Methanoic acid HCOOH is a weak acid present in the sting of some ants. It ionises in water according to HCOOH(aq) ⇌ H+(aq) + HCOO⁻(aq) K_a = 1.8 × 10⁻⁵ at 25°C a. Explain the meaning of the terms ‘weak acid’ and ‘strong acid’. b. Write the expression for the K_a of methanoic acid. c. Assuming a small degree of dissociation, calculate the concentrations of H+(aq) and HCOO⁻(aq) in 0.10 M methanoic acid at 25°C.

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Methanoic acid HCOOH is a weak acid present in the sting of some ants - VCE - SSCE Chemistry - Question 4 - 2003 - Paper 1

Question 4

Methanoic acid HCOOH is a weak acid present in the sting of some ants. It ionises in water according to HCOOH(aq) ⇌ H+(aq) + HCOO⁻(aq) K_a = 1.8 × 10⁻⁵ at 25°C ... show full transcript

Worked Solution & Example Answer:Methanoic acid HCOOH is a weak acid present in the sting of some ants - VCE - SSCE Chemistry - Question 4 - 2003 - Paper 1

Step 1

Explain the meaning of the terms ‘weak acid’ and ‘strong acid’.

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Answer

A weak acid is one that only partially ionizes in solution, meaning that only a small fraction of the acid molecules donate protons (H+) to the solution. This results in an equilibrium state where both the ionized and un-ionized forms of the acid are present.

In contrast, a strong acid completely ionizes in solution, meaning that almost all of the acid molecules donate their protons, leading to a high concentration of H+ ions and negligible amounts of the un-ionized acid.

Step 2

Write the expression for the K_a of methanoic acid.

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Answer

The expression for the acid dissociation constant (Ka) of methanoic acid can be written as:

$K_a = \frac{[HCOO^-][H^+]}{[HCOOH]}$

where [HCOO-] is the concentration of the methanoate ion, [H] is the concentration of hydrogen ions, and [HCOOH] is the concentration of methanoic acid.

Step 3

Assuming a small degree of dissociation, calculate the concentrations of H⁺(aq) and HCOO⁻(aq) in 0.10 M methanoic acid at 25°C.

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Answer

Let the concentration of H+ ions produced be represented as 'x'. For the dissociation

$HCOOH ⇌ H^+ + HCOO^-$

At equilibrium, the concentrations will be:

[H+] = x
[HCOO-] = x
[HCOOH] = 0.10 - x ≈ 0.10 (since x is small)

Substituting these into the Ka expression:

$K_a = \frac{x^2}{0.10}$

Given that Ka = 1.8 × 10-5, we have:

$1.8 \times 10^{-5} = \frac{x^2}{0.10}$

This simplifies to:

$x^2 = 1.8 \times 10^{-6}$

Taking the square root:

$x = \sqrt{1.8 \times 10^{-6}} \approx 1.34 \times 10^{-3} M$

Thus:

[H+] = 1.34 × 10-3 M
[HCOO-] = 1.34 × 10-3 M

Methanoic acid HCOOH is a weak acid present in the sting of some ants - VCE - SSCE Chemistry - Question 4 - 2003 - Paper 1

Worked Solution & Example Answer:Methanoic acid HCOOH is a weak acid present in the sting of some ants - VCE - SSCE Chemistry - Question 4 - 2003 - Paper 1

Explain the meaning of the terms ‘weak acid’ and ‘strong acid’.

Write the expression for the K_a of methanoic acid.

Assuming a small degree of dissociation, calculate the concentrations of H⁺(aq) and HCOO⁻(aq) in 0.10 M methanoic acid at 25°C.

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