Standardization of Na OH solution PDF

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Standardization of Na OH solution...

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Lab Chem200

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Standardization of NaOH solution

Section number: 05

 Name: Miriam Gallegos

Part A. Preparing a standard potassium hydrogen phthalate (KHP) solution. The molarity of the KHP in the 250 mL volumetric bottle. (Show your calculation processes) M1= 2.801 g of KHP V1= 250 mL (2.801g/ (204.22g/mol))/ (250mL/ (1000 mL/L)) = 0.0137156/0.25= 0.055 mol/L Part B. Preparing the sodium hydroxide (NaOH) solution. The molarity of the approximate concentration of NaOH in the 1L bottle. (Show your calculation processes ) 6.92 mL of 10M NaOH in 700. mL in a 1L bottle Molarity= ((10mol/L) x (6.92mL/1000mL/L))/ (700mL/ 1000mL/L)= 0.0692/0.7= 0.0989 mol/L Part C. Standardizing the NaOH solution Experimental data Initial volume of NaOH (mL): Final volume of NaOH (mL): Color of the titrated solution:

Trial One: 1.08

Trial Two: 16.89

Trial Three: 30.78

Trial Four: 2.34

Trial Five 16.22

Trial Six 30.12

16.89

30.78

44.69

16.22

30.12

49.42

Hot Pink

Pale Pink

Pale Pink

Pale Pink

Pale Pink

Hot Pink

Calculated data

Trial One: Trial Two: Moles of KHP (mol) = 1.375 mol Moles of NaOH (mol) = 1.375 mol Volume 15.81 13.89 of NaOH (mL)

Trial Three:

Trial Four:

Trial Five

Trial Six

13.91

13.88

13.90

19.30

Molarity of NaOH (M):

0.099M

0.099M

0.071M

0.071M

0.087M

0.099M

Average Molarity (6 trials) = 0.092M Standard deviation (6 trials) = 0.0105 Average Molarity (without bad data) = 0.09M Standard deviation (without bad data) = 0.011

Show your calculations for  volume of NaOH (trial one), molarity of NaOH (trial one). Average molarity for 6 trials and Standard deviation for 6 trials Indicate which data is bad data.

Moles of KHP= 0.055 mol/L x 25.00mL= 1.375 mol of KHP NaOH volume: Trial first: 16.89mL -1.08mL= 15.81mL Trial two: 30.78mL -16.89mL= 13.89mL Trial three: 44.69mL -30.78mL= 13.91mL Trial four: 16.22mL -2.34mL= 13.88mL Trial five: 30.12mL -16.22mL= 13.9mL Trial six: 49.42mL -30.12mL= 19.3mL Molarity of NaOH: Trial first: 1.375 mol/15.81mL= 0.087M Trial two: 1.375 mol/ 13.89mL= 0.099M Trial three: 1.375 mol/13.91mL= 0.099M Trial four: 1.375mol/ 13.88mL= 0.099M Trial five: 1.375mol/ 13.90mL= 0.099M Trial six: 1.375mol/ 19.30mL= 0.071M Average molarity: 0.554/6= 0.092M Standard deviation: 0.0105

Discussion/Questions: 1. Random errors can be affected by the environment or the measurements. Random error can be affected by the measurement such as adding too much liquid or not looking at the meniscus correctly. This also connects to systematic errors which also can be adding too much or too little of the liquids. 2. Part C, calculate the average value of the molarity of the NaOH solution. I believe I could improve my part with my solution. 3. In conclusion, I would not reject all my data in your calculation of the average molarity....