Question 7.1 [2 marks]
A Lowry-Brønsted base is defined as a substance that accepts a proton (H⁺) during a chemical reaction. In the context of acid-base reactions, a base is a species that can form its conjugate acid by gaining a proton.
Question 7.2 [6 marks]
To rewrite the given chemical reaction and indicate the conjugate acid-base pairs:
Reaction: \[ \text{HSO}_4^- + \text{HCO}_3^- \rightleftharpoons \text{SO}_4^{2-} + \text{H}_2\text{CO}_3 \]
Conjugate Acid-Base Pairs:
- Forward Reaction:
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Acid: \( \text{HSO}_4^- \) (donates a proton)
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Conjugate Base: \( \text{SO}_4^{2-} \)
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Base: \( \text{HCO}_3^- \) (accepts a proton)
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Conjugate Acid: \( \text{H}_2\text{CO}_3 \)
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Arrow indication:
- The reaction shows the evolution from the acid and base in the reactants to their respective conjugate pairs in the products.
Question 7.3 [2 marks]
The substance(s) considered ampholytes from the above reaction are:
- \( \text{HCO}_3^- \) (bicarbonate)
- \( \text{HSO}_4^- \) (hydrogen sulfate), as they can act as either an acid or a base depending on the reaction conditions.
Question 7.4 [3 marks]
The balanced chemical equation for the reaction when Magnesium Oxide (MgO) reacts with Nitric Acid (HNO₃) is:
\[ \text{MgO} + 2 \text{HNO}_3 \rightarrow \text{Mg(NO}_3\text{)}_2 + \text{H}_2\text{O} \]
Question 7.5
Question 7.5.1 [6 marks]
To calculate the percentage purity of calcium carbonate, we first need to determine the amount of calcium carbonate that reacts with a given amount of sulfuric acid.
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Calculate the moles of sulfuric acid (H₂SO₄): \[ \text{Concentration of H}_2\text{SO}_4 = 1.5 , \text{mol/dm}^3 \] \[ \text{Volume of H}_2\text{SO}_4 = 200 , \text{cm}^3 = 0.200 , \text{dm}^3 \] \[ \text{Moles of H}_2\text{SO}_4 = \text{Concentration} \times \text{Volume} = 1.5 \times 0.200 = 0.300 , \text{mol} \]
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The reaction between CaCO₃ and H₂SO₄ is: \[ \text{CaCO}_3 + \text{H}_2\text{SO}_4 \rightarrow \text{CaSO}_4 + \text{CO}_2 + \text{H}_2\text{O} \] From the reaction stoichiometry, 1 mole of CaCO₃ reacts with 1 mole of H₂SO₄.
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Thus, moles of CaCO₃ that reacted: \[ \text{Moles of CaCO}_3 = 0.300 , \text{mol} \]
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Calculate the mass of pure CaCO₃: \[ \text{Molar mass of CaCO}_3 = 40.08 + 12.01 + 3 \times 16.00 = 100.09 , \text{g/mol} \] \[ \text{Mass of CaCO}_3 = \text{Moles} \times \text{Molar mass} = 0.300 \times 100.09 = 30.027 , \text{g} \]
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Calculate the percentage purity of calcium carbonate: \[ \text{Percentage purity} = \left( \frac{\text{Mass of pure CaCO}_3}{\text{Mass of impure sample}} \right) \times 100 = \left( \frac{30.027}{50} \right) \times 100 \approx 60.05% \]
Question 7.5.2 [4 marks]
To find the volume of water required to dilute the concentrated sulfuric acid to a concentration of 1.5 mol/dm³, we can use the dilution equation: \[ C_1V_1 = C_2V_2 \]
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Given:
- \( C_1 = 9 , \text{mol/dm}^3 \)
- \( V_1 = 0.010 , \text{dm}^3 \) (10 cm³)
- \( C_2 = 1.5 , \text{mol/dm}^3 \)
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Calculate \( V_2 \): \[ V_2 = \frac{C_1V_1}{C_2} = \frac{9 \times 0.010}{1.5} = 0.060 , \text{dm}^3 \]
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Therefore, the total volume of diluted solution \( V_2 \) is 0.060 dm³. The volume of water \( V_{\text{water}} \) required to dilute is: \[ V_{\text{water}} = V_2 - V_1 = 0.060 , \text{dm}^3 - 0.010 , \text{dm}^3 = 0.050 , \text{dm}^3 \]
Thus, the volume of water required to obtain a solution of 1.5 mol/dm³ is 0.050 dm³.
1 answer