Answer :
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].
### 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].
