Question

10.00mL 5% acidity vinegar was used titrate with NaOH whose molarity is 0.1030M.

Exp. #	Concentration of NaOH	Initial volume of NaOH	Final volume of NaOH	Volume of NaOH used for titration
1	0.1030M	5.50mL	21.65mL	16.15mL
2	0.1030M	3.60mL	19.00mL	15.40mL
3	0.1030M	19.00mL	35.91mL	16.91mL

The vinegar was diluted to make 100.00mL final volume before trtration. If the molar mass of acetic acid is 60.05g/mole and density is 1.00 g/mL. Calculate the mass percent of acetic acid in the original vinegar solution.

Answer 1

Solution

Given data

10.00 ml 5% vinegar solution titrated with 0.1030 M NaOH

We have three trials in which volume of the NaOH needed for the titration was calculated. So lets first calculate the average of the NaOH solution needed for the titration.

Balanced reaction equation

NaOH + CH3COOH ---- > CH3COONa + H2O

Avg. NaOH used = (16.15 ml + 15.40 ml +16.91 ml )/3 =   16.16 ml NaOH

Now lets calculate the moles of NaOH reacted using its molarity and volume in liter

Moles = molarity * volume in liter

16.16 ml * 1 L / 1000 ml = 0.01616 L

Lets use this value in the above formula

Moles of NaOH = 0.1030 mol per L * 0.01616 L

                             = 0.001664 mol NaOH

Mole ratio of the acetic acid and NaOH is 1 : 1

Therefore moles of acetic acid reacted are same as moles of NaOH used

So moles of acetic acid reacted = 0.001664 mol

Now lets convert moles of the acetic acid to its mass

Formula to calculate the mass using the moles is as follows

Mass = moles * molar mass

Mass of acetic acid = 0.001664 mol * 60.05 g per mol

                                   = 0.0999 g acetic acid

Now lets calculate the mass percent of the acetic acid in vinegar

Volume of vinegar used = 10.00 ml

And density of vinegar = 1.00 g/ml

So the mass of solution = 10.00 g

Now lets calculate the percent by mass of the acetic acid in vinegar

% mass = (mass of acetic acid / mass of vinegar )*100%

            = (0.0999 g / 10.00 g)*100%

           = 0.999 % of acetic acid in vinegar.

So the mass percent of the acetic acid in vinegar = 0.999 %