### QUESTION 7
#### (START ON A NEW PAGE)
27 g of [tex]\( \text{Mg(OH)}_2 \)[/tex] is dissolved in 2 L of water at [tex]\( 25^\circ C \)[/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.
Magnesium hydroxide [tex]\( \text{Mg(OH)}_2 \)[/tex] is classified as a strong base because it fully dissociates in water to produce hydroxide ions ([tex]\( \text{OH}^- \)[/tex]). The dissociation can be represented by the following chemical equation:
[tex]\[ \text{Mg(OH)}_2 (s) \rightarrow \text{Mg}^{2+} (aq) + 2 \text{OH}^- (aq) \][/tex]
When [tex]\( \text{Mg(OH)}_2 \)[/tex] dissolves in water, every molecule of [tex]\( \text{Mg(OH)}_2 \)[/tex] produces one magnesium ion ([tex]\( \text{Mg}^{2+} \)[/tex]) and two hydroxide ions ([tex]\( \text{OH}^- \)[/tex]). The complete dissociation into ions significantly increases the concentration of hydroxide ions in the solution, which leads to a higher [tex]\( \text{pH} \)[/tex], making the solution basic.
#### 7.2 Calculate the pH of the [tex]\( \text{Mg(OH)}_2 \)[/tex] solution.
To calculate the pH of the solution, we first need to find several intermediate values such as the number of moles of [tex]\( \text{Mg(OH)}_2 \)[/tex], the concentration of hydroxide ions ([tex]\( \text{OH}^- \)[/tex]), the pOH, and finally the pH. Here is the step-by-step calculation:
1. Calculate the number of moles of [tex]\( \text{Mg(OH)}_2 \)[/tex]:
The molar mass of [tex]\( \text{Mg(OH)}_2 \)[/tex] is:
[tex]\[ \text{Molar mass} = 24.305 + 2 \times (16.00 + 1.008) = 58.32 \, \text{g/mol} \][/tex]
The mass of [tex]\( \text{Mg(OH)}_2 \)[/tex] given is 27 g:
[tex]\[ \text{Moles of } \text{Mg(OH)}_2 = \frac{\text{Mass}}{\text{Molar mass}} = \frac{27 \, \text{g}}{58.32 \, \text{g/mol}} = 0.46296 \, \text{mol} \][/tex]
2. Calculate the moles of [tex]\( \text{OH}^- \)[/tex] ions produced:
Since 1 mole of [tex]\( \text{Mg(OH)}_2 \)[/tex] produces 2 moles of [tex]\( \text{OH}^- \)[/tex]:
[tex]\[ \text{Moles of } \text{OH}^- = 2 \times 0.46296 \, \text{mol} = 0.92593 \, \text{mol} \][/tex]
3. Calculate the concentration of [tex]\( \text{OH}^- \)[/tex] ions (molarity):
The volume of the solution is 2 L:
[tex]\[ \text{Concentration of } \text{OH}^- = \frac{\text{Moles of } \text{OH}^-}{\text{Volume of solution}} = \frac{0.92593 \, \text{mol}}{2 \, \text{L}} = 0.46296 \, \text{M} \][/tex]
4. Calculate the pOH:
[tex]\[ \text{pOH} = -\log_{10} [ \text{OH}^- ] = -\log_{10} (0.46296) = 0.3345 \][/tex]
5. Calculate the pH:
Since [tex]\( \text{pH} + \text{pOH} = 14 \)[/tex]:
[tex]\[ \text{pH} = 14 - \text{pOH} = 14 - 0.3345 = 13.6655 \][/tex]
Hence, the pH of the [tex]\( \text{Mg(OH)}_2 \)[/tex] solution is approximately 13.6655.
