During the spoiling process, fruit juices often oxidize to create vinegar. A sample of a beverage with a mass of 2.578 grams reguires 18.62 ml of .095M NaOH for neutralization. What is the mass percent of acetic acid present, assuming that no other acids are present? Show the balanced equation in your calculations.

Sol: CH3COOH + NaOH <---> NaCH3COO +
H20

Convert: 18.62 mL = .01862 L

moles acetic acid: .095M NaOH * .01862L = .00177 moles because it is 1:1 ration.
.00177 moles * (60.05 g acetic acid/1 mol acetic aci) = 0.11 g acetic acid present.

The mass percent of acetic acid present, assuming that no other acids are present is 0.11 g. Is this correct? I have doubts because I imagine I must put grams in percent by multiplying by 100; also, I did nothing whatsoever wiht the given information of 2.578 grams required.

During the spoiling process, fruit juices often oxidize to create vinegar. A sample of a beverage with a mass of 2.578 grams reguires 18.62 ml of .095M NaOH for neutralization. What is the mass percent of acetic acid present, assuming that no other acids are present? Show the balanced equation in your calculations.

Sol: CH3COOH + NaOH <---> NaCH3COO +
H20
This step is OK.
Convert: 18.62 mL = .01862 L

moles acetic acid: .095M NaOH * .01862L = .00177 moles because it is 1:1 ration.
.00177 moles * (60.05 g acetic acid/1 mol acetic aci) = 0.11 g acetic acid present. OK to here EXCEPT that since I assume the molarity is really 0.0950 M (and you just omitted the last zero but I may be wrong), then you have three figures and the 0.11 should be carried out to 0.106 g. From here you did not convert grams to percent correctly. That is
%CH3COOH= ([g CH3COOH]/g sample) x 100 or % = (0.106/2.578) x 100 = ??


The mass percent of acetic acid present, assuming that no other acids are present is 0.11 g. Is this correct? I have doubts because I imagine I must put grams in percent by multiplying by 100; also, I did nothing whatsoever wiht the given information of 2.578 grams required.

One other point is that since the mL is to four places and the mass of the sample is to four places, I can't understand why the molarity is to only two places. If it really is 0.09500 then you should have four places in the grams CH3COOH, which I believe would be 0.1062. BUT if the 0.095 is correct, then your 0.11 is proper. I hope this helps.

if i have 100% acetic acid bottel how i can prepare 1ml 10mM of it??

I do believe this is the right annswer (.11g Acetic Acid present). When you you use the above equation for the mass percent of acetic acid (g CH3COOH/g sample) you end up with 4.3% Acetic Acid. That is a normal concentration of acetic acid [I looked at other sites for concentrations and in fruits acetic acid conc usually varies from 3 to 5.5%]

u fail

To prepare 1 mL of a 10 mM solution of acetic acid from a bottle of 100% acetic acid, you will need to dilute the concentrated acetic acid with a suitable solvent, such as water. Here's how you can do it:

1. Calculate the amount of acetic acid needed using the formula:
Mass (g) = Concentration (M) * Volume (L) * Molecular weight (g/mol)

Since you want a 10 mM solution, the concentration is 0.01 M. As you only need 1 mL, the volume is 0.001 L. The molecular weight of acetic acid is approximately 60.05 g/mol.

Mass (g) = 0.01 M * 0.001 L * 60.05 g/mol = 0.0006 g

2. Measure out 0.0006 g (or 0.6 mg) of 100% acetic acid using a precise balance.

3. Transfer the measured amount of acetic acid to a container, such as a small volumetric flask or a beaker.

4. Add a suitable amount of water to make up to 1 mL of total volume. You can use a graduated cylinder or a pipette to add the water.

5. Gently mix or swirl the solution to ensure thorough mixing of the acetic acid and water.

After following these steps, you will have a 1 mL solution containing approximately 10 mM of acetic acid.

To prepare 1 mL of a 10 mM solution of acetic acid from a 100% acetic acid bottle, you will need to dilute the concentrated acetic acid with a suitable solvent (usually water) to achieve the desired concentration.

Here's the step-by-step process:

1. Calculate the amount of acetic acid needed:
- 10 mM means 10 millimoles per liter (10 mmol/L).
- Since you want to prepare a 1 mL solution, you need 10 millimoles of acetic acid in that volume.
- 10 mmol/L * 0.001 L = 0.01 millimoles of acetic acid.

2. Convert the millimoles of acetic acid to grams:
- The molar mass of acetic acid is approximately 60.05 g/mol.
- 0.01 mmol * (1 mol / 1000 mmol) * 60.05 g/mol = 0.006005 grams (or approximately 6.005 mg) of acetic acid.

3. Add the calculated amount of acetic acid to a 1 mL volumetric flask or any small container.
- Be cautious when handling concentrated acetic acid as it can be corrosive. Follow safety precautions and use appropriate protective equipment.

4. Dilute the acetic acid with water up to the 1 mL mark on the volumetric flask or container.
- It is important to accurately measure the volume to ensure the desired concentration.

5. Mix the solution thoroughly to ensure proper distribution of the acetic acid.

Now, you have successfully prepared 1 mL of a 10 mM solution of acetic acid. Remember to label the container properly and handle the solution with care.