Answer :
To find the expected equilibrium concentration of [tex]\( \text{FeSCN}^{2+} \)[/tex] in the solution, we need to follow these steps:
1. Determine the initial moles of reactants:
- We are mixing 5.00 mL of 0.00200 M NaSCN solution with 5.00 mL of 0.500 M [tex]\( \text{Fe(NO}_3\text{)}_3 \)[/tex] solution.
- Calculate moles of NaSCN:
[tex]\[
\text{Moles of NaSCN} = \text{concentration} \times \text{volume} = 0.00200 \, \text{M} \times \frac{5.00 \, \text{mL}}{1000} = 1.00 \times 10^{-5} \, \text{moles}
\][/tex]
- Calculate moles of [tex]\( \text{Fe(NO}_3\text{)}_3 \)[/tex]:
[tex]\[
\text{Moles of Fe(NO}_3\text{)}_3 = 0.500 \, \text{M} \times \frac{5.00 \, \text{mL}}{1000} = 0.0025 \, \text{moles}
\][/tex]
2. Identify the limiting reactant:
- In this reaction, the product [tex]\( \text{FeSCN}^{2+} \)[/tex] is formed via [tex]\( \text{SCN}^- \)[/tex] reacting with [tex]\( \text{Fe}^{3+} \)[/tex] ions.
- Since the moles of [tex]\( \text{SCN}^- \)[/tex] (1.00 × 10⁻⁵ moles) are much less than the moles of [tex]\( \text{Fe}^{3+} \)[/tex] (0.0025 moles), [tex]\( \text{SCN}^- \)[/tex] is the limiting reactant.
3. Calculate the concentration of [tex]\( \text{FeSCN}^{2+} \)[/tex]:
- Since the reaction goes to completion, all the [tex]\( \text{SCN}^- \)[/tex] ions react with [tex]\( \text{Fe}^{3+} \)[/tex] ions to form [tex]\( \text{FeSCN}^{2+} \)[/tex].
- Therefore, the moles of [tex]\( \text{FeSCN}^{2+} \)[/tex] will be equal to the initial moles of [tex]\( \text{SCN}^- \)[/tex], which is [tex]\( 1.00 \times 10^{-5} \)[/tex] moles.
4. Calculate the concentration of [tex]\( \text{FeSCN}^{2+} \)[/tex] in the final solution:
- The total volume of the solution after mixing is 10.00 mL (5.00 mL + 5.00 mL).
- Convert the total volume into liters:
[tex]\[
\text{Final volume} = \frac{10.00 \, \text{mL}}{1000} = 0.0100 \, \text{L}
\][/tex]
- Calculate the concentration:
[tex]\[
\text{Concentration of } \text{FeSCN}^{2+} = \frac{1.00 \times 10^{-5} \, \text{moles}}{0.0100 \, \text{L}} = 0.00100 \, \text{M}
\][/tex]
So, the expected equilibrium concentration of [tex]\( \text{FeSCN}^{2+} \)[/tex] in the solution is [tex]\( 0.00100 \, \text{M} \)[/tex].
1. Determine the initial moles of reactants:
- We are mixing 5.00 mL of 0.00200 M NaSCN solution with 5.00 mL of 0.500 M [tex]\( \text{Fe(NO}_3\text{)}_3 \)[/tex] solution.
- Calculate moles of NaSCN:
[tex]\[
\text{Moles of NaSCN} = \text{concentration} \times \text{volume} = 0.00200 \, \text{M} \times \frac{5.00 \, \text{mL}}{1000} = 1.00 \times 10^{-5} \, \text{moles}
\][/tex]
- Calculate moles of [tex]\( \text{Fe(NO}_3\text{)}_3 \)[/tex]:
[tex]\[
\text{Moles of Fe(NO}_3\text{)}_3 = 0.500 \, \text{M} \times \frac{5.00 \, \text{mL}}{1000} = 0.0025 \, \text{moles}
\][/tex]
2. Identify the limiting reactant:
- In this reaction, the product [tex]\( \text{FeSCN}^{2+} \)[/tex] is formed via [tex]\( \text{SCN}^- \)[/tex] reacting with [tex]\( \text{Fe}^{3+} \)[/tex] ions.
- Since the moles of [tex]\( \text{SCN}^- \)[/tex] (1.00 × 10⁻⁵ moles) are much less than the moles of [tex]\( \text{Fe}^{3+} \)[/tex] (0.0025 moles), [tex]\( \text{SCN}^- \)[/tex] is the limiting reactant.
3. Calculate the concentration of [tex]\( \text{FeSCN}^{2+} \)[/tex]:
- Since the reaction goes to completion, all the [tex]\( \text{SCN}^- \)[/tex] ions react with [tex]\( \text{Fe}^{3+} \)[/tex] ions to form [tex]\( \text{FeSCN}^{2+} \)[/tex].
- Therefore, the moles of [tex]\( \text{FeSCN}^{2+} \)[/tex] will be equal to the initial moles of [tex]\( \text{SCN}^- \)[/tex], which is [tex]\( 1.00 \times 10^{-5} \)[/tex] moles.
4. Calculate the concentration of [tex]\( \text{FeSCN}^{2+} \)[/tex] in the final solution:
- The total volume of the solution after mixing is 10.00 mL (5.00 mL + 5.00 mL).
- Convert the total volume into liters:
[tex]\[
\text{Final volume} = \frac{10.00 \, \text{mL}}{1000} = 0.0100 \, \text{L}
\][/tex]
- Calculate the concentration:
[tex]\[
\text{Concentration of } \text{FeSCN}^{2+} = \frac{1.00 \times 10^{-5} \, \text{moles}}{0.0100 \, \text{L}} = 0.00100 \, \text{M}
\][/tex]
So, the expected equilibrium concentration of [tex]\( \text{FeSCN}^{2+} \)[/tex] in the solution is [tex]\( 0.00100 \, \text{M} \)[/tex].
