Using the average molarity of the NaOH solution determined from your results in part A

Using the average molarity of the NaOH solution determined from your results in part A, calculate the molarity and percent by mass of the acetic acid in your vinegar sample and report their average values. H2C2O4•2H2O(s) + 2NaOH (aq)= Na2C2O4 + 4H2O(l)  part A Mass of oxalic acid analyzed (g)- trial 1- 0.302  trial 2- 0.305 initial buret reading of NaOH (mL)- trial 1-0.31  trial 2- 23.42 final buret reading of NaOH (mL) trial 1- 23.42  trial 2 -46.73 Part B volume of vinegar analyzed (mL), trial 1- 5.00 , trial 2- 5.00 initial buret reading of NaOH (mL), trial 1 -0.00,  trial 2 – 22.73 final buret reading of NaOH (mL) trial 1- 22.73, trial 2- 45.50  There is no molarity of oxalic acid that was used during the experiment givenÂÂ
Using the average molarity of the NaOH solution determined from your results in part A, calculate the molarity and percent by mass of the acetic acid in your vinegar sample and report their average values.ÂÂ

H2C2O4•2H2O(s) + 2NaOH (aq)= Na2C2O4 + 4H2O(l)

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part AÂÂ

Mass of oxalic acid analyzed (g)- trial 1- 0.302  trial 2- 0.305

initial buret reading of NaOH (mL)- trial 1-0.31  trial 2- 23.42

final buret reading of NaOH (mL) trial 1- 23.42  trial 2 -46.73

Part B

volume of vinegar analyzed (mL), trial 1- 5.00 , trial 2- 5.00ÂÂ

initial buret reading of NaOH (mL), trial 1 -0.00,  trial 2 – 22.73

final buret reading of NaOH (mL) trial 1- 22.73, trial 2- 45.50ÂÂ

 There is no
molarity of oxalic acid that was used during the experiment givenÂÂ

The Correct Answer and Explanation is :

To calculate the molarity and percent by mass of acetic acid in the vinegar sample, we need to perform the following steps based on the information provided and the stoichiometric relationships from the balanced equation.

Part A: Determine the Molarity of NaOH

We can use the data from Part A to determine the molarity of NaOH. First, we need to calculate the volume of NaOH used in each trial:

Trial 1:

• Initial buret reading = 0.31 mL
• Final buret reading = 23.42 mL
• Volume of NaOH used = ( 23.42 \, \text{mL} – 0.31 \, \text{mL} = 23.11 \, \text{mL} )

Trial 2:

• Initial buret reading = 23.42 mL
• Final buret reading = 46.73 mL
• Volume of NaOH used = ( 46.73 \, \text{mL} – 23.42 \, \text{mL} = 23.31 \, \text{mL} )

Now, calculate the moles of oxalic acid used in each trial using the molar mass of oxalic acid dihydrate (H₂C₂O₄•2H₂O), which is approximately 126.07 g/mol.

Trial 1:

• Mass of oxalic acid = 0.302 g
• Moles of oxalic acid = ( \frac{0.302 \, \text{g}}{126.07 \, \text{g/mol}} = 0.00239 \, \text{mol} )

Trial 2:

• Mass of oxalic acid = 0.305 g
• Moles of oxalic acid = ( \frac{0.305 \, \text{g}}{126.07 \, \text{g/mol}} = 0.00242 \, \text{mol} )

The balanced equation is:

[
\text{H}_2\text{C}_2\text{O}_4 \cdot 2\text{H}_2\text{O} + 2\text{NaOH} \rightarrow \text{Na}_2\text{C}_2\text{O}_4 + 4\text{H}_2\text{O}
]

This indicates that 1 mole of oxalic acid reacts with 2 moles of NaOH. Therefore, the moles of NaOH required are:

Trial 1:

• Moles of NaOH = ( 2 \times 0.00239 \, \text{mol} = 0.00478 \, \text{mol} )

Trial 2:

• Moles of NaOH = ( 2 \times 0.00242 \, \text{mol} = 0.00484 \, \text{mol} )

Next, we can calculate the molarity of NaOH. The molarity ( M ) is given by:

[
M = \frac{\text{moles of NaOH}}{\text{volume of NaOH in L}}
]

Trial 1:

• Volume of NaOH = ( 23.11 \, \text{mL} = 0.02311 \, \text{L} )
• Molarity of NaOH = ( \frac{0.00478 \, \text{mol}}{0.02311 \, \text{L}} = 0.206 \, \text{M} )

Trial 2:

• Volume of NaOH = ( 23.31 \, \text{mL} = 0.02331 \, \text{L} )
• Molarity of NaOH = ( \frac{0.00484 \, \text{mol}}{0.02331 \, \text{L}} = 0.208 \, \text{M} )

The average molarity of NaOH is:

[
\text{Average molarity of NaOH} = \frac{0.206 + 0.208}{2} = 0.207 \, \text{M}
]

Part B: Calculate the Molarity and Percent by Mass of Acetic Acid in Vinegar

Now, we can use the average molarity of NaOH to calculate the molarity and percent by mass of acetic acid in the vinegar sample.

Trial 1:

• Volume of vinegar = 5.00 mL
• Volume of NaOH used = ( 22.73 \, \text{mL} – 0.00 \, \text{mL} = 22.73 \, \text{mL} = 0.02273 \, \text{L} )
• Moles of NaOH used = ( 0.207 \, \text{M} \times 0.02273 \, \text{L} = 0.00471 \, \text{mol} )

Since the balanced equation for acetic acid with NaOH is:

[
\text{CH}_3\text{COOH} + \text{NaOH} \rightarrow \text{CH}_3\text{COONa} + \text{H}_2\text{O}
]

This indicates a 1:1 molar ratio, so the moles of acetic acid are also 0.00471 mol.

Trial 2:

• Volume of vinegar = 5.00 mL
• Volume of NaOH used = ( 45.50 \, \text{mL} – 22.73 \, \text{mL} = 22.77 \, \text{mL} = 0.02277 \, \text{L} )
• Moles of NaOH used = ( 0.207 \, \text{M} \times 0.02277 \, \text{L} = 0.00471 \, \text{mol} )

Thus, the moles of acetic acid in each trial are 0.00471 mol. Now, we calculate the molarity of acetic acid:

Molarity of acetic acid:
[
\text{Molarity} = \frac{\text{moles of acetic acid}}{\text{volume of vinegar in L}} = \frac{0.00471 \, \text{mol}}{0.005 \, \text{L}} = 0.942 \, \text{M}
]

Now, to find the percent by mass of acetic acid:

1. The molar mass of acetic acid (CH₃COOH) is approximately 60.05 g/mol.
2. Mass of acetic acid = ( 0.00471 \, \text{mol} \times 60.05 \, \text{g/mol} = 0.283 \, \text{g} )
3. Percent by mass = ( \frac{0.283 \, \text{g}}{5.00 \, \text{g}} \times 100 = 5.66\% )

Conclusion:

• The average molarity of NaOH = 0.207 M
• The average molarity of acetic acid in vinegar = 0.942 M
• The average percent by mass of acetic acid in vinegar = 5.66%

This calculation assumes that the chemical reactions go to completion and that the NaOH solution was accurately prepared.

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