IDNLearn.com provides a collaborative environment for finding and sharing knowledge. Join our knowledgeable community to find the answers you need for any topic or issue.
Sagot :
### 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.
#### (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.
We appreciate your contributions to this forum. Don't forget to check back for the latest answers. Keep asking, answering, and sharing useful information. Your search for answers ends at IDNLearn.com. Thank you for visiting, and we hope to assist you again soon.
