The concentration of an unknown sample of sulfuric acid was determined by the method used in this experiment. In the first titration, the sodium hydroxide was standardized by titrating 0.1852g of oxalic acid dihydrate (molar mass= 126.07g/mol) with 32.30 mL of sodium hydroxide solution. In the second titration, 10.00 mL of the unknown sulfuric acid solution was titrated with 12.85mL of the sodium hydroxide solution. What was the concentration of the sulfuric acid?
The concentration of an unknown sample of sulfuric acid was determined by the method used in this experiment. In the first titration, the sodium hydroxide was standardized by titrating 0.1852g of oxalic acid dihydrate (molar mass= 126.07g/mol) with 32.30 mL of sodium hydroxide solution. In the second titration, 10.00 mL of the unknown sulfuric acid solution was titrated with 12.85mL of the sodium hydroxide solution. What was the concentration of the sulfuric acid?
Chemistry
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ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
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The concentration of an unknown sample of sulfuric acid was determined by the method used in this experiment. In the first titration, the sodium hydroxide was standardized by titrating 0.1852g of oxalic acid dihydrate (molar mass= 126.07g/mol) with 32.30 mL of sodium hydroxide solution. In the second titration, 10.00 mL of the unknown sulfuric acid solution was titrated with 12.85mL of the sodium hydroxide solution. What was the concentration of the sulfuric acid?

Transcribed Image Text:Experiment 11
Data to be completed and uploaded.
Only one trial!
(There is no part A)
Part B. Standardization of Sodium hydroxide solution
Balanced chemical equation:
Trial 1
Trial 2
Trial 3
Mass of oxalic acid
0.1314g
NA
NA
Initial burette reading
0.40mL
Final burette reading
20.60mL
-0.40 mz
20.20 mL
20.60mL_
Volume NaOH added
· 20.20m
moles of oxalic acid
0.00148m
0.1314 g
90g
moles of NaOH added
molcs
= 0.00145 m
0.00292m
2x 0.00146m=0.00292
Molarity of NaOH
0.14494
o-JoOHーolas
Volume
0.00292 m
20,20ML
molaviN
メ01445m

Transcribed Image Text:Experiment 11
Data to be completed and uploaded.
Only one trial!
(There is no part A)
Part B. Standardization of Sodium hydroxide solution
Balanced chemical equation:
Trial 1
Trial 2
Trial 3
Mass of oxalic acid
0.1314g
NA
NA
Initial burette reading
0.40mL
Final burette reading
20.60mL
-0.40 me
20.20mL
Volume NaOH added
20.60mL_
20.20m
moles of oxalic acid
a00143m
O.1314 g
90g
moles of NaOH added
molcs
= 0.00145 m
0.00292m
2x 0.00146m=0.00292
Molarity of NaOH
o-NaOHーolas
Volume
0.00292 m
20,20ML
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