Answer to Question #147514 in General Chemistry for Gusty

Question

2. A 22.0 mL volume of 0.095 M NaOH is required to reach the phenolphthalein endpoint in the titration of a 3.05 g sample of vinegar.

a. Calculate the number of moles of acetic acid in the vinegar sample.

b. Calculate the mass of acetic acid in the vinegar sample.

c. Calculate the percent by mass of acetic acid in the vinegar sample. Assume the density of the vinegar is 1.0 g/mL

Answer 1

Q147514

2. A 22.0 mL volume of 0.095 M NaOH is required to reach the phenolphthalein endpoint in the titration of a 3.05 g sample of vinegar.

a. Calculate the number of moles of acetic acid in the vinegar sample.

b. Calculate the mass of acetic acid in the vinegar sample.

c. Calculate the percent by mass of acetic acid in the vinegar sample. Assume the density of the vinegar is 1.0 g/mL.

Solution:

Our first step will be to write the reaction of NaOH and acetic acid.

The chemical formula of acetic acid is CH₃COOH.

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

H+ of acid and OH^- from base combines to form H₂O (l) .

a) Calculate the number of moles of acetic acid in the vinegar sample.

We are given the volume and concentration of NaOH required for the neutralization of acetic acid.

Convert 22.0 mL to ‘L’ and then plug in the volume and molarity of NaOH in the formula of molarity and find the moles of NaOH.

22.0 mL in ‘L’ = 22.0 mL * 1L/1000mL = 0.022 L  

Molarity = moles of NaOH / volume of NaOH in ‘L’  ;

0.095M = moles of NaOH /0.022 L ;

multiply both the side by 0.022L, we have  

0.095mol/L * 0.022L = moles of NaOH/0.022L * 0.022L ; (M = mol/L)  

0.00209mol = moles of NaOH;  

The mole to mole ratio of CH₃COOH and NaOH in the reaction is 1:1. So the mole of CH₃COOH will also be equal to 0.00209 moles .

Moles of CH₃COOH = 0.0209 mol NaOH * 1mol CH₃COOH / 1mol NaOH

= 0.00209 mol CH₃COOH.

b. Calculate the mass of acetic acid in the vinegar sample.

We will convert the moles of acetic acid obtained in Part A to grams using molar mass of acetic acid.

Molar mass of CH₃COOH = 2 * atomic mass of C + 2 * atomic mass of + 4 * atomic mass of H

= 2 * 12.011g/mol + 2* 15.999 g/mol + 4 * 1.0079 g/mol

= 60.0516 g/mol

using molar mass 60.0516g/mol, convert 0.0209 mol CH₃COOH to grams.

Grams of CH₃COOH = 0.00209 mol CH₃COOH * 60.0516 g CH₃COOH / 1 mol CH₃COOH

= 0.1255 grams of CH₃COOH.

Hence the mass of acetic acid in the vinegar sample is 0.1255 grams.  

In question the quantity with the least numbe of significant figure is 0.095M. It is in 2 significant figure.

So in correct significant figure the answer is 0.13 grams.

c. Calculate the percent by mass of acetic acid in the vinegar sample. Assume the density of the vinegar is 1.0 g/mL.

So, now we know that the mass of acetic acid in the given sample of vinegar is 1.255grams

and the mass of vinegar is 3.05grams.

Mass % = mass of acetic acid / mass of vinegar sample * 100

= 0.1255g/ 3.05 g * 100 = 4.12 %

which in correct significant figure is 4.1 %

Hence

a) number of moles of acetic acid in the vinegar sample = 0.0021 mol.

b) mass of acetic acid in the vinegar sample = 0.13 grams

c) percent by mass of acetic acid in the vinegar sample = 4.1 %