Acid-Base Titration Lab Report
Lab Report: Acid-Base Titration Brianna Morrison
Chemistry 111 October 11, 2010
Aim:
To standardize a solution of the base sodium hydroxide using oxalic acid dihydrate as primary standard acid. Also to determine the amount of sodium hydroxide it takes to titrate a weighted sample of an unknown acid.
Procedure:
As outlined in instructions provided, no changes were made to the procedure.
Data:
Part A: Trial 1: Amount of H C O 2H O: 0.96 g Buret before titration: 3.6 mL Buret after titration: 32.5 mL
Trial 2: Amount of H C O 2H O: 0.095 g Buret before titration: 3.0 mL Buret after titration: 31.9 mL
Part B: Unknown acid # 310 Trial 1: Amount of unknown: 1.001 g Buret before titration: 1.4 mL …show more content…
Buret after titration: 40.4 mL Trial 2: Amount of unknown: 1.001 g Buret before titration: 1.9 mL Buret after titration: 41.1 mL
Observations: When the base was added to the acid, the indicator in the acid turned the solution pink when we reached the end point of titration.
Results:
Calculations:
Part A:
Determine the concentration of NaOH:
Trial 1: .96g H C O 2 H )
Mol H C O 2H O= 126 g/mol = .0076 mol
Mol of NaOH= (.0076 mol H C O 2H O) ( 2NaOH ) = .0152 mol 1 H C O 2H O
M NaOH= 0.0152 mol NaOH = .530M NaOH 0.02870 L NaOH
Trial 2:
Mol H C O 2H O= 0.9g H C O 2 H O = .0075 mol 126 g/mol mol of NaOH= (0.0075mol H C O 2H O) ( 2NaOH ) = 0.0151 mol 1 H C O 2H O
M of NaOH= 0.0151 mol NaOH = 0.522M
NaOH 0.0289 L NaOH
Part B:
Molar Mass of unknown acid
Trial 1:
Mol NaOH= ( .0302 L NaOH) (0.530 M NaOH)= .0160 mol NaOH
Mol unknown= ( 0.0160 mol NaOH) (1 mol unknown)= .0008 mol 2 moles NaOH
molar mass= 1.01 g = 126.3 g .008 mol unknown
Trial 2:
Mol NaOH= (.0303L NaOH)( .0522 M NaOH)= 0.0158 mol NaOH
Mol unknown= (.0158 mol NaOH) ( 1 unknown )= .0079 mol unknown 2 moles NaOH
molar mass= 1.00g = 126.6 g .0079 mol unknown
Conclusion: Molar mass of unknown 310 is 126.5 g
Chemistry 111 October 11, 2010
Aim:
To standardize a solution of the base sodium hydroxide using oxalic acid dihydrate as primary standard acid. Also to determine the amount of sodium hydroxide it takes to titrate a weighted sample of an unknown acid.
Procedure:
As outlined in instructions provided, no changes were made to the procedure.
Data:
Part A: Trial 1: Amount of H C O 2H O: 0.96 g Buret before titration: 3.6 mL Buret after titration: 32.5 mL
Trial 2: Amount of H C O 2H O: 0.095 g Buret before titration: 3.0 mL Buret after titration: 31.9 mL
Part B: Unknown acid # 310 Trial 1: Amount of unknown: 1.001 g Buret before titration: 1.4 mL …show more content…
Buret after titration: 40.4 mL Trial 2: Amount of unknown: 1.001 g Buret before titration: 1.9 mL Buret after titration: 41.1 mL
Observations: When the base was added to the acid, the indicator in the acid turned the solution pink when we reached the end point of titration.
Results:
Calculations:
Part A:
Determine the concentration of NaOH:
Trial 1: .96g H C O 2 H )
Mol H C O 2H O= 126 g/mol = .0076 mol
Mol of NaOH= (.0076 mol H C O 2H O) ( 2NaOH ) = .0152 mol 1 H C O 2H O
M NaOH= 0.0152 mol NaOH = .530M NaOH 0.02870 L NaOH
Trial 2:
Mol H C O 2H O= 0.9g H C O 2 H O = .0075 mol 126 g/mol mol of NaOH= (0.0075mol H C O 2H O) ( 2NaOH ) = 0.0151 mol 1 H C O 2H O
M of NaOH= 0.0151 mol NaOH = 0.522M
NaOH 0.0289 L NaOH
Part B:
Molar Mass of unknown acid
Trial 1:
Mol NaOH= ( .0302 L NaOH) (0.530 M NaOH)= .0160 mol NaOH
Mol unknown= ( 0.0160 mol NaOH) (1 mol unknown)= .0008 mol 2 moles NaOH
molar mass= 1.01 g = 126.3 g .008 mol unknown
Trial 2:
Mol NaOH= (.0303L NaOH)( .0522 M NaOH)= 0.0158 mol NaOH
Mol unknown= (.0158 mol NaOH) ( 1 unknown )= .0079 mol unknown 2 moles NaOH
molar mass= 1.00g = 126.6 g .0079 mol unknown
Conclusion: Molar mass of unknown 310 is 126.5 g