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Ethanoic acid (CH₃COOH) is a weak acid in water - VCE - SSCE Chemistry - Question 7 - 2005 - Paper 1

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Ethanoic acid (CH₃COOH) is a weak acid in water. a. Write an equation showing the ionisation of ethanoic acid in water. ___________________________________________... show full transcript

Worked Solution & Example Answer:Ethanoic acid (CH₃COOH) is a weak acid in water - VCE - SSCE Chemistry - Question 7 - 2005 - Paper 1

Step 1

a. Write an equation showing the ionisation of ethanoic acid in water.

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Answer

The ionisation of ethanoic acid ( CH₃COOH) in water can be represented by the following equilibrium equation:

CH3COOH(aq)CH3COO(aq)+H+(aq)CH₃COOH(aq) \rightleftharpoons CH₃COO^-(aq) + H^+(aq)

Step 2

b.i. Calculate the hydrogen ion concentration in a 0.100 M solution of ethanoic acid.

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Answer

To find the hydrogen ion concentration [H+][H^+], we can use the formula for pH:

pH=log[H+]pH = -\log[H^+]

Given that the pH is 2.88, we can rearrange this to:

[H+]=10pH=102.881.32×103 M[H^+] = 10^{-pH} = 10^{-2.88} \approx 1.32 \times 10^{-3} \text{ M}

Step 3

b.ii. Calculate the acidity constant of ethanoic acid at 25°C.

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Answer

The acidity constant (Ka) can be calculated using the formula:

Ka=[CH3COO][H+][CH3COOH]K_a = \frac{[CH₃COO^-][H^+]}{[CH₃COOH]}

Substituting the values:

  • Initial concentration of CH₃COOH = 0.100 M
  • [H+]ext,calculatedfrompartb.i,1.32×103 M[H^+] ext{, calculated from part b.i, } \approx 1.32 \times 10^{-3} \text{ M}

At equilibrium:

  • [CH3COO]=[H+]=1.32×103extM[CH₃COO^-] = [H^+] = 1.32 \times 10^{-3} ext{ M}
  • [CH3COOH]extatequilibrium=0.1001.32×1030.100extM[CH₃COOH] ext{ at equilibrium } = 0.100 - 1.32 \times 10^{-3} \approx 0.100 ext{ M} (since 1.32 x 10^-3 is negligible)

Now substituting:

Ka=(1.32×103)(1.32×103)0.1001.74×105K_a = \frac{(1.32 \times 10^{-3})(1.32 \times 10^{-3})}{0.100} \approx 1.74 \times 10^{-5}

Step 4

c.i. Comparing two 0.10 M solutions of methanoic and ethanoic acids, which solution would have the higher pH? Give a simple qualitative explanation for your answer.

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Answer

Methanoic acid (HCOOH) has a higher acidity constant compared to ethanoic acid (CH₃COOH), as stated in the question. Since methanoic acid dissociates more completely in solution due to its higher acidity, it produces a greater concentration of H+H^+ ions. Consequently, the solution of ethanoic acid will have a higher pH than that of methanoic acid because a lower concentration of H+H^+ ions corresponds to a higher pH.

Step 5

c.ii. Equal volumes of both solutions were titrated against a 0.10 M solution of NaOH. Which of the solutions, if either, would require the greater volume of the NaOH solution for complete neutralisation? Explain your conclusion.

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Answer

In the titration of a weak acid with NaOH, the weaker the acid, the more volume of titrant is usually required for neutralisation. As ethanoic acid is weaker than methanoic acid, it will require a greater volume of 0.10 M NaOH to reach the equivalence point compared to the methanoic acid solution. This is because the lower dissociation of the weak acid means there are fewer H+H^+ ions available to react with NaOH, requiring more NaOH to achieve neutralisation.

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