3) Calculate average volume of NaOH used in titrations of acetic acid and corres

Question

3) Calculate average volume of NaOH used in titrations of acetic acid and corresponding concentration of vinegar sample (before dilution) both in terms of molarity and %(V/V).

a) Plot the titration curve(s) for the titration of acetic acid, determine the equivalence point and concentration of vinegar.

Data: Obtain a sample of vinegar. Pipet 25mL of this sample into a clean 100mL volumetric flask and fill it with deionized water to the mark. Then, pipet 10 mL of this diluted acetic acid solution into each of the 3 titration flasks and add 90mL of deionized water.

Titration 1: 10.20 mL of NaOH

Titration 2: 10.19 mL of NaOH

Titration 3: 9.99 mL of NaOH

Explanation / Answer

3. a) average volume of NaOH= (10.20+10.10.19+9.99)l3

=10.12ml

concentration of vinegar

mass of NaOH = density*vol

=2.13* 10.12

=21.55g

moles = 21.55/40= 0.53

NaOH+ CH3COOH ---> CH3COONa + H2O

so, the stochiometry of the reaction is 1:1

so o.53 moles of NaOH for 0.53 moles of vinegar

so vinegar was 0.53 moles in the original solution

mass= moles* Molar mass

= 0.53 * 60

=31.8g

in volume = 31.8/1.05= 30.3ml

b) i cant do this for u since i will not be able to upload the graph. sorry!

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