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QUESTION 7 [23 marks]
Consider the reaction below:
HSO4
- + HCO3
- → SO4
2- + H2CO3
Question 7.1 [2 marks]
Define a Lowry-Brønsted base.
Question 7.2 [6 marks]
Rewrite the chemical equation and indicate with the help of lines and arrows the
conjugated acid-base pairs. Clearly indicate the conjugated acid and base pairs
for the forward and reverse reactions. You may use colour to help distinguish
the labels.
Question 7.3 [2 marks]
Identify the substance(s) that are considered ampholytes from the above
reaction.
Question 7.4 [3 marks]
Write down the balanced chemical equation for the reaction when Magnesium
Oxide (MgO) reacts with Nitric Acid.
Question 7.5
50 g of IMPURE calcium carbonate reacts with 200 cm3 of a dilute
sulphuric acid with a concentration of 1,5 mol∙dm-3 All the calcium carbonate
and sulphuric acid react completely leaving the impurities unreacted at thecompletion of the reaction.
CaCO3
(s) + H2SO4
(aq) → CaSO4
(s) + CO2
(g) + H2O (l)
Question 7.5.1 [6 marks]
Calculate the percentage purity of the calcium carbonate.
Question 7.5.2 [4 marks]
To create the sulphuric acid (1,5 mol∙dm-3
) solution that was used in the above
reaction. 10 cm3 of a concentrated sulphuric acid solution of concentration 9
mol∙dm-3
is added to water.
Calculate the volume (in dm3
) of water required to dilute this solution in order to
obtain the concentration of 1.5 mol.dm-3 which was used in the reaction.

Answer 1

### 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:**
1. **Forward Reaction:**
- **Acid:** \( \text{HSO}_4^- \) (donates a proton)
- **Conjugate Base:** \( \text{SO}_4^{2-} \)

- **Base:** \( \text{HCO}_3^- \) (accepts a proton)
- **Conjugate Acid:** \( \text{H}_2\text{CO}_3 \)

**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.

1. Calculate the moles of sulfuric acid (H₂SO₄):
\[
\text{Concentration of H}_2\text{SO}_4 = 1.5 \, \text{mol/dm}^3
\]
\[
\text{Volume of H}_2\text{SO}_4 = 200 \, \text{cm}^3 = 0.200 \, \text{dm}^3
\]
\[
\text{Moles of H}_2\text{SO}_4 = \text{Concentration} \times \text{Volume} = 1.5 \times 0.200 = 0.300 \, \text{mol}
\]

2. 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.

3. Thus, moles of CaCO₃ that reacted:
\[
\text{Moles of CaCO}_3 = 0.300 \, \text{mol}
\]

4. Calculate the mass of pure CaCO₃:
\[
\text{Molar mass of CaCO}_3 = 40.08 + 12.01 + 3 \times 16.00 = 100.09 \, \text{g/mol}
\]
\[
\text{Mass of CaCO}_3 = \text{Moles} \times \text{Molar mass} = 0.300 \times 100.09 = 30.027 \, \text{g}
\]

5. 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:
\[ C_1V_1 = C_2V_2 \]

1. Given:
- \( C_1 = 9 \, \text{mol/dm}^3 \)
- \( V_1 = 0.010 \, \text{dm}^3 \) (10 cm³)
- \( C_2 = 1.5 \, \text{mol/dm}^3 \)

2. Calculate \( V_2 \):
\[
V_2 = \frac{C_1V_1}{C_2} = \frac{9 \times 0.010}{1.5} = 0.060 \, \text{dm}^3
\]

3. 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**.