Gravimetric and Volumetric Analysis PDF

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Siddhant Nayak Prof. Feudale Chemistry I 4 April 2020 Gravimetric and Volumetric Analysis ● Abstract:- Two methods, gravimetric and volumetric analyses, were performed to determine the concentration of

H 2 S O4

in two solutions that were prepared by

dilution. In gravimetric analysis, a precipitation reaction is used to separate the ions from the solution. For the first method, gravimetric analysis, 10 mL of 6M of

H 2 S O 4 was

pipetted into a 100 mL flask and diluted with distilled water, referred to as the stock solution. In a second 100 mL flask, 25 mL of the stock solution was diluted with distilled water, referred to as the final solution. Separately, in two beakers 5 mL of

BaC l 2 and

5mL of the final solution with 5 mL of HCl was heated for approximately 10 minutes. Once the two solutions were heated to satisfactory, the 5 mL of

BaC l 2

was mixed

with the stock and HCl solution. After the mixture was slightly cooled, it was filtered through filter paper and a funnel while flushing the precipitate with distilled water. The precipitate was then placed and dried in an oven. The mass of the precipitate is used to determine the number of ionic compounds in the original solutions. Two trials were performed for an average Molarity of the original solution of 6.174 M. The second method, volumetric analysis, was used to determine the concentration of a solution through titration. In this experiment, a solution of an unknown concentration, final solution, is added to another solution with a known concentration, NaOH. By identifying the volume of NaOH, the concentration of the final solution can be

determined since the concentration of NaOH is known. This experiment was performed by adding 5mL of the final solution mixed with 3 drops of the indicator, phenolphthalein, a metal rod and enough distilled water to cover the metal rod to a flask. While the final solution is stirred, 0.100 M of NaOH in a buret was titrated until the endpoint was achieved. The endpoint is identified when the mixed solution remains pink for about 30 seconds. After the titration, the molarity of the final solution can be calculated from the measured data. This experiment was performed four times to ensure precision. The average Molarity of the original solution is 5.685 M. ● Pre-lab:○ What is the principal purpose of this experiment? ■ The purpose of this experiment is to employ both methods of analysis to determine the concentration of

H 2 S O 4 in solutions that have been

prepared by dilution from a sulfuric acid solution of definite, but unknown, concentration. ○ Write a balanced equation for the reaction used in the gravimetric analysis part of this experiment. ■

H 2 S O 4 (aq) + BaC l2 (aq) →

BaS O 4 (s) + 2 HCl

(aq)

○ What must be done with the solid barium sulfate waste? ■ Put precipitate and filter paper into a waste bottle labeled for solids. ○ 10.00 mL of approximately 6 M sulphuric acid is transferred to a 100-mL volumetric flask and diluted to the mark with distilled water and mixed. Then 10.00 mL of this solution is diluted further to 100 mL. What is the molarity of this last solution?

■ Dilution formula = M1*V1 = M2*V2 M1 = 6 M; M2 = unknown; V1 = 0.010 L; V2 = 0.10 L 6*(0.010) = M2*(0.10) M2 = 6*(0.010) / (0.10) M2 = 0.6 M ○ 10.00 mL of the final acid solution is reacted with excess barium chloride to produce a precipitate of barium sulfate (FW = 233.4 g/mol). The dry solid weighs 0.397g. Use this mass and the dilution volumes to calculate the actual molarity of the sulfuric acid in the initial solution. ■

H 2 S O 4 (aq) + BaC l2 (aq) →

■ 0.397g

BaS O 4 *

BaS O 4 (s) + 2 HCl

1 mol BaS O 4 * 233.4 g BaS O 4

(aq)

1 mol H 2 S O 4 = 0.00170 mol 1 mol BaS O 4

H 2 S O4 .

■ M=

mole = volume

0.00170mol = 0.170 M 0.010 L

■ Dilution formula = M1*V1 = M2*V2 M1 = unknown; M2 = 0.170M; V1 = 0.010 L; V2 = 0.10 L M1*(0.010) = 0.170*(0.10) M1 = 0.170*(0.10) / (0.010) M1 = 1.7 M ■ We plug in the molarity found above in the same equation. We find the molarity of the original solution to be 17M. Dilution formula = M1*V1 = M2*V2 M1 = unknown; M2 = 1.7M; V1 = 0.010 L; V2 = 0.10 L

M1*(0.010) = 1.7*(0.10) M1 = 1.7*(0.10) / (0.010) M1 = 17 M ● Data Collected:Table 1: Gravimetric Determination of Sulfate Ion Trial #1

Sample Calc

Mass of BaS O 4 and filter paper (g)

1.2399

N/A

Mass of the filter paper (g)

1.0495

N/A

BaS O 4 precipitate (g)

0.1904

1.3115- 1.1319

Millimoles of BaS O 4 (Mmol)

0.8158

(0.1904/233.392)*1000

0.8158

N/A

5

N/A

H 2 S O 4 , final solution (M)

0.1632

0.8158 / 5

H 2 S O 4 , stock solution (M)

0.6528

0.1632 * 4

H 2 S O 4 , original solution (M)

6.528

0.6528 * 10

Mass of

Millimoles of H 2 S O 4 reacted (Mmol)

that

Volume of final solution reacted, (mL)

Table 2: Volumetric Determination of Acid Concentration Trial #1

Trial #2

Trial #3

Trial #4

Sample Calc

Initial NaOH buret reading, (mL)

0.100

15.21

30.39

31.30

N/A

Final NaOH buret reading, (mL)

15.21

30.39

45.51

42.81

N/A

Volume of NaOH (aq) used, (mL)

15.11

15.18

15.12

15.25

15.21-0.100

Millimoles of NaOH(Mmol)

1.511

1.518

1.512

1.525

0.100 * 15.11

Millimoles of H 2 S O4 titrated

0.7076

0.7095

0.7150

0.7120

0.1412 * 5

Volume of final solution titrated, mL

5

5

5

5

N/A

H 2 S O4 , final solution (M)

0.1511

0.1518

0.1512

0.1525

1.511 * 0.100

H 2 S O4 , stock solution (M)

0.6044

0.6072

0.6048

0.6100

0.1511 * 4

H 2 S O4 , original solution (M)

6.044

6.072

6.048

6.100

0.6044 * 10

Average H 2 S O4 , original solution, (M)

6.066

● Post lab:○ Write the balanced molecular equation that occurs for the complete reaction of sulfuric acid with sodium hydroxide. ■

H 2 S O 4 (aq) + 2NaOH

→

N a2 S O4 (aq) + 2 H 2 O

○ To determine the molarity of your final sulfuric acid solution, you titrate the final solution with a standardized sodium hydroxide solution to the second equivalence point. If 26.32 mL of 0.1000 M sodium hydroxide was required for complete reaction with 10.00 mL of the final sulfuric acid solution, determine the molarity of the final sulfuric acid solution precisely to four significant figures. ■

0.1000mol NaOH 1 L NaOH

*

0.02632

1 0.010 L H 2 S O 4

L

NaOH

*

1 mol H 2 S O 4 * 2mol NaOH

= 0.1316 M of H 2 S O 4

○ If the same dilution factors used in the lab were also used to prepare this final sulfuric acid solution, determine the molarity of both the stock solution and the initial solution of sulfuric acid. ■ Using the molarity found above. We find the molarity of the original and stock solution to be. Final Solution- 0.1316 M (¼ of Stock Solution) Stock Solution- 0.5264 M (1/10 of Original Solution) 0.1316= ¼ * X, X= Stock Solution Original Solution- 5.264 M

0.5264= 1/10 * X, X= Original Solution ○ A 0.158–gram sample of salt, comprised of barium and one of the halide ions (BaX2), was dissolved in water, and an excess of sulfuric acid was added to form barium sulfate, BaSO4. After filtering and drying, the solid BaSO4 weighed 0.124 grams. Determine the identity of the halide ion (X) and the formula of the barium halide salt. You must have the correct supporting work to receive full credit for this problem. ■

Ba X 2 (s) + 1 mol of

H 2 S O 4 (aq) → BaS O 4 (s) + 2HX (aq)

Ba X 2

------------- 1 mol of BaS O 4 Ba X 2 ------------ 233.3 g of

BaS O 4

Ba X 2 --------------------- 0.124 g of

BaS O 4

(137.3 + 2X) g of 0.158 g of

■ By Cross multiplication, we get:0.124 * (137.3 + 2X) = 0.158 * 233.3 17.0252 + 0.248X = 36.8614 0.248X = 19.8363 X = 19.8363 / 0.248 X ≈ 80 g ■ By looking at the periodic table, the element that has a molar mass closest to 80 g/mol in group 7 is Bromine (Br). So the salt has formula

BaB r 2 ....