For the aqueous [Ag (CN)₂] complex K₁=1.26 × 10²¹ at 25 °C. Suppose equal volumes of 0.0022M AgNO3 solution and 0.48M KCN solution are mixed. Calculate the equilibrium molarity of aqueous Agion. Round your answer to 2 significant digits. M x10
For the aqueous [Ag (CN)₂] complex K₁=1.26 × 10²¹ at 25 °C. Suppose equal volumes of 0.0022M AgNO3 solution and 0.48M KCN solution are mixed. Calculate the equilibrium molarity of aqueous Agion. Round your answer to 2 significant digits. M x10
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![### Equilibrium Calculation for [Ag(CN)₂]⁻ Complex
**Problem Statement:**
- For the aqueous \( [\text{Ag}(\text{CN})_2]^{-} \) complex, the formation constant \( K_f = 1.26 \times 10^{21} \) at 25 °C.
- Suppose equal volumes of 0.0022 M AgNO₃ solution and 0.48 M KCN solution are mixed. Calculate the equilibrium molarity of aqueous \( \text{Ag}^+ \) ion.
- Round your answer to 2 significant digits.
**Answer Format:**
- The answer should be provided in molarity (M).
**Diagram Explanation:**
The image contains a rectangular box where the equilibrium molarity of \( \text{Ag}^+ \) ion should be entered. Next to it, there is a small control panel with options to input the response, such as multiplication by 10, a check mark, and a help button.
This setup appears to be from an online educational platform, where students can input their calculated answer and possibly access hints or verify their response.
### Detailed Steps for Solving the Problem:
1. **Mixing Solutions:**
- Initial concentration of AgNO₃ = 0.0022 M
- Initial concentration of KCN = 0.48 M
- When mixed, the concentration of each will halve due to equal volumes mixed:
- \([ \text{Ag}^+ ]_{\text{initial}} = \frac{0.0022 \text{ M}}{2} = 0.0011 \text{ M}\)
- \([ \text{CN}^- ]_{\text{initial}} = \frac{0.48 \text{ M}}{2} = 0.24 \text{ M}\)
2. **Formation of Complex Ion:**
- The reaction for the formation of the complex:
\[ \text{Ag}^+ + 2 \text{CN}^- \leftrightharpoons [ \text{Ag}(\text{CN})_2 ]^- \]
- The formation constant expression:
\[ K_f = \frac{[ \text{Ag}(\text{CN})_2^- ]}{[\text{Ag}^+ ][ \text{CN}^- ]^](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fc0633024-a4bc-4f66-b9e9-fd2e8bdcc354%2F66c65b30-adcf-40cb-b582-4b57ca4f71e0%2Fejs2ktt_processed.png&w=3840&q=75)
Transcribed Image Text:### Equilibrium Calculation for [Ag(CN)₂]⁻ Complex
**Problem Statement:**
- For the aqueous \( [\text{Ag}(\text{CN})_2]^{-} \) complex, the formation constant \( K_f = 1.26 \times 10^{21} \) at 25 °C.
- Suppose equal volumes of 0.0022 M AgNO₃ solution and 0.48 M KCN solution are mixed. Calculate the equilibrium molarity of aqueous \( \text{Ag}^+ \) ion.
- Round your answer to 2 significant digits.
**Answer Format:**
- The answer should be provided in molarity (M).
**Diagram Explanation:**
The image contains a rectangular box where the equilibrium molarity of \( \text{Ag}^+ \) ion should be entered. Next to it, there is a small control panel with options to input the response, such as multiplication by 10, a check mark, and a help button.
This setup appears to be from an online educational platform, where students can input their calculated answer and possibly access hints or verify their response.
### Detailed Steps for Solving the Problem:
1. **Mixing Solutions:**
- Initial concentration of AgNO₃ = 0.0022 M
- Initial concentration of KCN = 0.48 M
- When mixed, the concentration of each will halve due to equal volumes mixed:
- \([ \text{Ag}^+ ]_{\text{initial}} = \frac{0.0022 \text{ M}}{2} = 0.0011 \text{ M}\)
- \([ \text{CN}^- ]_{\text{initial}} = \frac{0.48 \text{ M}}{2} = 0.24 \text{ M}\)
2. **Formation of Complex Ion:**
- The reaction for the formation of the complex:
\[ \text{Ag}^+ + 2 \text{CN}^- \leftrightharpoons [ \text{Ag}(\text{CN})_2 ]^- \]
- The formation constant expression:
\[ K_f = \frac{[ \text{Ag}(\text{CN})_2^- ]}{[\text{Ag}^+ ][ \text{CN}^- ]^
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