Question
Cobalt ions form complex ions with water and chloride as shown in the reaction. The left side of the reaction is pink, and the right side of the reaction is blue.
Co(H2O)62+ + 4Cl-minus + heat ⇄ CoCl42-minus+ 6H2O
Which statement about the system at equilibrium is correct?
(1 point)
Responses
Heating a blue solution will turn it purple.
Heating a blue solution will turn it purple.
Adding water to a purple solution will turn it blue.
Adding water to a purple solution will turn it blue.
Cooling a purple solution will turn it pink.
Cooling a purple solution will turn it pink.
Adding hydrogen chloride (HCl) to a blue solution will turn it purple.
Co(H2O)62+ + 4Cl-minus + heat ⇄ CoCl42-minus+ 6H2O
Which statement about the system at equilibrium is correct?
(1 point)
Responses
Heating a blue solution will turn it purple.
Heating a blue solution will turn it purple.
Adding water to a purple solution will turn it blue.
Adding water to a purple solution will turn it blue.
Cooling a purple solution will turn it pink.
Cooling a purple solution will turn it pink.
Adding hydrogen chloride (HCl) to a blue solution will turn it purple.
Answers
GPT-4o mini
To analyze the chemical equilibrium you've provided, we can use Le Chatelier's principle, which states that if a system at equilibrium is subjected to a change in concentration, temperature, or pressure, the system will adjust to counteract that change.
The equilibrium is established as follows:
\( \text{Co(H}_2\text{O)}_6^{2+} + 4\text{Cl}^- + \text{heat} \rightleftharpoons \text{CoCl}_4^{2-} + 6\text{H}_2\text{O} \)
1. **Heating**:
- The forward reaction (formation of \(\text{CoCl}_4^{2-}\)) is endothermic, meaning it absorbs heat.
- Heating the solution shifts the equilibrium to the right (toward the blue side). Therefore, the initially blue solution will become more intensely blue, not purple.
2. **Adding water**:
- Adding more water will dilute the solution, shifting the equilibrium to the left (to the pink side). Therefore, it will not turn blue when water is added to a purple solution.
3. **Cooling**:
- Cooling the solution will shift the equilibrium to the left (toward the pink side), making it pink when starting from a purple solution.
4. **Adding hydrogen chloride (HCl)**:
- Adding HCl provides more chloride ions. This would shift the equilibrium to the right (toward the blue side), not toward purple.
Based on this analysis, the correct statement is:
**Cooling a purple solution will turn it pink.**
The equilibrium is established as follows:
\( \text{Co(H}_2\text{O)}_6^{2+} + 4\text{Cl}^- + \text{heat} \rightleftharpoons \text{CoCl}_4^{2-} + 6\text{H}_2\text{O} \)
1. **Heating**:
- The forward reaction (formation of \(\text{CoCl}_4^{2-}\)) is endothermic, meaning it absorbs heat.
- Heating the solution shifts the equilibrium to the right (toward the blue side). Therefore, the initially blue solution will become more intensely blue, not purple.
2. **Adding water**:
- Adding more water will dilute the solution, shifting the equilibrium to the left (to the pink side). Therefore, it will not turn blue when water is added to a purple solution.
3. **Cooling**:
- Cooling the solution will shift the equilibrium to the left (toward the pink side), making it pink when starting from a purple solution.
4. **Adding hydrogen chloride (HCl)**:
- Adding HCl provides more chloride ions. This would shift the equilibrium to the right (toward the blue side), not toward purple.
Based on this analysis, the correct statement is:
**Cooling a purple solution will turn it pink.**