**Problem Statement:**What is the approximate concentration of free Fe²⁺ ion at equilibrium when 1.90×10⁻² mol iron(II) nitrate is added to 1.00 L of solution that is 1.480 M in CN⁻? For [Fe(CN)₆]⁴⁻, K₆ = 1.0×10³⁵.**Solution Box:**[Fe²⁺] = ______ M**Explanation:**This problem involves calculating the concentration of free Fe²⁺ ions in solution after an equilibrium is established. Iron(II) nitrate is added to a cyanide solution, and the formation constant (K₆) for the complex ion [Fe(CN)₆]⁴⁻ is provided. Here’s how to approach the problem:1. Identify the initial concentrations: - Initial moles of Fe²⁺ = 1.90×10⁻² mol - Volume of solution = 1.00 L - Therefore, initial concentration of Fe²⁺ = 1.90×10⁻² M - Concentration of CN⁻ = 1.480 M2. Use the formation constant (K₆) value and set up the equilibrium expression: - K₆ = [Fe(CN)₆]⁴⁻ / ([Fe²⁺] [CN⁻]⁶) - Solve for [Fe²⁺] knowing K₆ is very large, implying that most Fe²⁺ is complexed.3. Calculate the equilibrium concentration of Fe²⁺ by considering the shift in equilibrium due to complex formation.This process requires understanding of equilibrium, stoichiometry, and calculations involving large formation constants to predict the extent of reaction and the concentration of ions in the solution at equilibrium.

Given, Moles of Iron(ll) nitrate =1.90×10-2 mol Volume of solution = 1.00 L Concentration of CN-=…

What is the approximate concentration of free Fe2+ ion at equilibrium when 1.90x102 mol iron(II) nitrate is added to 1.00 L of solution that is 1.480 M in CN. For [Fe(CN)6l*, Ke= 1.0x1035, (Fe*) =| M

What is the approximate concentration of free Fe2+ ion at equilibrium when 1.90x102 mol iron(II) nitrate is added to 1.00 L of solution that is 1.480 M in CN. For [Fe(CN)6l, Ke= 1.0x1035, (Fe) =| M

Chemistry

10th Edition

ISBN:9781305957404

Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Chapter1: Chemical Foundations

Section: Chapter Questions

Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...

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Question

**Problem Statement:**

What is the approximate concentration of free Fe²⁺ ion at equilibrium when 1.90×10⁻² mol iron(II) nitrate is added to 1.00 L of solution that is 1.480 M in CN⁻? For [Fe(CN)₆]⁴⁻, K₆ = 1.0×10³⁵.

**Solution Box:**

[Fe²⁺] = ______ M

**Explanation:**

This problem involves calculating the concentration of free Fe²⁺ ions in solution after an equilibrium is established. Iron(II) nitrate is added to a cyanide solution, and the formation constant (K₆) for the complex ion [Fe(CN)₆]⁴⁻ is provided. Here’s how to approach the problem:

1. Identify the initial concentrations:
- Initial moles of Fe²⁺ = 1.90×10⁻² mol
- Volume of solution = 1.00 L
- Therefore, initial concentration of Fe²⁺ = 1.90×10⁻² M
- Concentration of CN⁻ = 1.480 M

2. Use the formation constant (K₆) value and set up the equilibrium expression:
- K₆ = [Fe(CN)₆]⁴⁻ / ([Fe²⁺] [CN⁻]⁶)
- Solve for [Fe²⁺] knowing K₆ is very large, implying that most Fe²⁺ is complexed.

3. Calculate the equilibrium concentration of Fe²⁺ by considering the shift in equilibrium due to complex formation.

This process requires understanding of equilibrium, stoichiometry, and calculations involving large formation constants to predict the extent of reaction and the concentration of ions in the solution at equilibrium.

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