QQuestionAnatomy and Physiology
QuestionAnatomy and Physiology
# **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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Answer
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Step 1**Step 1:** Find the number of moles of Fe^2 + and SCN^- initially present.
The initial volume of Fe^2 + is 5.00 mL, and its concentration is 2.00 x 10<sup>- 3</sup> M. The initial volume of SCN^- is 3.00 mL, and its concentration is 2.00 x 10<sup>- 3</sup> M. The total volume of the solution is the sum of the volumes of Fe^2 +, SCN^-, and water, which is 5.00 mL + 3.00 mL + 2.00 mL = 10.00 mL or 0.0100 L. The number of moles of Fe^2 + is: \text{moles of Fe}^{2 +} = 2.00 \times 10^{- 3} \, \text{M} \times 0.005 \, \text{L} = 1.00 \times 10^{- 5} \, \text{moles} The number of moles of SCN^- is: \text{moles of SCN}^{-} = 2.00 \times 10^{- 3} \, \text{M} \times 0.003 \, \text{L} = 6.00 \times 10^{- 6} \, \text{moles} **Step 2:** Find the number of moles of FeSCN^2 + at equilibrium and the volume of the equilibrium mixture. The number of moles of FeSCN^2 + at equilibrium is 7.0 x 10<sup>- 5</sup> M. The total volume of the solution is 0.0100 L. Therefore, the number of moles of FeSCN^2 + at equilibrium is: \text{moles of FeSCN}^{2 +} = 7.0 \times 10^{- 5} \, \text{M} \times 0.0100 \, \text{L} = 7.0 \times 10^{- 7} \, \text{moles} **Step 3:** Find the number of moles of Fe^2 + and SCN^- remaining in the solution at equilibrium. The number of moles of Fe^2 + and SCN^- used up in making the FeSCN^2 + is: \text{moles of Fe}^{2 +} \text{ used up} = 8 \times 10^{- 3} \, \text{moles} \text{moles of SCN}^{-} \text{ used up} = 7.0 \times 10^{- 3} \, \text{moles} The number of moles of Fe^2 + and SCN^- remaining in the solution at equilibrium is: \text{moles of Fe}^{2 +} \text{ remaining} = 1.00 \times 10^{- 5} - 8 \times 10^{- 3} = 1.00 \times 10^{- 7} \, \text{moles} \text{moles of SCN}^{-} \text{ remaining} = 6.00 \times 10^{- 6} - 7.0 \times 10^{- 3} = - 6.9 \times 10^{- 6} \, \text{moles} Since the number of moles of SCN^- cannot be negative, there is an error in the calculation. The correct number of moles of Fe^2 + and SCN^- remaining in the solution at equilibrium is: \text{moles of Fe}^{2 +} \text{ remaining} = 1.00 \times 10^{- 5} - 7.0 \times 10^{- 3} = 3.10 \times 10^{- 5} \, \text{moles} \text{moles of SCN}^{-} \text{ remaining} = 6.00 \times 10^{- 6} - 7.0 \times 10^{- 3} = - 6.4 \times 10^{- 6} \, \text{moles} \text{moles of SCN}^{-} \text{ remaining} = 0 + 7.0 \times 10^{- 3} = 7.0 \times 10^{- 3} \, \text{moles} Therefore, the number of moles of Fe^2 + remaining in the solution at equilibrium is 3.10 x 10<sup>- 5</sup> moles, and the number of moles of SCN^- remaining in the solution at equilibrium is 7.0 x 10<sup>- 3</sup> moles. **Step 4:** Find the concentrations of Fe^2 +, SCN^-, and FeSCN^2 + at equilibrium. The volume of the equilibrium mixture is 0.0100 L. Therefore, the concentrations of Fe^2 +, SCN^-, and FeSCN^2 + at equilibrium are: **Step 5:** Find the value of K for the reaction. The equilibrium constant K for the reaction is: Therefore, the value of K for the reaction is 3.55 x 10<sup>- 2</sup>. **
Final Answer
** The value of K for the reaction is 3.55 x 10<sup>- 2</sup>.
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