Titration part 1- standardization of NaOH #1) How much 6M NaOH is required to make 500 mL of 0.1 M NaOH? How much DI water is required? Show all your calculations. # 2) Calculate the number of grams of oxalic acid dihydrate (M.W. = 126.1 g/mol) necessary to react with 35 mL of the 0.1 M NaOH solution. Show all your calculations. Note: 35 mL is the approximate volume of NaOH solution that needs to be used in the titration in order to obtain the precision desired in this experiment.

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Chapter9: Acids, Bases, And Salts
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Titration part 1- standardization of NaOH
#1) How much 6M NaOH is required to make 500 mL of 0.1 M NaOH? How much DI water is
required? Show all your calculations.
# 2) Calculate the number of grams of oxalic acid dihydrate (M.W. = 126.1 g/mol) necessary to
react with 35 mL of the 0.1 M NaOH solution. Show all your calculations. Note: 35 mL is the
approximate volume of NaOH solution that needs to be used in the titration in order to obtain the
precision desired in this experiment.
Transcribed Image Text:Titration part 1- standardization of NaOH #1) How much 6M NaOH is required to make 500 mL of 0.1 M NaOH? How much DI water is required? Show all your calculations. # 2) Calculate the number of grams of oxalic acid dihydrate (M.W. = 126.1 g/mol) necessary to react with 35 mL of the 0.1 M NaOH solution. Show all your calculations. Note: 35 mL is the approximate volume of NaOH solution that needs to be used in the titration in order to obtain the precision desired in this experiment.
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