of methylene chloride BA + NaOH --> sodium benzoate + H2O BA = 0.0205 M NaOH = 0.0189 M NaOH volume used = 17.4 mL Moles of NaOH: 0.0189 mol/L x 17.4 mL x 1L/1000 mL = 0.00032886 moles of NaOH BA:NaOH = 1:1, therefore there is 0.00032886 moles of BA remaining in aqueous phase 0.00032886 m
Am I calculating this correctly?
Single 10 mL Extraction and Determination of Distribution Coefficient
1) Calculation of initial amount of benzoic acid in grams
Volume of benzoic acid solution: 50.00 mL
Molarity of benzoic acid solution: 0.0205 M
50.00 mL x 1 L/1000 mL L x 0.0205 mol/L x 122.122 g/mol = 0.12517505 g ≈ 0.125 g of benzoic acid
A2. Calculation of amount of benzoic acid (g) remaining in aqueous solution after extraction with 10 mL of methylene chloride
BA + NaOH --> sodium benzoate + H2O
BA = 0.0205 M
NaOH = 0.0189 M
NaOH volume used = 17.4 mL
Moles of NaOH:
0.0189 mol/L x 17.4 mL x 1L/1000 mL = 0.00032886 moles of NaOH
BA:NaOH = 1:1, therefore there is 0.00032886 moles of BA remaining in aqueous phase
0.00032886 moles BA x 122.122 g/mol = 0.04016106 g
0.12517505 g BA (original) - 0.04016106 g BA (remaining) = 0.08158944 g BA in 10 mL of dichloromethane
If everything is done correctly, how would I calculate of the amount of benzoic acid (g) extracted into the methylene chloride layer?
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