Question:
# **Experiment 23**
## **Advance Study Assignment: Determination of the Equilibrium Constant for a Chemical Reaction**
1. A student mixes 5.00 mL 2.00 × 10<sup>-3</sup> M Fe(NO<sub>3</sub>)<sub>3</sub> in 1 M HNO<sub>3</sub> with 3.00 mL 2.00 × 10<sup>-3</sup> M KSCN and 2.00 mL of water. She finds that in the equilibrium mixture the concentration of FeSCN<sup>2+</sup> is 7.0 × 10<sup>-5</sup> M. Find *K* for the reaction Fe<sup>2+</sup>(aq) + SCN<sup>-</sup>(aq) with Fe(SCN)<sup>2+</sup>(aq).
**Step 1** Find the number of moles Fe<sup>2+</sup> and SCN<sup>-</sup> initially present. (Use Eq. 3.)
$$
\begin{aligned}
2.00 \times 10^{-3} \, \text{mM} & \times 0.005L + 1.00 \times 10^{-2} \, \text{mM} \\
1L & \times 0.005L + 4.00 \times 10^{-2} \, \text{mM} & \times 3.10^{-2} \, \text{moles Fe}^{2+} \\
10.00 \, \text{mL} & \times 0.005L + 7.00 \times 10^{-3} \, \text{mM} & \times 3.10^{-2} \, \text{moles Fe}^{2+} \\
10.00 \, \text{mL} & \times 4.00^{-2} \, \text{moles FeSCN}^{2+}
\end{aligned}
$$
**Step 2** How many moles of FeSCN<sup>2+</sup> are in the mixture at equilibrium? What is the volume of the equilibrium mixture? (Use Eq. 3.)
$$
\begin{aligned}
2.00 \times 10^{-3} \, \text{mM} & \times 0.005L + 7.00 \times 10^{-3} \, \text{mM} \\
1L & \times 0.005L + 4.00 \times 10^{-3} \, \text{mM} \\
10.00 \, \text{mL} & \times 3.10^{-2} \, \text{moles FeSCN}^{2+}
\end{aligned}
$$
How many moles of Fe<sup>2+</sup> and SCN<sup>-</sup> are used up in making the FeSCN<sup>2+</sup>?
$$
\begin{aligned}
8 \, \text{mM} \, \text{Fe}^{2+} + \text{SCN}^{2+} & + 7.00 \times 10^{-3} \, \text{mM} \\
7.00 \, \text{mM} & + 7.00 \times 10^{-3} \, \text{mM} \\
7.00 \, \text{moles Fe}^{2+} & + 7.00 \times 10^{-3} \, \text{moles SCN}^{2+}
\end{aligned}
$$
**Step 3** How many moles of Fe<sup>2+</sup> and SCN<sup>-</sup> remain in the solution at equilibrium? (Use Eq. 4 and the results of Steps 1 and 2.)
$$
\begin{aligned}
(1.00 \times 10^{-5} - 7.00 \times 10^{-5}) & = 7.00 \times 10^{-4} \, \text{mM} \\
(4.00 \times 10^{-4} - 7.00 \times 10^{-5}) & = 5.30 \times 10^{-4} \, \text{mM} \\
1.00 \times 10^{-4} & = 5.30 \times 10^{-4} \, \text{moles Fe}^{2+} \\
1.00 \times 10^{-4} & = 5.30 \times 10^{-4} \, \text{moles SCN}^{2+}
\end{aligned}
$$
**Step 4** What are the concentrations of Fe<sup>2+</sup>, SCN<sup>-</sup>, and FeSCN<sup>2+</sup> at equilibrium? (Use Eq. 3 and the results of Step 2 and Step 3.)
$$
\begin{aligned}
9.30 \times 10^{-2} & = 5.30 \times 10^{-2} + 5.30 \times 10^{-1} \\
0.01L & = 5.30 \times 10^{-2} + 5.30 \times 10^{-1} \\
1.010^{-2} & = 5.30 \times 10^{-2} + 5.30 \times 10^{-1}
\end{aligned}
$$
$$
\begin{aligned}
7.00 \times 10^{-3} & = 7.00 \times 10^{-3} + 5.30 \times 10^{-2} \\
10.00 \times 10^{-3} & = 5.30 \times 10^{-3} + 5.30 \times 10^{-1}
\end{aligned}
$$
**Step 5** What is the value of *K* for the reaction? (Use Eq. 2 and the results of Step 4.)
$$
\begin{aligned}
K_0 &= \frac{142}{142} \\
&\quad \text{(continued on following page)}
\end{aligned}
$$
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