility. Titrimetric methods based upon silver nitrate are sometimes termed as argentometric methods. This experiment also involves the use of the Mohr method. For concentrations lower than 0.1 M, an indicator blank may also be run to compensate for the overconsumption of reagent and for the acuity of the analyst in detecting the appearance of color. II. DATA AND RESULTS Standardization of Silver Nitrate Solution Primary Standard used: NaCl Formula mass of 1O standard: 58.44 g/mol % Purity of 1O standard: 99.5% Trials Mass of NaCl (g) 1 2 3 0.1990 0.1850 0.1769 Final Reading AgNO3 (ml) 37.50 31.80 34.66 Initial Reading AgNO3 (ml) 6.21 1.19 5.45 Volume AgNO3 used (ml) 31.29 30.79 29.21 30.82 31.33 34.19 Corrected Volume of AgNO3, mL Molarity of AgNO3 0.1105 Average Molarity of AgNO3 Titer NaCl (mg/mL) 0.1010 0.0885 0.1000 ± 0.110 6.457 Average Titer NaCl (mg/mL) 5.905 5.174 5.841 ± 0.6436 Indicator Blank Trials 1 2 3 Final volume AgNO3 (ml) 32.80 33.00 33.40 Initial volume AgNO3 (ml) 32.0 32.80 33.00 Net Volume AgNO3 (ml) 0.80 0.20 0.40 Average volume of AgNO3, (ml) 0.47 ± 0.305 Analysis of Unknown Trials Mass of the sample (g) 1 2 3 0.1597 0.1664 0.1567 Final volume AgNO3 (ml) 30.19 44.51 17.85 Initial volume AgNO3 (ml) 17.85 30.19 5.50 Net Volume AgNO3 (ml) 12.34 14.32 12.35 Corrected volume of AgNO3, (ml) 11.87 13.85 11.88 mg Cl- of unknown 42.08 49.10 42.11 % Cl- of unknown 26.35% 29.51% 26.87% Average % Cl- of unknown III. CALCULATIONS 27.58 ± 1.694 ππππβπ‘ ππ πΆπ π₯ %ππ’πππ‘π¦ = π ππ π΄πππ3 × π ππ π΄πππ3 πΈππ’ππ£ππππππ ππππβπ‘ 0.1990 × 0.995 58.4428 Trial 1 = π ππ π΄πππ3 π₯ 0.03082 π ππ π΄πππ3 = 0.1099 0.1850 × 0.995 = π ππ π΄πππ3 π₯ 0.03133 58.4428 Trial 2 π ππ π΄πππ3 = 0.10053 0.1769 × 0.995 = π ππ π΄πππ3 π₯ 0.03419 58.4428 Trial 3 π ππ π΄πππ3 = 0.0881 πΆππππππ‘ππ ππππ’ππ = ππππ’ππ π΄πππ3 π’π ππ − π΄π£π. πππ. ππ π΄πππ3 (πΌππππππ‘ππ π΅ππππ) Trial 1 1.29 mL– 0.47 mL = 30.82 mL Trial 2 31.80 mL – 0.47 mL = 31.33 mL Trial 3 34.66 mL – 0.47 mL = 34.19 mL ππππβπ‘ πππΆπ = π ππ π΄πππ3 × πΆππππππ‘ππ ππππ’ππ ππ π΄πππ3 (ππ πΏ) ππππππ’πππ ππππβπ‘ πππΆπ Trial 1 0.1990 π = π × 0.03082 πΏ 58.4428 π = 0. 1105 π Trial 2 0.1850 π = π × 0.03133 πΏ 58.4428 π = 0.1010 π Trial 3 0.1769 π = π × 0.03419 πΏ 58.4428 π = 0.0885 π π΄π£πππππ πππππππ‘π¦ ππ π΄πππ3 = 0.1105 + 0.1010 + 0.0885 = 0.1000 π 3 ππ ππππβπ‘ πππΆπ (ππ ππ) πππ‘ππ πππΆπ ( ) = ππΏ πΆππππππ‘ππ ππππ’ππ π΄πππ3 Trial 1 1000 ππ 1π πππ‘ππ πππΆπ = 30.82 ππ = 6.457 ππΏ Trial 2 1000 ππ 1π πππ‘ππ πππΆπ = 31.33 ππ = 5.905 ππΏ Trial 3 1000 ππ 1π πππ‘ππ πππΆπ = 34.19 ππΏ ππ = 5.174 ππΏ 0.1990 π × 0.1850 π × 0.1769 π × π΄π£πππππ πππ‘ππ = 6.457 + 5.905 + 5.174 = 5.845 ππ/ππΏ 3 πΆππππππ‘ππ ππππ’ππ = πππ‘ ππππ’ππ π΄πππ3 − π΄π£π. πππ. ππ π΄πππ3 (πΌππππππ‘ππ π΅ππππ) Trial 1 12.34 mL– 0.80 mL = 11.87 mL Trial 2 14.32 mL – 0.20 mL = 13.85 mL Trial 3 12.35 mL – 0.40 mL = 11.88 mL πππππππππ πΆπ = π ππ π΄πππ3 × πΆππππππ‘ππ ππππ’ππ ππ πππΆπ (ππΏ) Trial 1 πππππ πΆπ = π΄π£π. πππππππ‘π¦ π΄πππ3 × ππππ’ππ π΄πππ3 = 0.1000 π × 11.87 ππΏ = 1.187 ππππππ Trial 2 πππππ πΆπ = π΄π£π. πππππππ‘π¦ π΄πππ3 × ππππ’ππ π΄πππ3 = 0.1000 π × 13.85 ππΏ = 1.385 ππππππ Trial 3 πππππ πΆπ = π΄π£π. πππππππ‘π¦ π΄πππ3 × ππππ’ππ π΄πππ3 = 0.1000 π × 11.88 ππΏ = 1.188 ππππππ ππ πΆπ = πππππ πΆπ × ππππππ’πππ ππππβπ‘ πΆπ (ππ) 1ππππ ππ πΆπ = 1.187 ππππππ × Trial 1 35.45 ππ πΆπ 1 ππππ = 42.08 ππ ππ πΆπ = 1.385 ππππππ × Trial 2 35.45 (ππ) 1 ππππ = 49.10 ππ ππ πΆπ = 1.188 ππππππ × Trial 3 35.45 (ππ) 1 ππππ = 42.11 ππ %πΆπ ππ π’πππππ€π π πππππ = ππ πΆπ × 100 πππ π π πππππ (ππ) Trial 1 %πΆπ = 42.08 ππ × 100 = 26.35% 159.7 ππ Trial 2 %πΆπ = 49.10 ππ × 100 = 24.51% 166.4 ππ Trial 3 %πΆπ = 42.11 ππ × 100 = 26.87% 156.7 ππ π΄π£πππππ % πΆπ ππ π’πππππ€π = 26.35 + 24.51 + 26.87 = 25.91% 3 IV. DISCUSSION AND INTERPRETATION OF RESULTS The main goal of this precipitation titration experiment involved the determination of the chloride in the sample with silver nitrate as the indicator. First, 250 ml of 0.1 M AgNO3 solution was prepared from 1.0 M solution. To know the volume of silver nitrate needed the equation: 1.0 π × π = 0.1 π × 250 ππ π = 25.0 ππ π΄πππ3 Next, silver nitrate was standardized using Mohr method. Three samples of NaCl weighing at least 0.15 g to 0.20 g was measured and then dissolved in 50 ml water. After which, 2 ml of potassium chromate was added in each Erlenmeyer flask containing the dissolved salt. Potassium chromate here served as the indicator for the argentometric determination of chloride ions by reacting to the silver ions to form a brick-red silver chromate precipitate in the equivalence point region. Mohr method is the method which uses chromate ions as an indicator in the titration of chloride ions with a silver nitrate standard solution. After all the chloride has been precipitated as white silver chloride, the first excess of titrant results in the formation of a silver chromate precipitate, which signals the endpoint. The reactions involved here are: Titration reaction π΄π+ + πΆπ − ↔ π΄ππΆπ(π ) Indicator reaction 2π΄π+ + πΆππ4 2− ↔ π΄π2 πΆππ4 (π ) An indicator blank was run first. A small amount of calcium chromate was dissolved in 100 ml water added with 2 ml potassium chromate indicator. Based on records, the average volume of silver nitrate used for the 3 trials was 0.47 ml. From the recorded data, we get the molarity of AgNo3 for trial 1 to be 0.1105, 0.1010 for trial 2, 0.0885 for trial 3 with an average of 0.1000 and a deviation of 0.110. Furthermore, based on the amount of salt used per trial and the corrected volume of silver nitrate solution and also taking into consideration the percent purity of the primary standard used, we get titer NaCl for trial 1 to be 6.457, 5.905 for trial 2, and 5.174 for trial 3. The Average Titer NaCl is 5.841 with a deviation of 0.6436. Another set up was then prepared for the determination of chloride in the sample. From the dried sample, at least 0.15 g to 0.20 g was measured for the three samples. Each was then dissolved in 50 ml water and a 2 ml 0.1 M potassium chromate solution was then added. Using the data recorded, the total calculated chloride in the sample for trial 1 was 42.08, for trial 2 it was 49.10, and 42.11 for trial 3. The percent chloride for trial 1 was 26.35%, for trial 2 was 29.51%, and for trial 3 it was 26.87%. V. CONCLUSION AND/OR RECOMMENDATION(S) The solution of sodium chloride with the potassium chromate as the color indicator, when titrated to the silver nitrate solution forms silver chloride and shifts from a yellow solution to a reddish brown. This color change and clump formation of silver chloride is a sign that the end point is near. Indicator blank is used to correct the volume of silver nitrate used. Also, the average volume of silver nitrate calculated in the running of indicator blank also helped in calculating the milligrams of chloride and the percent chloride used in each trial. The use of Mohr method in this experiment is important in that it helped in our understanding of the chemical reactions that occur during each titration process. VI. REFERENCES ο· Skoog, Douglas A., et al. Fundamentals of Analytical Chemistry. 8th ed. Singapore: Thomson Learning Asia, 2004 ο· Korkmaz, Deniz. Precipitation Titration: Determination of Chloride by the Mohr Method. <http://academic.brooklyn.cuny.edu/esl/gonsalves/tutorials/Writing_a_Lab_Report/xPrecipitati on%20Titration%20edited%203.pdf> ο· University of Canterbury. Determination of Chloride by Precipitation Titration with Silver Nitrate - Mohr's Method. < http://www.outreach.canterbury.ac.nz/chemistry/chloride_mohr.shtml>
