Question

Prelab Activity:

Titrations Continued – Titration of Household Products

1. A 2.40 g sample of vinegar was added to an Erlenmeyer flask along with 100 mL of deionized water and 3 drops of phenolphthalein indicator. It took 22.15 mL of 0.0981 M NaOH (aq) to reach the faint pink endpoint. The following balanced chemical equation represents the chemistry of the titration:

NaOH (aq) + CH₃COOH (aq) à CH₃COO^–Na⁺ (aq) + H₂O (l)

1. Calculate the mass % of acetic acid (CH₃COOH) in the vinegar sample.
2. Two subsequent trials found the mass % of acetic acid in vinegar to be 5.54% and 5.53%, respectively. Using the mass % found in part a, along with the values for subsequent trials, report an average value and standard deviation of mass % for the three trials. Report answer in the format: AVG ± S.D.

2.4.0 mL of 3.00 M HCl reacts completely with the CaCO₃ (active ingredient) in an antacid tablet. The remaining solution (excess HCl) is then titrated with 0.0981 M NaOH to the bromothymol blue endpoint.

1. Write the balanced chemical equation for the reaction of CaCO₃ (s) with HCl (aq).

2. If the volume of NaOH delivered at the endpoint is 22.20 mL, determine the amount (in milligrams) of CaCO₃ that was contained in the antacid tablet.

Answer 1

1-

The given neutralization reaction is-

NaOH (aq) + CH3COOH (aq) ---------> CH3COO–Na+ (aq) + H2O (l)

That means 1 mole of NaOH is nedded to neutralize 1 mole of CH3COOH

Now given

volume of NaOH used for the reaction = 22.15 mL

Concentration of NaOH used for the reaction = 0.0981 M

That means moles of NaOH used for the reaction = concentration * volume

= 0.0981 M * 22.15 mL

= 0.0981 mole / 1000 mL * 22.15 mL

= 0.00217 moles

That means moles of CH3COOH neutralized = 0.00217 moles

So we can say in the taken vinegar sample, we have 0.00217 moles of CH3COOH

So mass of CH3COOH present = moles * molar mass of CH3COOH

= 0.00217 moles * 60 g/mole

= 0.13 g

So mass % of acetic acid (CH3COOH) in the vinegar sample

= (mass of acetic acid present / mass of vinegar sample) * 100

= (0.13 g / 2.40 g) * 100

= 5.43 %

b-

Now given in subsequent 2 trials, % of acetic acid (CH3COOH) found = 5.54% and 5.53%

So average % of acetic acid (CH3COOH) found = (5.54% + 5.53% + 5.43 % ) / 3

= 5.5 %

Now we can calcuclate standard deviation as-

i- Caluclate the mean = 5.5

ii- substract the mean from each value i.e

5.54 - 5.5 = 0.04

5.53 - 5.5 = 0.03

5.43 - 5.5 = -0.07

iii- square each of the deviation values and add them i.e

(0.04)2 +  (0.03)2 +  (-0.07)2 = 0.0074‬

iv- substract 1 from total number of data sets. We have total number of data sets = 3 (i.e 5.54, 5.43)

So here 3-1 = 2

v- devide the sum of squares by this value found in step iv i,e

0.0074‬ / 2 = 0.0037‬

vi- Tkae the square root of the numebr i.e

sq. rt (0.0037) = 0.061

This is the standard deviation.

So the mass % of acetic acid (CH3COOH) found = Avg ± SD

= 5.5 % ± 0.061