When 12.0 mL of a 6.95 x 10-4 M magnesium nitrate solution is combined with 22.0 mL of a 7.27 x 10-4 M potassium hydroxide solution does a precipitate form? (Ksp (Mg(OH)2) = 1.5 × 10-¹¹) O Yes, the precipitate forms. O No, the precipitate doesn't form. For these conditions the Reaction Quotient, Q, is equal to
When 12.0 mL of a 6.95 x 10-4 M magnesium nitrate solution is combined with 22.0 mL of a 7.27 x 10-4 M potassium hydroxide solution does a precipitate form? (Ksp (Mg(OH)2) = 1.5 × 10-¹¹) O Yes, the precipitate forms. O No, the precipitate doesn't form. For these conditions the Reaction Quotient, Q, is equal to
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![### Precipitation Reaction and Reaction Quotient (Q)
**Problem Statement:**
When 12.0 mL of a \(6.95 \times 10^{-4} \,M\) magnesium nitrate solution is combined with 22.0 mL of a \(7.27 \times 10^{-4} \,M\) potassium hydroxide solution, does a precipitate form?
\[ K_{sp} \left( \text{Mg(OH)}_2 \right) = 1.5 \times 10^{-11} \]
- ◯ Yes, the precipitate forms.
- ◯ No, the precipitate doesn't form.
For these conditions, the Reaction Quotient, \( Q \), is equal to \(\_\_\_\_\_\_\_\_\).
**Explanation:**
To determine whether a precipitate forms, you need to compare the reaction quotient \( Q \) to the solubility product constant \( K_{sp} \).
- If \( Q > K_{sp} \), a precipitate forms.
- If \( Q < K_{sp} \), no precipitate forms.
- If \( Q = K_{sp} \), the solution is at equilibrium and no further precipitate will form under current conditions.
**Steps to Calculate \( Q \):**
1. **Calculate the initial moles of each ion:**
- Moles of \( \text{Mg}^{2+} \) from magnesium nitrate:
\[
\text{Moles of } \text{Mg}^{2+} = 12.0 \, \text{mL} \times \frac{6.95 \times 10^{-4} \,M}{1000 \, \text{mL}}
\]
- Moles of \( \text{OH}^- \) from potassium hydroxide:
\[
\text{Moles of } \text{OH}^- = 22.0 \, \text{mL} \times \frac{7.27 \times 10^{-4} \,M}{1000 \, \text{mL}}
\]
2. **Determine the final concentrations after mixing (taking into account the total volume):**
- Total volume \( = 12.0 \, \text{mL} + 22.0 \, \text{mL} = 34](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F2ad64212-e120-4fe2-97f2-e4e87c5c792a%2F0ac38459-b699-492d-926c-6a4580ae5e0e%2F7xppgvx6_processed.png&w=3840&q=75)
Transcribed Image Text:### Precipitation Reaction and Reaction Quotient (Q)
**Problem Statement:**
When 12.0 mL of a \(6.95 \times 10^{-4} \,M\) magnesium nitrate solution is combined with 22.0 mL of a \(7.27 \times 10^{-4} \,M\) potassium hydroxide solution, does a precipitate form?
\[ K_{sp} \left( \text{Mg(OH)}_2 \right) = 1.5 \times 10^{-11} \]
- ◯ Yes, the precipitate forms.
- ◯ No, the precipitate doesn't form.
For these conditions, the Reaction Quotient, \( Q \), is equal to \(\_\_\_\_\_\_\_\_\).
**Explanation:**
To determine whether a precipitate forms, you need to compare the reaction quotient \( Q \) to the solubility product constant \( K_{sp} \).
- If \( Q > K_{sp} \), a precipitate forms.
- If \( Q < K_{sp} \), no precipitate forms.
- If \( Q = K_{sp} \), the solution is at equilibrium and no further precipitate will form under current conditions.
**Steps to Calculate \( Q \):**
1. **Calculate the initial moles of each ion:**
- Moles of \( \text{Mg}^{2+} \) from magnesium nitrate:
\[
\text{Moles of } \text{Mg}^{2+} = 12.0 \, \text{mL} \times \frac{6.95 \times 10^{-4} \,M}{1000 \, \text{mL}}
\]
- Moles of \( \text{OH}^- \) from potassium hydroxide:
\[
\text{Moles of } \text{OH}^- = 22.0 \, \text{mL} \times \frac{7.27 \times 10^{-4} \,M}{1000 \, \text{mL}}
\]
2. **Determine the final concentrations after mixing (taking into account the total volume):**
- Total volume \( = 12.0 \, \text{mL} + 22.0 \, \text{mL} = 34
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