Question

Vinegar is a dilute solution ethanoic acid in water (see the question above for the structure of ethanoic acid). In order to test the strength of a vinegar, a 10.00 gram sample was titrated with sodium hydroxide (0.1001 mole/kg of solution). The mass of the. burette before the titration is 131.44 g and upon reaching the endpoint the burette weighted 48.11 g. How many moles of acetic acid are in the sample?

Answer 1

Answer:

Explanation:

Below is the equation of the chemical reaction:

CH3COOH (aq) + NaOH (aq) ----------> CH3COONa (aq) + H2O (l)

The mole ratio of acid and base in the balanced equation is:

Ma = 1, Mb = 1

Cb = 0.1001 mole/Kg

Ca = ???

Va = 10 grams

Vb = 131.44 - 48.11 = 83.33 grams

From the acid-base mole ration formula

$\frac{Ma}{Mb} = \frac{Ca x Va }{Cb * Vb}$

$\frac{1}{1} = \frac{Ca x 10 }{0.1001 * 83.33}$

$Ca = \frac{0.1001 x 83.33}{10.00} = 0.8341333 mole/kg$

The number of moles of acetic acid in the sample = $0.8341333 * \frac{10}{1000} = 0.008341333 moles$

Therefore 0.008341333 moles of acetic acid are in the sample.