Answered

Welcome to Westonci.ca, the place where your questions find answers from a community of knowledgeable experts. Our platform provides a seamless experience for finding reliable answers from a network of experienced professionals. Our platform provides a seamless experience for finding reliable answers from a network of experienced professionals.

Consider the reaction between [tex]Fe^{3+}[/tex] and [tex]NCS^{-}[/tex] ions. The equilibrium constant for this reaction is 620.4. Calculate the equilibrium concentration of NCS ion in a solution that contains:

- [tex]10.0 \, mL[/tex] of [tex]0.05 \, M[/tex] ferric nitrate in [tex]1 \, M \, HNO_3[/tex]
- [tex]2.0 \, mL[/tex] of [tex]5.0 \times 10^{-4} \, M \, NaNCS[/tex]
- [tex]8.0 \, mL[/tex] of distilled water

Assume that all the NCS is converted to [tex]FeNCS^{2+}[/tex].

A. [tex]1.00 \times 10^{-6} \, M[/tex]
B. [tex]6.07 \times 10^{-8} \, M[/tex]
C. [tex]3.04 \times 10^{-6} \, M[/tex]
D. [tex]0 \, M[/tex]

Sagot :

GinnyAnswer
To find the equilibrium concentration of \( \text{NCS}^{-} \) ions, we need to use the given data and follow these steps:

### Step 1: Determine the Total Volume of the Solution
We are given:
- Volume of ferric nitrate solution: \( 10.0 \, \text{mL} \)
- Volume of \( \text{NaNCS} \) solution: \( 2.0 \, \text{mL} \)
- Volume of distilled water: \( 8.0 \, \text{mL} \)

Adding these volumes together gives the total volume of the solution:
[tex]\[ \text{Total Volume} = 10.0 \, \text{mL} + 2.0 \, \text{mL} + 8.0 \, \text{mL} = 20.0 \, \text{mL} \][/tex]

### Step 2: Calculate Initial Moles of \( \text{Fe}^{3+} \) and \( \text{NCS}^{-} \)
- Concentration of ferric nitrate solution: \( 0.05 \, \text{M} \)
- Volume of ferric nitrate solution: \( 10.0 \, \text{mL} \) \( = 0.010 \, \text{L} \)

[tex]\[ \text{Moles of } \text{Fe}^{3+} = 0.05 \, \text{M} \times 0.010 \, \text{L} = 0.0005 \, \text{moles} \][/tex]

- Concentration of \( \text{NaNCS} \) solution: \( 5.0 \times 10^{-4} \, \text{M} \)
- Volume of \( \text{NaNCS} \) solution: \( 2.0 \, \text{mL} \) \( = 0.002 \, \text{L} \)

[tex]\[ \text{Moles of } \text{NCS}^{-} = 5.0 \times 10^{-4} \, \text{M} \times 0.002 \, \text{L} = 1.0 \times 10^{-6} \, \text{moles} \][/tex]

### Step 3: Calculate Initial Concentrations in the Final Mixture
- Total volume of the solution: \( 20.0 \, \text{mL} \) \( = 0.020 \, \text{L} \)

[tex]\[ \text{Initial concentration of } \text{Fe}^{3+} = \frac{0.0005 \, \text{moles}}{0.020 \, \text{L}} = 0.025 \, \text{M} \][/tex]

[tex]\[ \text{Initial concentration of } \text{NCS}^{-} = \frac{1.0 \times 10^{-6} \, \text{moles}}{0.020 \, \text{L}} = 5.0 \times 10^{-5} \, \text{M} \][/tex]

### Step 4: Establish the Equilibrium Condition
Given that all the \( \text{NCS}^- \) is converted to \( \text{FeNCS}^{2+} \), we have:
[tex]\[ [\text{FeNCS}^{2+}] = [\text{NCS}^{-}]_\text{initial} \][/tex]

Thus,
[tex]\[ [\text{FeNCS}^{2+}] = 5.0 \times 10^{-5} \, \text{M} \][/tex]

### Step 5: Equilibrium Concentration of \( \text{NCS}^{-} \)
At equilibrium, since all \( \text{NCS}^- \) is consumed to form \( \text{FeNCS}^{2+} \), the concentration of \( \text{NCS}^{-} \) at equilibrium is:
[tex]\[ [\text{NCS}^{-}]_\text{eq} = 0 \, \text{M} \][/tex]

Thus, the equilibrium concentration of the [tex]\( \text{NCS}^{-} \)[/tex] ion in the solution is [tex]\( \boxed{0 \, \text{M}} \)[/tex].
Thank you for visiting. Our goal is to provide the most accurate answers for all your informational needs. Come back soon. We hope our answers were useful. Return anytime for more information and answers to any other questions you have. Westonci.ca is here to provide the answers you seek. Return often for more expert solutions.