Question

A sample of acetic acid (weak acid) was neutralized with .05M NaOH solution by titration. 34 mL of NaOH had been used. Show your work.

CH3COOH + NaOH-----CH3COONa + H2O

a) Calculate how many moles of NaOH were used?

b) How many moles of aspirin were in a sample?

c) Calculate how many grams of acetic acid were in the sample

d) When acetic acid is titrated with NaOH solution what is the pH at the equivalence point? Circle the right answer.

pH>7

pH=7

pH<7

Answer 1

a) Moles of NaOH used =

Molarity = Moles of NaCL / Liter of solution

therefore ,

Moles of NaCl = Molarity x liter of solution

= 0.05 x 0.034 ( 34 ml = 0.034 liter )

= 0.0017 Moles of NaCl

b) Moles of Acetic acid =

In the reaction the ratio of Acetic acid to NaCl is 1;1

therefore moles of acetic acid is also 0.0017

c ) Grams of acetic acid =

We know that No. of moles = Grams / Mol. Weight

therefore ,

Grams = Mol. Weight x No. of moles

= 60.05 x 0.0017

= 0.102 grams of acetic acid.

d ) At the equivalence point the pH is greater then 7 because all of the acetic acid has been converted to its conjugate base CH3COO- by the addition of NaOH and now the equilibrium moves backwards towards Acetic acid and produces hydroxide, Therefore,

pH > 7.