eral calculations with the data you obtain in the titrations lab. Below is an ex

Question

eral calculations with the data you obtain in the titrations lab. Below is an exercise to You will perfor helpprep (a) If your titration solution is 0.363 M in NaOH, and the endpoint occurs at 12.11 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 HC2H302 grams HC2H302 (c) How many grams of acetic acid is this? (d) If the mass of analyte is 11.44 grams, what is the mass % of acetic acid in the analyte?

Explanation / Answer

Ans :

a) Number of mmoles = volume of solution in mL x Molarity

= 12.11 x 0.363

= 4.396 mmol NaOH

b) The reaction is given as :

HC2H3O2 + NaOH = CH3COONa + H2O

So equimolar amounts of both the reagents are required in the reaction.

Then the number of mmol = 4.396 mmol HC2H3O2

c) The grams of acetic acid = molar mass x number of moles

= 60.05 x 4.396 x 10-3

= 0.264 grams

d) The mass % of acetic acid in analyte = (0.264 / 11.44) x 100

= 2.31 %

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