Acid Base Stoichiometry - Formal Report PDF

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Adic Base Stoichiometry Lab Report...

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Josselyn Zetino Mendoza November 8, 2020 Acid Base Stoichiometry – Formal Report Purpose: The objective of this experiment is to prepare a 01. M NaOH and solution and standardize it with sulfuric acid. Then use the standardize solution to determine molarity of an unknown sulfuric acid solution. Theory Principles: On the following we will first calculate the volume of three molar sodium hydroxide needed to prepare one liter of 0.1 molar sodium hydroxide. After obtaining all data for NaOH we will calculate the molarity of NaOH. Then after obtaining the data for the unknown H 2 SO 4 we will calculate its molarity. Experimental Procedures: First rinse a battle with distilled water. Then obtain a 3.0 molar sodium hydroxide and pour it into a beaker. After that measure 33 mL of sodium hydroxide. Filled the bottle halfway with distilled water and then add the 33 mL of the 3 molar sodium hydroxide. Then weight 1 gram of sulfuric acid on a weighting paper using a centigram balance. Then divide into three parts the sulfuric acid and transfer it to three different papers and repeat weight. Next use 3 250 mL Erlenmeyer flasks and transfer each sample to its corresponding flaks. To each flask ass approximately 25 mL of distilled water and 2-3 drops of phenolphthalein. Then place one of the flasks containing the indicator and sulfamic acid under the buret and slowly start adding the sodium hydroxide until you start seeing a pale pink end point. Before doing this we had to record the initial volume of NaOH and at the end of titration record final volume level of NaOH. After obtaining this data we will be able to calculate molarity of NaOH. Then we will fill a buret with sodium hydroxide and we will repeat the same procedure to each flaks when adding phenolphthalein with the unknown solution. After we had done this we will be able to calculate the molarity of unknown sulfuric acid solution. Data Tables:

Mass of weighting paper + Sulfamic acid, g Mass of weighting paper, g Initial Buret Reading, mL Final Buret Reading, mL

10.1 g 0.3912 g 0 mL 38 mL

Sulfuric Acid Initial Buret Reading, mL Sulfuric Acid Final Buret Reading, mL NaOH Initial Buret Reading, mL NaOH Final Buret Reading, mL

Trial 1

Trial 2

0

20

20

40

0

0

24

24

Results: Molarity of NaOH

Mass of sulfomic acid=mass paper + sample−mass paper Mass of sulfomic acid=10.1 g−0.3912 g=9.7088 g

V t =V final −V initial V t =38 mL−0 mL=38 mL

Ratio of Titrant=

Vt mass sulfomic acid

Ratioof Titrant=

3 8 mL =3. 91397495 g /mL 9.7088 g

Moles of NaOH=Moles of Sulfuric Acid

Moles of H 2 SO 4=

9.7088 g =0.09998 764 mol 97.1 g ∙mol

Moles of NaOH=

39.997 g =0.41191555 mol 97.1 g ∙ mol

1L ( 1000mL ) 1L 3 8 mL−0 mL( =0.03 8 L 1000 mL ) V Final NaOH −V I nitial NaOH

M NaOH = M NaOH =

masssulfuric acid ( 97.1 g ∙mol ) V t

9.7088 g =2.63125373 mol ( 97.1 g ∙mol ) 0.038

H 2 SO 4

Molarity of 1st Trial

initial H 2 SO 4 V Final H SO −V ¿ ¿ 2

¿

4

SO ( 12 HNaOH )

[ NaOH ](V final NaOH −V initial NaOH )

[ H 2 SO4] =

2

¿

H SO ( 21 NaOH ) 1 ( 97.1 g ∙mol ) [ NaOH ]=24 mL−0 mL ( 2 ( 39.997 g ) )

[ NaOH ]=V Final NaOH −V I nitial NaOH

2

4

( 79.994 g )

[ NaOH ]=24 mL 97.1 g ∙ mol

[ NaOH ]=29.13 mL /mol

[ H 2 SO4] =V Final H SO −V I nitial H SO 2

4

2

4

[ H 2 SO4] =20 mL−0 mL=20 mL Molarity [ H 2 SO 4 ]=

2sd Trial

29.13 mL∙ mol =1.4565 mol 20 mL

4

initial H 2 SO 4 V Final H SO −V ¿ ¿ 2

¿

4

SO ( 12 HNaOH )

[ NaOH ](V final NaOH −V initial NaOH )

[ H 2 SO4] =

2

4

¿

[ NaOH ]=V Final NaOH −V I nitial NaOH [ NaOH ]=24 mL−0 mL

(

(

1 H 2 SO 4 2 NaOH

1 ( 97.1 g ∙mol ) 2 ( 39.997 g )

)

)

( 79.994 g )

[ NaOH ]=24 mL 97.1 g ∙ mol

[ NaOH ]=29.13 mL /mol

[ H 2 SO4] =V Final H SO −V I nitial H SO 2

4

2

4

[ H 2 SO4 ] =4 0 mL−2 0 mL=20 mL Molarity [ H 2 SO 4 ]=

29.13 mL∙ mol =1.4565 mol 20 mL

Averge Molarity=1.4565 mol +1.4565 mol =2.913 moles

Discussions:

H 2 SO 4 in 150 mL of unknown solution. 0.0 2 × 98. 07=1.9614 g PO 4 ¿ 2+ 6 H 2 O 2. Balanced reaction 2 H 3 P O 4 + 3 Ba(OH )2 → Ba3 ¿ M 1 V 1=M 2 V 2 3.

1. Grams of

1.00 ×10−2 V 1=25.0 mL 1.5 ×10−2=37.5 mL 4. a. The effect of Sulfamic acid in Part A contained water because someone left the bottle open would be that the sulfamic acid would react with water. b. If the NaOH solution contains a small amount of KOH contaminant, then there wouldn’t be much a difference as both are bases.

c. If the buret was dirty in part A, but we cleaned it for part B then our results for part A wouldn’t be correct because of the contamination. d. If we added 75 mL of water to the sulfamic acid in part A then there wouldn’t be a reaction between the sulfamic acid and water as sulfamic acid is a strong acid.

Conclusion: By standardizing NaOH solution and using it to help us indicate the end point for both the sulfamic acid and the sodium hydroxide to determine its co concentration. References: Harold Goldwhite; Wayne Tikkanen. Acid Base Stoichiometry. Experiments in General Chemistry, Fifth Edition ; Department of Chemistry California State University, Los Angeles, 2018....