Question

I am supposed to find the individual and average weight percents of acetic acid in vinegar and the relative percent error between my average results and the manufacturers listed value.

Data

6ml vinegar @ 5% acetic acid

3.585g vinegar

42.2ml of NaOH @ 0.100M

I'm not sure how to do this at all and want to understand. I know we need to use the molarity equation somewhere.

If anyone can list the steps I need to take to answer the above question that would really help me understand what the question requires.

Any additional explanation would be greatly appreciated.

This is based of a titration experiment

Answer 1

Solution-

Manufacturers listed value Vinegar solution amount= 6 ml , with 5% acetic acid

Vinegar in gm =3.585g

NaOH solution= 42.2 mL , having strength 0.100M

Let's start exercise for estimation of weight percent of vinegar in 3.585 g

Molar mass of acetic acid=60.05 g/mol

100 %acetic acid in 3.585g of vinegar=

= (3.585x5)/100

=0.1792gm of acetic acid

Let see balance reaction between acetic acid and sodium hydroxide

CH₃COOH + NaOH CH3COONa + H₂O

That means one mole of acetic acid required one mole of sodium hydroxide.

Moles of acetic acid(n)= mass of acetic acid (m)/molar mass of acetic acid

= 0.1792/60.05

= 0.002985 g mol

Based on chemical reaction , moles of NaOH required = 0.002985 g mol, but we will see your used NaOH moles by following calculation-

molarity of NaOH = number of moles / per liter of solution

Molarity of NaOH=0.100

mL of NaOH=42.2 mL ~ 42.2/1000 = 0.0422 Liter

So moles of NaOH(n)=Molarity x per liter of solution

=0.100 x 0.0422

= 0.00422 g mol used for titration

By moles of NaOH used are greater than moles of acetic acid, so excess NaOH moles used= 0.00422-0.002985

= 0.001135 g mole

So percent error= (0.001235/0.002985)x100

= +41.37%

Now your acetic acid should be='0.002985x 60.05

=0.1792 g of acetic acid

You should use 0.100'M NaOH=?

= 0.002985 x 40

= 0.1194g of 100% NaOH required but you where using 0.100M NaOH, so amount of NaOH in mL.

Molarity= number of moles(n)/ per liter of solution

Per liter of solution= (0.002985)/0.100

=0.02985 Liter ~ 29.85 mL

So for particular sample you have to use only 29.85 mL of 0.100M NaOH.