Standardization of Sodium Hydroxide Against Potassium Hydrogen - Lab 4 | CHEM 213, Lab Reports of Quantitative Techniques

Download Standardization of Sodium Hydroxide Against Potassium Hydrogen - Lab 4 | CHEM 213 and more Lab Reports Quantitative Techniques in PDF only on Docsity! Lab #4 (36B-3&6), STANDARDIZATION OF SODIUM HYDROXIDE AGAINST POTASSIUM HYDROGEN PHTHALATE THE DETERMINATION OF POTASSIUM HYDROGEN PHTHALATE IN AN IMPURE SAM- PLE #343 RADU PURTUC QUANTITATIVE ANALYSIS LAB, CHEM 213 Group # 5 Radu Purtuc Lavonne Salis Tunde Tamas Experiment Start Date: June 22, 2006 Experiment End Date: June 24, 2006 Report Due Date: June 28, 2006 PAGE 1 ANALYSIS The purpose of this experiment was to accomplish standardization titration of Sodium Hydroxide against Potassium Hydrogen Phthalate and the determination of Potassium Hydrogen Phthalate in an impure compound sample (mass percentage of KHP in sample). The unknown sample was # 343 The indicator used throughout the experiment was the Phenolphthalein, which resulted in formation of pink tint color from original colorless marking the end point. PROCEDURE Clean and dry three Erlenmeyer flasks, three Burettes, two 200 mL beakers, and two 500 mL beakers, a spatula, and a glass rod. Record Lot Number of compounds used in this experiment: NaOH Lot# 001819.24 KHP Lot# 100.03 Unknown Sample # 343 ** note “Ф” is the equivalent of a mole. 1. Mix 5.00 mL of NaOH with 995 mL of di H2O to give 0.1 M NaOH solution. " M1V1=M2V2 19.1M(V1)=(0.1M)(1000mL) => V1=5.20 mL 2. Weigh three individual 0.4080g samples KHP into 250-mL conical flasks. 3. Dissolve each in about 50-mL of di H2O. 4. Add 3 drops of Phenolphthalein. 5. Titrate with dilute NaOH until the solution just begins to change from colorless to pink (and pink color remains constant for 30 seconds before fading). 6. Correct reagent weights for the blank (-0.05mL) 10. Titrate unknown # 343 " a. 1.1100g  20.70 mL -0.05 = 20.65 mL " b. 1.1106g  20.80 mL -0.05 = 20.75 mL " c. 1.1104g  20.80 mL -0.05 = 20.75 mL 11. Report the percentage of KHP in the unknown compound. 12. Dispose titrant and reagents as directed by the instructor. 13. Clean the desk and desk accessories; wash all glassware and safely place them back. 1. Standardization of 0.1 M NaOH Chemical Reaction C8H5O4K + OH- >> C8H4O42- + H2O + K+ Mix 5.00 mL of NaOH into 995 mL di H2O for a total volume of 1000 mL. Mix well. Determine how much KHP is needed to give about 20 mL of reagent NaOH. PAGE 2 Table 2: Calculation of KHP mass that participated in reaction 1 Ф KHP 0.1 Ф OH- 204.222 g KHP 20.00 x E^-3 L OH- = 0.4080 g KHP 1 Ф OH- 1 L OH- 1 Ф KHP Table 3: Calculation of OH- molarity that participated in reaction 1 Ф OH- 1 Ф KHP 0.4080 g KHP = 0.1041 M OH- 1 Ф KHP 204.222 g KHP 19.20 x E^-3 L OH- Mix 5.00 mL of NaOH into 995 mL di H2O for a total volume of 1000 mL. Mix well. Part B The exact mass of KHP needed for each Flask was determined. Determine how much KHP is needed to give about 20 mL of reagent NaOH. 1 Ф KHP 0.1 Ф OH- 204.222 g KHP 20.00 x E^-3 L OH- = 0.4080 g KHP 1 Ф OH- 1 L OH- 1 Ф KHP Add approximately 0.4080 g KHP into three Erlenmeyer flasks + 50 mL di H2O + indicator and titrate against OH- stock solution. Next, calculate the actual concentration (x.xxxx M). The results showed that 0.1 M of titrant is in fact a mean of 0.1046 M OH-, called standard solution. My own set of data gave a 0.1047 M OH-. 1 Ф OH- 0.1 Ф KHP 0.4080 g KHP 17.007 g OH- = 0.0340 g OH- 1 Ф KHP 204.222 g KHP 1 Ф OH- 0.0340 g OH- 1 Ф OH- = 0.1041 M OH- 17.007 g OH- 19.20 x E^-3 L OH- II. Determination of KHP composition in Unknown # 343 C8H5O4K + OH- >> C8H4O42- + H2O + K+ PAGE 5 Table 4: Titrant volumes molarity of OH-, and mass of KHP in the Unknown compound Analyst M OH- Mass (g) OH- Volume (mL) OH- Blank Ad- justment ( - 0.05 mL ) Mass % KHP Mass (g) unknown compound Mass of KHP pre- sent in compound Radu 0.1041 0.0340 20.70 20.65 39.69 1.1100 0.4406 Lavonne 0.1044 0.0340 20.80 20.75 39.91 1.1106 0.4431 Tunde 0.1047 0.0340 20.80 20.75 39.91 1.1104 0.4431 Mean 39.84 1.1103 0.4423 Standard Dev 0.10 0.0003 0.0011 RSD 0.25 0.03 0.25 Mean % = Σ xi / N = (39.69 + 39.91 + 39.91) / 3 = 39.84 (±0.10) Standard deviation (s) = √(Σ(xi - )2 / N-1) = 0.10 RSD = (s) / Mean x100 = [0.10/39.84] x 100 = 0.25 % To calculate the unknown’s titration volume, we have done a quick titration experiment using 0.1000 g of unknown which yield a 1.80 mL (blank adjusted) total volume titrant used. The calculated volume as- sumed the final titrant volume in the range of 20.00 mL. X g Unknown 20.00 mL x 0.1000 g = 1.1100 g Unknown 1.80 mL KHP compound Finding mass of OH- that reacted with KHP from the Unknown composition 0.1041 Ф OH- 20.65 E^-3 L OH- 17.007 g HCl = 0.0367 g OH- 1 L OH- 1 Ф HCl Finding mass of KHP present in Unknown 1 Ф KHP 1 Ф OH- 0.0367 g OH- 204.222 g KHP = 0.4406 g KHP 1 Ф OH- 17.007 g OH- 1 Ф KHP PAGE 6 Finding the mass percentage of KHP in the unknown KHP compound 0.4406 g KHP x 100 = 39.69 % KHP 1.1100 g Unknown # 343 CONCLUSIONS The experiment calculated what was the percent of KHP present in unknown compound. Following a number of titrations, mass measurements, a percent by mass mean was determined as 39.84 (±0.10). Possible errors are as follows: Instrument errors: Calibration of scales and volumetric ware (graduated cylinders). Method errors: Not delivering the entire amount of crystals measured into flasks. Personal errors: Reading of the meniscus in graduated cylinders. PAGE 7