Question

a) If your titration solution is 0.409 M in NaOH, and the endpoint occues at 13.00...

a) If your titration solution is 0.409 M in NaOH, and the endpoint occues at 13.00 mL of titrant, how many mmol of NaOH are required to reach the endpoint?

____ mmol NaOH

b) How many mmol of acetic acid (HC2H3O2) are required to react with the NaOH?

____ mmol HC2H3O2

c) How many grams of acetic acid is this?

____ grams HC2H3O2

d) If the mass of analyte is 10.15 grams, what is the mass % of acetic acid in the analyte?

____ %

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Homework Answers

Answer #1

(a) Calculate the number of millimoles present in 13.00 mL of 0.409 M NaOH.

Thus, 5.317 mmol of NaOH are present.

(b) The reaction between acetic acid and sodium acetate is as given below.

Hence, 1 mole of acetic acid will react with 1 mole of NaOH.

Hence, 5.317 mmol of NaOH will react with 5.317 mmol of acetic acid.

(c) Calculate the mass of acetic acid present.

Thus, 0.319 g of acetic acid are present.

(d) Calculate the percentage of acetic acid in the analyte.

%

Hence, the percentage of acetic acid in the analyte is 3.14 %

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