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Discuss the shortcomings of using old sodium hydroxide solution and how to mitigate them, and how to assess the accuracy of vinegar can labels.

Chemistry

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Answer

b

0.034 M\text{0.034 M}

Step-by-step solution

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Part a

Shortcomings of using old sodium hydroxide solution and mitigation measures

Shortcomings:
Sodium hydroxide (NaOH) is a hygroscopic substance, meaning it readily absorbs moisture from the air, which dilutes the solution. Additionally, it reacts with atmospheric carbon dioxide (CO2\text{CO}_2) to form sodium carbonate (Na2CO3\text{Na}_2\text{CO}_3), a weaker base.

2NaOH(aq)+CO2(g)→Na2CO3(aq)+H2O(l)2\text{NaOH(aq)} + \text{CO}_2\text{(g)} \rightarrow \text{Na}_2\text{CO}_3\text{(aq)} + \text{H}_2\text{O(l)}

These processes mean that the actual concentration of the sodium hydroxide solution stored for over a month will be lower and less accurate than its initial stated concentration, making it an unreliable standard for titration.

Mitigation measures:

  • Standardize the NaOH solution: Before using it to titrate the vinegar, the sodium hydroxide solution should first be standardized against a primary standard (e.g., potassium hydrogen phthalate or oxalic acid) to determine its exact current concentration.
  • Proper storage: Store sodium hydroxide solutions in tightly sealed, airtight containers to minimize exposure to atmospheric moisture and carbon dioxide.

Part b

0.034 M\text{0.034 M}

Assessment of the accuracy of the labels on the vinegar cans

To assess the accuracy, we will calculate the actual concentration of the ethanoic acid (vinegar) using the titration data and compare it to the labeled concentration.

The balanced chemical equation for the reaction between ethanoic acid (CH3COOH\text{CH}_3\text{COOH}) and sodium hydroxide (NaOH\text{NaOH}) is:

CH3COOH(aq)+NaOH(aq)→CH3COONa(aq)+H2O(l)\text{CH}_3\text{COOH(aq)} + \text{NaOH(aq)} \rightarrow \text{CH}_3\text{COONa(aq)} + \text{H}_2\text{O(l)}

The stoichiometry is 1:1.

Given data:
Volume of NaOH\text{NaOH} solution (VbV_b) = 25.0 cm3=0.0250 L25.0 \text{ cm}^3 = 0.0250 \text{ L}
Concentration of NaOH\text{NaOH} solution (CbC_b) = 0.025 M0.025 \text{ M}
Volume of vinegar solution (VaV_a) = 18.60 cm3=0.01860 L18.60 \text{ cm}^3 = 0.01860 \text{ L}
Labeled concentration of vinegar = 0.05 M0.05 \text{ M}

  1. Calculate the moles of NaOH\text{NaOH} used.

    Moles of NaOH=Cb×Vb\text{Moles of NaOH} = C_b \times V_b Moles of NaOH=0.025 M×0.0250 L=0.000625 mol\text{Moles of NaOH} = 0.025 \text{ M} \times 0.0250 \text{ L} = 0.000625 \text{ mol}
  2. Determine the moles of ethanoic acid reacted.

    Since the reaction is 1:1, the moles of ethanoic acid are equal to the moles of NaOH\text{NaOH}.

    Moles of CH3COOH=0.000625 mol\text{Moles of CH}_3\text{COOH} = 0.000625 \text{ mol}
  3. Calculate the actual concentration of the ethanoic acid (vinegar).

    Concentration of CH3COOH(Ca)=Moles of CH3COOHVa\text{Concentration of CH}_3\text{COOH} (C_a) = \frac{\text{Moles of CH}_3\text{COOH}}{V_a} Ca=0.000625 mol0.01860 L=0.03360215... MC_a = \frac{0.000625 \text{ mol}}{0.01860 \text{ L}} = 0.03360215... \text{ M}

    Rounding to two significant figures (limited by the concentration of NaOH\text{NaOH}):

    Ca≈0.034 MC_a \approx \boxed{\text{0.034 M}}
  4. Assess the accuracy of the label.

    The calculated concentration of the vinegar is 0.034 M0.034 \text{ M}.
    The labeled concentration is 0.05 M0.05 \text{ M}.

    To assess accuracy, we can calculate the percentage difference:

    Percentage Difference=∣Calculated Concentration−Labeled Concentration∣Labeled Concentration×100%\text{Percentage Difference} = \frac{|\text{Calculated Concentration} - \text{Labeled Concentration}|}{\text{Labeled Concentration}} \times 100\% Percentage Difference=∣0.034 M−0.05 M∣0.05 M×100%\text{Percentage Difference} = \frac{|0.034 \text{ M} - 0.05 \text{ M}|}{0.05 \text{ M}} \times 100\% Percentage Difference=0.016 M0.05 M×100%=0.32×100%=32%\text{Percentage Difference} = \frac{0.016 \text{ M}}{0.05 \text{ M}} \times 100\% = 0.32 \times 100\% = 32\%

    The calculated concentration (0.034 M0.034 \text{ M}) is significantly lower than the labeled concentration (0.05 M0.05 \text{ M}), with a difference of 32%32\%. Therefore, the labels on the vinegar cans are inaccurate.

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