Answer :
Let's break down the problem step-by-step for clarity and to understand how the final answer was achieved.
Given Data:
1. The equilibrium constant for the reaction, \( K_{eq} = 620.4 \).
2. Volume of ferric nitrate solution, \( V_{Fe(NO_3)_3} = 10.0 \) mL.
3. Concentration of ferric nitrate, \( [Fe(NO_3)_3] = 0.05 \) M.
4. Volume of NaNCS solution, \( V_{NaNCS} = 2.0 \) mL.
5. Concentration of NaNCS, \( [NaNCS] = 5.0 \times 10^{-4} \) M.
6. Volume of distilled water added, \( V_{H_2O} = 8.0 \) mL.
7. Assume all NCS\(^-\) is converted to FeNCS\(^{2+}\).
Step-by-Step Solution:
1. Calculate the Total Volume of the Solution:
[tex]\[ V_{total} = V_{Fe(NO_3)_3} + V_{NaNCS} + V_{H_2O} \][/tex]
[tex]\[ V_{total} = 10.0 \, \text{mL} + 2.0 \, \text{mL} + 8.0 \, \text{mL} = 20.0 \, \text{mL} \][/tex]
2. Calculate the Initial Moles of Ferric Nitrate, \(Fe^{3+}\):
[tex]\[ \text{moles}_{Fe(NO_3)_3} = \text{concentration} \times \text{volume (in L)} \][/tex]
Since \(1 \, \text{mL} = 0.001 \, \text{L}\):
[tex]\[ \text{moles}_{Fe(NO_3)_3} = 0.05 \, \text{M} \times \left( \frac{10.0 \, \text{mL}}{1000} \right) \][/tex]
[tex]\[ \text{moles}_{Fe(NO_3)_3} = 0.05 \, \text{M} \times 0.01 \, \text{L} = 0.0005 \, \text{moles} \][/tex]
3. Calculate the Initial Moles of NCS\(^-\):
[tex]\[ \text{moles}_{NCS^-} = \text{concentration} \times \text{volume (in L)} \][/tex]
[tex]\[ \text{moles}_{NCS^-} = 5.0 \times 10^{-4} \, \text{M} \times \left( \frac{2.0 \, \text{mL}}{1000} \right) \][/tex]
[tex]\[ \text{moles}_{NCS^-} = 5.0 \times 10^{-4} \, \text{M} \times 0.002 \, \text{L} = 0.000001 \, \text{moles} = 1.0 \times 10^{-6} \, \text{moles} \][/tex]
4. Equilibrium Concentration of NCS\(^-\):
Given that all NCS\(^-\) is converted to FeNCS\(^{2+}\), the concentration of NCS\(^-\) at equilibrium is zero.
Therefore, the equilibrium concentration of NCS\(^-\) is:
[tex]\[ [ NCS^- ]_{eq} = 0 \, \text{M} \][/tex]
In conclusion, the equilibrium concentration of the NCS[tex]\(^-\)[/tex] ion in this solution is [tex]\(0 \, \text{M}\)[/tex].
Given Data:
1. The equilibrium constant for the reaction, \( K_{eq} = 620.4 \).
2. Volume of ferric nitrate solution, \( V_{Fe(NO_3)_3} = 10.0 \) mL.
3. Concentration of ferric nitrate, \( [Fe(NO_3)_3] = 0.05 \) M.
4. Volume of NaNCS solution, \( V_{NaNCS} = 2.0 \) mL.
5. Concentration of NaNCS, \( [NaNCS] = 5.0 \times 10^{-4} \) M.
6. Volume of distilled water added, \( V_{H_2O} = 8.0 \) mL.
7. Assume all NCS\(^-\) is converted to FeNCS\(^{2+}\).
Step-by-Step Solution:
1. Calculate the Total Volume of the Solution:
[tex]\[ V_{total} = V_{Fe(NO_3)_3} + V_{NaNCS} + V_{H_2O} \][/tex]
[tex]\[ V_{total} = 10.0 \, \text{mL} + 2.0 \, \text{mL} + 8.0 \, \text{mL} = 20.0 \, \text{mL} \][/tex]
2. Calculate the Initial Moles of Ferric Nitrate, \(Fe^{3+}\):
[tex]\[ \text{moles}_{Fe(NO_3)_3} = \text{concentration} \times \text{volume (in L)} \][/tex]
Since \(1 \, \text{mL} = 0.001 \, \text{L}\):
[tex]\[ \text{moles}_{Fe(NO_3)_3} = 0.05 \, \text{M} \times \left( \frac{10.0 \, \text{mL}}{1000} \right) \][/tex]
[tex]\[ \text{moles}_{Fe(NO_3)_3} = 0.05 \, \text{M} \times 0.01 \, \text{L} = 0.0005 \, \text{moles} \][/tex]
3. Calculate the Initial Moles of NCS\(^-\):
[tex]\[ \text{moles}_{NCS^-} = \text{concentration} \times \text{volume (in L)} \][/tex]
[tex]\[ \text{moles}_{NCS^-} = 5.0 \times 10^{-4} \, \text{M} \times \left( \frac{2.0 \, \text{mL}}{1000} \right) \][/tex]
[tex]\[ \text{moles}_{NCS^-} = 5.0 \times 10^{-4} \, \text{M} \times 0.002 \, \text{L} = 0.000001 \, \text{moles} = 1.0 \times 10^{-6} \, \text{moles} \][/tex]
4. Equilibrium Concentration of NCS\(^-\):
Given that all NCS\(^-\) is converted to FeNCS\(^{2+}\), the concentration of NCS\(^-\) at equilibrium is zero.
Therefore, the equilibrium concentration of NCS\(^-\) is:
[tex]\[ [ NCS^- ]_{eq} = 0 \, \text{M} \][/tex]
In conclusion, the equilibrium concentration of the NCS[tex]\(^-\)[/tex] ion in this solution is [tex]\(0 \, \text{M}\)[/tex].
