Answer :
### QUESTION 7
27 g of [tex]\( \text{Mg(OH)}_2 \)[/tex] is dissolved in 2 L of water at [tex]\( 25^{\circ} \text{C} \)[/tex]. A drop of bromothymol blue is added to the solution and it turns blue.
#### 7.1 Explain why [tex]\( \text{Mg(OH)}_2 \)[/tex] is classified as a strong base.
[tex]\( \text{Mg(OH)}_2 \)[/tex] or magnesium hydroxide is classified as a strong base because it dissociates completely in water to release [tex]\( \text{OH}^- \)[/tex] (hydroxide) ions. Strong bases are substances that can fully ionize in solution, which significantly increases the concentration of [tex]\( \text{OH}^- \)[/tex] ions, thereby increasing the solution's pH.
#### 7.2 Calculate the pH of the [tex]\( \text{Mg(OH)}_2 \)[/tex] solution
To calculate the pH of the solution, let's go through the following steps in detail:
Step 1: Calculate the moles of [tex]\( \text{Mg(OH)}_2 \)[/tex].
Given data:
- Mass of [tex]\( \text{Mg(OH)}_2 \)[/tex]: 27 g
- Molar mass of [tex]\( \text{Mg(OH)}_2 \)[/tex]: 58.32 g/mol
Using the formula:
[tex]\[ \text{moles of } \text{Mg(OH)}_2 = \frac{\text{mass of } \text{Mg(OH)}_2}{\text{molar mass of } \text{Mg(OH)}_2} \][/tex]
[tex]\[ \text{moles of } \text{Mg(OH)}_2 = \frac{27 \text{ g}}{58.32 \text{ g/mol}} \][/tex]
[tex]\[ \text{moles of } \text{Mg(OH)}_2 = 0.46296296296296297 \text{ mol} \][/tex]
Step 2: Calculate the concentration of [tex]\( \text{Mg(OH)}_2 \)[/tex].
[tex]\[ \text{concentration of } \text{Mg(OH)}_2 = \frac{\text{moles of } \text{Mg(OH)}_2}{\text{volume of solution in liters}} \][/tex]
[tex]\[ \text{concentration of } \text{Mg(OH)}_2 = \frac{0.46296296296296297 \text{ mol}}{2 \text{ L}} \][/tex]
[tex]\[ \text{concentration of } \text{Mg(OH)}_2 = 0.23148148148148148 \text{ M} \][/tex]
Step 3: Each mole of [tex]\( \text{Mg(OH)}_2 \)[/tex] gives 2 moles of [tex]\( \text{OH}^- \)[/tex]. Thus, calculate the concentration of [tex]\( \text{OH}^- \)[/tex] ions.
[tex]\[ \text{concentration of } \text{OH}^- = 2 \times \text{concentration of } \text{Mg(OH)}_2 \][/tex]
[tex]\[ \text{concentration of } \text{OH}^- = 2 \times 0.23148148148148148 \text{ M} \][/tex]
[tex]\[ \text{concentration of } \text{OH}^- = 0.46296296296296297 \text{ M} \][/tex]
Step 4: Calculate the pOH of the solution.
[tex]\[ \text{pOH} = -\log[ \text{OH}^-] \][/tex]
[tex]\[ \text{pOH} = -\log(0.46296296296296297) \][/tex]
[tex]\[ \text{pOH} = 0.3344537511509309 \][/tex]
Step 5: Calculate the pH of the solution.
Since at [tex]\( 25^{\circ} \text{C} \)[/tex], the sum of pH and pOH of any aqueous solution is 14:
[tex]\[ \text{pH} = 14 - \text{pOH} \][/tex]
[tex]\[ \text{pH} = 14 - 0.3344537511509309 \][/tex]
[tex]\[ \text{pH} = 13.665546248849068 \][/tex]
Hence, the pH of the [tex]\( \text{Mg(OH)}_2 \)[/tex] solution is 13.67.
### QUESTION 8 (Start on a new page.)
27 g of [tex]\( \text{Mg(OH)}_2 \)[/tex] is dissolved in 2 L of water at [tex]\( 25^{\circ} \text{C} \)[/tex]. A drop of bromothymol blue is added to the solution and it turns blue.
#### 7.1 Explain why [tex]\( \text{Mg(OH)}_2 \)[/tex] is classified as a strong base.
[tex]\( \text{Mg(OH)}_2 \)[/tex] or magnesium hydroxide is classified as a strong base because it dissociates completely in water to release [tex]\( \text{OH}^- \)[/tex] (hydroxide) ions. Strong bases are substances that can fully ionize in solution, which significantly increases the concentration of [tex]\( \text{OH}^- \)[/tex] ions, thereby increasing the solution's pH.
#### 7.2 Calculate the pH of the [tex]\( \text{Mg(OH)}_2 \)[/tex] solution
To calculate the pH of the solution, let's go through the following steps in detail:
Step 1: Calculate the moles of [tex]\( \text{Mg(OH)}_2 \)[/tex].
Given data:
- Mass of [tex]\( \text{Mg(OH)}_2 \)[/tex]: 27 g
- Molar mass of [tex]\( \text{Mg(OH)}_2 \)[/tex]: 58.32 g/mol
Using the formula:
[tex]\[ \text{moles of } \text{Mg(OH)}_2 = \frac{\text{mass of } \text{Mg(OH)}_2}{\text{molar mass of } \text{Mg(OH)}_2} \][/tex]
[tex]\[ \text{moles of } \text{Mg(OH)}_2 = \frac{27 \text{ g}}{58.32 \text{ g/mol}} \][/tex]
[tex]\[ \text{moles of } \text{Mg(OH)}_2 = 0.46296296296296297 \text{ mol} \][/tex]
Step 2: Calculate the concentration of [tex]\( \text{Mg(OH)}_2 \)[/tex].
[tex]\[ \text{concentration of } \text{Mg(OH)}_2 = \frac{\text{moles of } \text{Mg(OH)}_2}{\text{volume of solution in liters}} \][/tex]
[tex]\[ \text{concentration of } \text{Mg(OH)}_2 = \frac{0.46296296296296297 \text{ mol}}{2 \text{ L}} \][/tex]
[tex]\[ \text{concentration of } \text{Mg(OH)}_2 = 0.23148148148148148 \text{ M} \][/tex]
Step 3: Each mole of [tex]\( \text{Mg(OH)}_2 \)[/tex] gives 2 moles of [tex]\( \text{OH}^- \)[/tex]. Thus, calculate the concentration of [tex]\( \text{OH}^- \)[/tex] ions.
[tex]\[ \text{concentration of } \text{OH}^- = 2 \times \text{concentration of } \text{Mg(OH)}_2 \][/tex]
[tex]\[ \text{concentration of } \text{OH}^- = 2 \times 0.23148148148148148 \text{ M} \][/tex]
[tex]\[ \text{concentration of } \text{OH}^- = 0.46296296296296297 \text{ M} \][/tex]
Step 4: Calculate the pOH of the solution.
[tex]\[ \text{pOH} = -\log[ \text{OH}^-] \][/tex]
[tex]\[ \text{pOH} = -\log(0.46296296296296297) \][/tex]
[tex]\[ \text{pOH} = 0.3344537511509309 \][/tex]
Step 5: Calculate the pH of the solution.
Since at [tex]\( 25^{\circ} \text{C} \)[/tex], the sum of pH and pOH of any aqueous solution is 14:
[tex]\[ \text{pH} = 14 - \text{pOH} \][/tex]
[tex]\[ \text{pH} = 14 - 0.3344537511509309 \][/tex]
[tex]\[ \text{pH} = 13.665546248849068 \][/tex]
Hence, the pH of the [tex]\( \text{Mg(OH)}_2 \)[/tex] solution is 13.67.
### QUESTION 8 (Start on a new page.)
