Answer :
Certainly! Let's answer the questions step by step:
### Question 7.1.1: Why is [tex]\( \text{H}_2\text{SO}_4 \)[/tex] considered a STRONG ACID?
Explanation:
Sulfuric acid ([tex]\( \text{H}_2\text{SO}_4 \)[/tex]) is considered a strong acid because it completely dissociates in water. A strong acid ionizes almost completely, releasing a high concentration of hydrogen ions ([tex]\( \text{H}^+ \)[/tex]) when dissolved in water. Specifically, each molecule of [tex]\( \text{H}_2\text{SO}_4 \)[/tex] can dissociate to release two hydrogen ions:
[tex]\[ \text{H}_2\text{SO}_4 \rightarrow 2\text{H}^+ + \text{SO}_4^{2-} \][/tex]
This complete dissociation characteristic is a hallmark of strong acids, in contrast to weak acids, which only partially dissociate in solution. Due to its complete ionization and high hydrogen ion concentration, [tex]\( \text{H}_2\text{SO}_4 \)[/tex] exhibits the typical properties of strong acids, such as a low [tex]\( \text{pH} \)[/tex], high conductivity, and vigorous reactions with bases.
### Question 7.1.2: Calculate the concentration of the final solution
To determine the concentration of magnesium sulfate ([tex]\( \text{MgSO}_4 \)[/tex]) in the final solution, we need to follow these steps:
1. Calculate the moles of [tex]\( \text{Mg(OH)}_2 \)[/tex] and [tex]\( \text{H}_2\text{SO}_4 \)[/tex] used:
- Molecular mass of [tex]\( \text{Mg(OH)}_2 \)[/tex]:
[tex]\[ \text{Mg} = 24.305 \, \text{g/mol}, \quad \text{O} = 16 \, \text{g/mol}, \quad \text{H} = 1.008 \, \text{g/mol} \][/tex]
[tex]\[ \text{Molar mass of } \text{Mg(OH)}_2 = 24.305 + 2 \times (16 + 1.008) = 58.319 \, \text{g/mol} \][/tex]
[tex]\[ \text{Moles of } \text{Mg(OH)}_2 = \frac{\text{Mass}}{\text{Molar Mass}} = \frac{2 \, \text{g}}{58.319 \, \text{g/mol}} \approx 0.0343 \, \text{mol} \][/tex]
- Volume of [tex]\( \text{H}_2\text{SO}_4 \)[/tex] solution:
[tex]\[ \text{Volume} = 30 \, \text{cm}^3 = 0.03 \, \text{dm}^3 \][/tex]
- Concentration of [tex]\( \text{H}_2\text{SO}_4 \)[/tex]:
[tex]\[ 1.5 \, \text{mol/dm}^3 \][/tex]
[tex]\[ \text{Moles of } \text{H}_2\text{SO}_4 = \text{Concentration} \times \text{Volume} = 1.5 \, \text{mol/dm}^3 \times 0.03 \, \text{dm}^3 = 0.045 \, \text{mol} \][/tex]
2. Identify the limiting reactant:
From the balanced chemical equation:
[tex]\[ \text{H}_2\text{SO}_4(aq) + \text{Mg(OH)}_2(aq) \rightarrow 2\text{H}_2\text{O}(l) + \text{MgSO}_4(aq) \][/tex]
The reaction ratio is 1:1. Therefore, we compare the moles:
- Moles of [tex]\( \text{Mg(OH)}_2 \)[/tex]: [tex]\( 0.0343 \, \text{mol} \)[/tex]
- Moles of [tex]\( \text{H}_2\text{SO}_4 \)[/tex]: [tex]\( 0.045 \, \text{mol} \)[/tex]
Since [tex]\( \text{Mg(OH)}_2 \)[/tex] has fewer moles, it is the limiting reactant.
3. Determine the moles of [tex]\( \text{MgSO}_4 \)[/tex] produced:
According to the reaction stoichiometry:
[tex]\[ \text{Moles of } \text{MgSO}_4 = \text{Moles of the limiting reactant } = 0.0343 \, \text{mol} \][/tex]
4. Calculate the total volume of the final solution:
Since the volume of [tex]\( \text{Mg(OH)}_2 \)[/tex] added is negligible, the total volume is essentially the same as the volume of the [tex]\( \text{H}_2\text{SO}_4 \)[/tex] solution:
[tex]\[ \text{Total volume} = 0.03 \, \text{dm}^3 \][/tex]
5. Calculate the concentration of the [tex]\( \text{MgSO}_4 \)[/tex] in the final solution:
[tex]\[ \text{Concentration of } \text{MgSO}_4 = \frac{\text{Moles of } \text{MgSO}_4}{\text{Total volume}} = \frac{0.0343 \, \text{mol}}{0.03 \, \text{dm}^3} \approx 1.143 \, \text{mol/dm}^3 \][/tex]
So, the final concentration of [tex]\( \text{MgSO}_4 \)[/tex] in the solution is approximately [tex]\( 1.143 \, \text{mol/dm}^3 \)[/tex].
### Question 7.1.1: Why is [tex]\( \text{H}_2\text{SO}_4 \)[/tex] considered a STRONG ACID?
Explanation:
Sulfuric acid ([tex]\( \text{H}_2\text{SO}_4 \)[/tex]) is considered a strong acid because it completely dissociates in water. A strong acid ionizes almost completely, releasing a high concentration of hydrogen ions ([tex]\( \text{H}^+ \)[/tex]) when dissolved in water. Specifically, each molecule of [tex]\( \text{H}_2\text{SO}_4 \)[/tex] can dissociate to release two hydrogen ions:
[tex]\[ \text{H}_2\text{SO}_4 \rightarrow 2\text{H}^+ + \text{SO}_4^{2-} \][/tex]
This complete dissociation characteristic is a hallmark of strong acids, in contrast to weak acids, which only partially dissociate in solution. Due to its complete ionization and high hydrogen ion concentration, [tex]\( \text{H}_2\text{SO}_4 \)[/tex] exhibits the typical properties of strong acids, such as a low [tex]\( \text{pH} \)[/tex], high conductivity, and vigorous reactions with bases.
### Question 7.1.2: Calculate the concentration of the final solution
To determine the concentration of magnesium sulfate ([tex]\( \text{MgSO}_4 \)[/tex]) in the final solution, we need to follow these steps:
1. Calculate the moles of [tex]\( \text{Mg(OH)}_2 \)[/tex] and [tex]\( \text{H}_2\text{SO}_4 \)[/tex] used:
- Molecular mass of [tex]\( \text{Mg(OH)}_2 \)[/tex]:
[tex]\[ \text{Mg} = 24.305 \, \text{g/mol}, \quad \text{O} = 16 \, \text{g/mol}, \quad \text{H} = 1.008 \, \text{g/mol} \][/tex]
[tex]\[ \text{Molar mass of } \text{Mg(OH)}_2 = 24.305 + 2 \times (16 + 1.008) = 58.319 \, \text{g/mol} \][/tex]
[tex]\[ \text{Moles of } \text{Mg(OH)}_2 = \frac{\text{Mass}}{\text{Molar Mass}} = \frac{2 \, \text{g}}{58.319 \, \text{g/mol}} \approx 0.0343 \, \text{mol} \][/tex]
- Volume of [tex]\( \text{H}_2\text{SO}_4 \)[/tex] solution:
[tex]\[ \text{Volume} = 30 \, \text{cm}^3 = 0.03 \, \text{dm}^3 \][/tex]
- Concentration of [tex]\( \text{H}_2\text{SO}_4 \)[/tex]:
[tex]\[ 1.5 \, \text{mol/dm}^3 \][/tex]
[tex]\[ \text{Moles of } \text{H}_2\text{SO}_4 = \text{Concentration} \times \text{Volume} = 1.5 \, \text{mol/dm}^3 \times 0.03 \, \text{dm}^3 = 0.045 \, \text{mol} \][/tex]
2. Identify the limiting reactant:
From the balanced chemical equation:
[tex]\[ \text{H}_2\text{SO}_4(aq) + \text{Mg(OH)}_2(aq) \rightarrow 2\text{H}_2\text{O}(l) + \text{MgSO}_4(aq) \][/tex]
The reaction ratio is 1:1. Therefore, we compare the moles:
- Moles of [tex]\( \text{Mg(OH)}_2 \)[/tex]: [tex]\( 0.0343 \, \text{mol} \)[/tex]
- Moles of [tex]\( \text{H}_2\text{SO}_4 \)[/tex]: [tex]\( 0.045 \, \text{mol} \)[/tex]
Since [tex]\( \text{Mg(OH)}_2 \)[/tex] has fewer moles, it is the limiting reactant.
3. Determine the moles of [tex]\( \text{MgSO}_4 \)[/tex] produced:
According to the reaction stoichiometry:
[tex]\[ \text{Moles of } \text{MgSO}_4 = \text{Moles of the limiting reactant } = 0.0343 \, \text{mol} \][/tex]
4. Calculate the total volume of the final solution:
Since the volume of [tex]\( \text{Mg(OH)}_2 \)[/tex] added is negligible, the total volume is essentially the same as the volume of the [tex]\( \text{H}_2\text{SO}_4 \)[/tex] solution:
[tex]\[ \text{Total volume} = 0.03 \, \text{dm}^3 \][/tex]
5. Calculate the concentration of the [tex]\( \text{MgSO}_4 \)[/tex] in the final solution:
[tex]\[ \text{Concentration of } \text{MgSO}_4 = \frac{\text{Moles of } \text{MgSO}_4}{\text{Total volume}} = \frac{0.0343 \, \text{mol}}{0.03 \, \text{dm}^3} \approx 1.143 \, \text{mol/dm}^3 \][/tex]
So, the final concentration of [tex]\( \text{MgSO}_4 \)[/tex] in the solution is approximately [tex]\( 1.143 \, \text{mol/dm}^3 \)[/tex].
