Question

1. include the titration reactions of naoh and oxalic acid and naoh and acetic acid in the introduction of your lab manual.
2. what is the mean molarity of naoh?
3. what is the percent acetic acid in your vinegar sample?
4. identify your unknown vinegar sample based on its percent acetic acid and calculate the percent error (theoretical percent acetic acid of the potential unknowns are the following: white- 6%, and apple cider- 5%)?
5. compare your vinegar sample with the rest of the class. how different is the percent acetic acid in each of the brands (white, rice and apple cider)?
6. what are the possible sources of error and how would they impact your results? along with any other errors, please address the following questions:
7. will the concentration of solid naoh increase or decrease if the solid is exposed to air for prolonged periods? briefly explain.
8. will the concentration of naoh solution increase or decrease if the solution is exposed to air for prolonged periods? briefly explain.
9. will the concentration of the $\ce{h_{2}c_{2}o_{4}}$ solution increase or decrease if the solution is stored in a clear bottle in a lighted room? briefly explain.
10. a vinegar sample contained 4.60% acetic acid. how many ml of 0.450 m sodium hydroxide would be required to titrate 25.00 ml of the vinegar sample? assume the density of the sample to be 1.00 g/ml.

Explanation:

Question 6 Solution (Chemistry, a subfield of Natural Science)

Step 1: Find moles of acetic acid

The mass of the vinegar sample is \( m = \text{volume} \times \text{density} = 25.00\ \text{mL} \times 1.00\ \text{g/mL} = 25.00\ \text{g} \).
The mass of acetic acid (\( \ce{CH3COOH} \)) is \( 4.60\% \) of \( 25.00\ \text{g} \):
\( m_{\ce{CH3COOH}} = 0.0460 \times 25.00\ \text{g} = 1.15\ \text{g} \).
Molar mass of \( \ce{CH3COOH} \) is \( 60.05\ \text{g/mol} \), so moles of \( \ce{CH3COOH} \):
\( n_{\ce{CH3COOH}} = \frac{1.15\ \text{g}}{60.05\ \text{g/mol}} \approx 0.01915\ \text{mol} \).

Step 2: Stoichiometry of reaction

Reaction: \( \ce{CH3COOH + NaOH -> CH3COONa + H2O} \).
Mole ratio \( \ce{CH3COOH : NaOH} = 1:1 \), so \( n_{\ce{NaOH}} = n_{\ce{CH3COOH}} = 0.01915\ \text{mol} \).

Step 3: Calculate volume of NaOH

Molarity of \( \ce{NaOH} \) is \( 0.450\ \text{M} = \frac{n}{V} \), so \( V = \frac{n}{M} \).
\( V_{\ce{NaOH}} = \frac{0.01915\ \text{mol}}{0.450\ \text{mol/L}} \approx 0.04256\ \text{L} = 42.6\ \text{mL} \).

Answer:

\( \approx 42.6\ \text{mL} \)