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?

Principles of Modern Chemistry
8th Edition
ISBN:9781305079113
Author:David W. Oxtoby, H. Pat Gillis, Laurie J. Butler
Publisher:David W. Oxtoby, H. Pat Gillis, Laurie J. Butler
Chapter11: Solutions
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Problem 25P
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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?
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 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
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
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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