Acid Base Titration

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Lab Report: Acid-Base TitrationBrianna Morrison
Chemistry 111October 11, 2010

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.

As outlined in instructions provided, no changes were made to the procedure.

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
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.

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...
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