Chapter 22, Problem 43QP

Interpretation Introduction

Interpretation:

The formation constant for ${\left[\text{Fe}{\left({\text{H}}_2\text{O}\right)}_5\text{NCS}\right]}^{2+}$ for the given conditions is to be calculated.

Concept introduction:

The formation constant describes the formation of a complex ion that takes place from its central ion and attached ligands. Its symbol is $K_{\text{f}}$.

Answer to Problem 43QP

Solution: $1.4\times {10}^2{\text{ M}}^{-1}$

Explanation of Solution

Given information:

Volume of $\text{Fe}{\left({\text{NO}}_3\right)}_3$ is $1.0\text{ mL}$.

Concentration of $\text{Fe}{\left({\text{NO}}_3\right)}_3$ is $0.20\text{ M}$.

Volume of $\text{KSCN}$ is $1.0\text{ mL}$.

Concentration of $\text{KSCN}$ is $\text{1}\text{.0}\times {\text{10}}^{-3}\text{ M}$.

Volume of ${\text{HNO}}_3$ is $8.0\text{ mL}$.

Concentration of ${\left[\text{Fe}{\left({\text{H}}_2\text{O}\right)}_5\text{NCS}\right]}^{2+}$ is $7.3\times {\text{10}}^{-5}\text{ M}$.

The given chemical equation is as:

${\left[\text{Fe}{\left({\text{H}}_2\text{O}\right)}_6\right]}^{3+}+{\text{SCN}}^{-}\rightleftharpoons {\text{H}}_2\text{O}+{\left[\text{Fe}{\left({\text{H}}_2\text{O}\right)}_5\text{NCS}\right]}^{2+}$

According to the above equation, the expression for theformation constant $\left(K_{\text{f}}\right)$ is as:

$K_{\text{f}}=\frac{{\left[\text{Fe}{\left({\text{H}}_2\text{O}\right)}_5\text{NCS}\right]}^{2+}}{{\left[\text{Fe}{\left({\text{H}}_2\text{O}\right)}_6\right]}^{3+}\left[{\text{SCN}}^{-}\right]}$

According to the given information, the original volumes of both $\text{Fe}\left(\text{III}\right)$ and ${\text{SCN}}^{-}$ were $1.0\text{ mL}$ and the final volume is $10.0\text{ mL}$. This difference represents the ten-fold dilution. So, the concentrations of $\text{Fe}\left(\text{III}\right)$ and ${\text{SCN}}^{-}$ will become:

$\begin{array}{c}\text{Fe}\left(\text{III}\right)=\frac{0.20\text{ M}}{10}\\ =0.020\text{ M}\\ {\text{SCN}}^{-}=\frac{\text{1}\text{.0}\times {\text{10}}^{-3}\text{ M}}{10}\\ =\text{1}\text{.0}\times {\text{10}}^{-4}\text{ M}\end{array}$

Now, make an ICE table as:

$\begin{array}{l}{\left[\text{Fe}{\left({\text{H}}_2\text{O}\right)}_6\right]}^{3+}+{\text{SCN}}^{-}\rightleftharpoons {\text{H}}_2\text{O}+{\left[\text{Fe}{\left({\text{H}}_2\text{O}\right)}_5\text{NCS}\right]}^{2+}\\ \text{Initial}\left(\text{M}\right)\text{ 0}\text{.020 1}\text{.0}\times {\text{10}}^{-4}\text{ 0 0}\\ \text{Change}\left(\text{M}\right)-7.3\times {\text{10}}^{-5}-7.3\times {\text{10}}^{-5}x7.3\times {\text{10}}^{-5}\\ \text{Equilibrium}\left(\text{M}\right)\text{ 0}\text{.020 2}\text{.7}\times {\text{10}}^{-5}x7.3\times {\text{10}}^{-5}\end{array}$

From the ICE table, the expression of $K_{\text{f}}$ will be as:

$\begin{array}{c}{K}_{\text{f}}=\frac{7.3\times {\text{10}}^{-5}\cancel{\text{M}}}{\left(0.020\cancel{\text{M}}\right)\left(2.7\times {10}^{-5}\text{M}\right)}\\ \approx 1.4\times {10}^2{\text{ M}}^{-1}\end{array}$

Conclusion

Hence, the formation constant of ${\left[\text{Fe}{\left({\text{H}}_2\text{O}\right)}_5\text{NCS}\right]}^{2+}$ is $1.4\times {10}^2{\text{ M}}^{-1}$.