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

Material Type: Lab; Class: Quantitative Analysis; Subject: Chemistry; University: Northeastern Illinois University; Term: Summer 2006;

Typology: Lab Reports

Pre 2010

Uploaded on 08/04/2009

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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
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Lab #4 (36B-3&6), STANDARDIZATION OF SODIUM HYDROXIDE AGAINST POTASSIUM HYDROGEN PHTHALATE THE DETERMINATION OF POTASSIUM HYDROGEN PHTHALATE IN AN IMPURE SAM- PLE # 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

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. KHP Lot# 100. 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.

WORK MANAGEMENT

Tunde was the group leader for this experiment. Lavonne and Radu alternated between tasks so that one does a measurement while the other visually checks the results to ensure work efficiency while keeping determinant error to its lowest possible value. There were no incidents reported throughout the experi- ment. Each of group member wore eyewear very diligently and practiced good lab techniques. Following the end of the experiment, each member shared in the responsibility of cleaning, drying, and returning to stock each of the utensils and glassware used. RESULTS I. Standardization of Titrant Part A C8H5O4K + OH-^ >> C8H4O42-^ + H2O + K+ Table 1 : Titrant volumes and molar concentration of HCl Analyst Volume (mL) OH- Blank Adjustment ( - 0.05 mL ) Mass KHP M OH- Radu 19.25 19.20 0.4076 0. Lavonne 19.20 19.15 0.4081 0. Tunde 19.15 19.10 0.4083 0. Mean 0.4080 0. Standard Dev 0.0004 0. RSD % 0.09 0.

Mean Molarity = Σ xi / N = (0.1041 + 0.1044 + 0.1047) / 3 = 0.1044 (±0.0003)

Standard deviation (s) = √(Σ( xi - )^2 / N-1) = 0.

RSD = (s) / Mean x100 = [0.0003/0.1044] x 100 = 0.28 %

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+

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.