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1 (a) Hydrogen fluoride, HF, behaves as a weak acid in water, with K_a = 5.6 × 10⁻⁴ mol dm⁻³ - CIE - A-Level Chemistry - Question 1 - 2011 - Paper 1
Question 1
1 (a) Hydrogen fluoride, HF, behaves as a weak acid in water, with K_a = 5.6 × 10⁻⁴ mol dm⁻³. Calculate the pH of a 0.050 mol dm⁻³ solution of HF. (b) Gaseous ammo... show full transcript
Worked Solution & Example Answer:1 (a) Hydrogen fluoride, HF, behaves as a weak acid in water, with K_a = 5.6 × 10⁻⁴ mol dm⁻³ - CIE - A-Level Chemistry - Question 1 - 2011 - Paper 1
Step 1
Calculate the pH of a 0.050 mol dm⁻³ solution of HF.
Answer
To determine the pH of a 0.050 mol dm⁻³ solution of hydrogen fluoride (HF), we need to find the concentration of hydrogen ions ( [H^+] ) in the solution. HF is a weak acid and only partially dissociates in water:
ightleftharpoons H^+ + F^- $$ We can use the ionization constant (K_a) to establish the equilibrium concentration. The expression for K_a is: $$ K_a = \frac{[H^+][F^-]}{[HF]} $$ Assuming that the initial concentration of HF is 0.050 mol dm⁻³, let's denote the change in concentration during dissociation as x. At equilibrium: - [H^+] = x, - [F^-] = x, - [HF] = 0.050 - x Substituting into the K_a expression: $$ 5.6 imes 10^{-4} = \frac{x^2}{0.050-x} $$ For simplicity, we can assume that x is small compared to 0.050 and thus: $$ K_a \approx \frac{x^2}{0.050} $$ From this we find: $$ x^2 = 5.6 imes 10^{-4} \times 0.050 \implies x^2 = 2.8 \times 10^{-5} $$ $$ x = \sqrt{2.8 \times 10^{-5}} \approx 5.29 \times 10^{-3} \text{ mol dm}^{-3} $$ Now substituting back to find the pH: $$ pH = -\log(5.29 \times 10^{-3}) \approx 2.3 $$Step 2
What type of reaction is this?
Answer
This is a Brønsted-Lowry acid-base reaction, where ammonia (NH₃) acts as a base and hydrogen fluoride (HF) acts as an acid, transferring a proton to form ammonium fluoride (NH₄F). Additionally, it can be classified as an exothermic reaction due to the negative enthalpy change (ΔH = -147 kJ mol⁻¹).
Step 3
Draw dot-and-cross diagrams (outer shells only) describing the bonding in the three compounds involved in this reaction.
Answer
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For NH₃ (Ammonia):
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The nitrogen atom (N) forms three covalent bonds with three hydrogen atoms (H) using three of its outer shell electrons, sharing one electron from each hydrogen atom.
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Diagram:
H H \ / N | H -
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For HF (Hydrogen Fluoride):
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The fluorine atom (F) shares its seven outer shell electrons with one of the hydrogen's single outer shell electron to form a covalent bond.
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Diagram:
H | F -
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For NH₄F (Ammonium Fluoride):
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The ammonium ion (NH₄⁺) has covalent bonds between nitrogen and four hydrogens, and the fluoride ion (F⁻) is ionically bonded to NH₄⁺.
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Diagram:
H | H--N--H | H F -
Step 4
There are three types of bonding in NH₄F. Give the names of each of the three types, and state where in the compound each type occurs.
Answer
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Covalent bonding: This occurs between nitrogen (N) and hydrogen (H) atoms within the NH₄⁺ ion, where electrons are shared.
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Dative (co-ordinate) bonding: This type occurs between the nitrogen (NH₄⁺) and one of its hydrogen atoms, where the nitrogen donates a pair of electrons to the hydrogen.
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Ionic bonding: This occurs between the ammonium ion (NH₄⁺) and the fluoride ion (F⁻), where there is a transfer of electrons, resulting in opposite charges attracting each other.