7.1 Sulphuric acid is a strong acid present in acid rain - NSC Physical Sciences - Question 7 - 2018 - Paper 2
Question 7
7.1 Sulphuric acid is a strong acid present in acid rain. It ionises in two steps as follows:
I: H₂SO₄(aq) ⇌ H₃O⁺(aq) + HSO₄⁻(aq)
II: HSO₄⁻(aq) ⇌ H₃O⁺(aq) + SO₄²⁻(... show full transcript
Worked Solution & Example Answer:7.1 Sulphuric acid is a strong acid present in acid rain - NSC Physical Sciences - Question 7 - 2018 - Paper 2
Step 1
7.1.1 Define an acid in terms of the Lowry-Brønsted theory.
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Answer
An acid is defined as a proton donor according to the Lowry-Brønsted theory. It is a substance that can donate a proton (H⁺) to another substance.
Step 2
7.1.2 Write down the FORMULA of the conjugate base of H₃O⁺.
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Answer
The conjugate base of H₃O⁺ is H₂O.
Step 3
7.1.3 Write down the FORMULA of the substance that acts as an ampholyte in the ionisation of sulphuric acid.
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The substance acting as an ampholyte is H₂O.
Step 4
7.2.1 Define hydrolysis of a salt.
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Hydrolysis of a salt is the reaction between salt ions and water, resulting in the formation of acidic or basic solutions from the dissociation of the salt.
Step 5
7.2.2 Explain, with the aid of the relevant HYDROLYSIS reaction, how limestone can neutralise the acid.
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Limestone (CaCO₃) undergoes hydrolysis in water to form calcium ions (Ca²⁺) and bicarbonate ions (HCO₃⁻), which can react with hydronium ions (H₃O⁺) to neutralise the acid:
ightleftharpoons ext{Ca}^{2+} + ext{HCO}_3^- + ext{OH}^-$$
The hydroxide ions (OH⁻) produced neutralise excess hydronium ions, thus reducing acidity.
Step 6
7.3.1 Calculate the concentration of the hydronium ions in the water.
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Answer
To find the concentration of hydronium ions (H₃O⁺), we use the formula: extpH=−extlog[H3O+]
Given that the
pH = 5, we have:
7.3.2 If the final amount of hydronium ions is 1.26 × 10³ moles, calculate the mass of lime that was added to the lake.
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To calculate the mass of CaO added, we first convert moles to grams:
From the balanced equation, for every mole of CaO, 2 moles of H₃O⁺ are produced:
ext{moles of CaO} = rac{1.26 imes 10^3}{2} = 6.3 imes 10^2 ext{ moles}
Using the molar mass of CaO (56 g/mol), we find:
extmassofCaO=6.3imes102extmolesimes56extg/mol=3.528imes104extg
