# Formation Constant

A complex ion contains a metal center with surrounding molecules. The equilibrium constant for the formation of a complex ion, Kf, is called a formation constant and tend to be very large.

$\mathrm{metal}(aq) + \mathrm{ligands}(aq) \rightleftharpoons \mathrm{complex~ion}(aq) \qquad K_f = \dfrac{[\mathrm{complex~ion}]}{[\mathrm{metal}][\mathrm{ligands}]}$

We can determine the equilibrium concentration of a complex ion (or dissociated ions) from a Kf.

Suppose we have a 10.0 L aqueous solution that initially contains 0.200 M [AgCl2] complex ion.

$\mathrm{Ag^+}(aq) + \mathrm{2Cl^-}(aq) \rightleftharpoons \mathrm{[AgCl_2]^+}(aq) \qquad K_{\mathrm{f}} = 1.8\times 10^{5}$

We will answer the following questions:

1. What the Ag+ concentration (in M), mass concentration (g mol–1), and mass (in g) at equilibrium?
2. What is the mass concentration (in g L–1) of AgCl2+ complex ion and mass (in g) that dissociated in solution?

Begin by writing out an ICE table.

 Ag+(aq) 2Cl– (aq) $$\rightleftharpoons$$ AgCl2+(aq) I 0 0 0.200 C +x +2x –x E x 2x 0.200 -x

Solve for [Ag+].

\begin{align*} \dfrac{[\mathrm{AgCl_2^+}]}{[\mathrm{Ag^+}]\mathrm{[Cl^-]^2}} &= 1.8\times 10^{5}\\ \dfrac{0.200-x}{(x)(2x)^2} &= 1.8\times 10^{5}\\ \dfrac{0.200}{4x^3} &= 1.8\times 10^{5}\\ 0.200 &= 7.2\times 10^{5}x^3\\ x &= 6.52\times 10^{-3}\\[2ex] [\mathrm{Ag^+}] &= 6.52\times 10^{-3}~M \end{align*}

From here we will address question 1.

Using molar mass (107.87 g mol–1), we can get the mass concentration of Ag+ dissolved. \begin{align*} \rho_{\mathrm{Ag^+}} &= \dfrac{6.52\times 10^{-3}~\mathrm{mol~Ag^{+}}}{\mathrm{L}} \left ( \dfrac{107.87~\mathrm{g~mol^{-1}}}{\mathrm{mol~Ag^+}} \right ) = \dfrac{7.03\times 10^{-1}~\mathrm{g}}{\mathrm{L}} \end{align*}

Using volume, we can find the total mass of Ag+ in solution. \begin{align*} m_{\mathrm{Ag^+}} = 7.03\times 10^{-1}~\mathrm{g~L^{-1}} \left ( 10.0~\mathrm{L} \right ) = 7.03~\mathrm{g} \end{align*}

We now address question 2 where we find how much AgCl2+ complex ion dissociated in solution.

Using stoichiometry and molar mass (143.32 g mol–1), we can get the mass concentration of AgCl2 dissolved.

\begin{align*} \rho_{\mathrm{AgCl_2}} &= \dfrac{6.52\times 10^{-3}~\mathrm{mol~Ag^{+}}}{\mathrm{L}} \left ( \dfrac{1~\mathrm{mol~AgCl_2}}{1~\mathrm{mol~Ag^{+}}} \right ) \left ( \dfrac{143.32~\mathrm{g~mol^{-1}}}{\mathrm{mol~AgCl_2}} \right ) = \dfrac{9.34\times 10^{-1}~\mathrm{g}}{\mathrm{L}} \end{align*}

Using volume, we can find the total mass of AgCl2+ that actually dissociated in solution. \begin{align*} m_{\mathrm{AgCl_2^+}} = 9.34\times 10^{-1}~\mathrm{g~L^{-1}} \left ( 10.0~\mathrm{L} \right ) = 9.34~\mathrm{g} \end{align*}

## Practice Problems

### Problem 1

An initial 0.100 M [Cu(CN)2] aqueous solution is created. What is the concentration (in M) and mass concentration (in g L–1) of Cu+? Also, what is the mass of Cu+ (in g) dissolved if the final volume of the solution was 10.0 L?

$\mathrm{Cu^+}(aq) + \mathrm{2CN^-}(aq) \rightleftharpoons \mathrm{[Cu(CN)_2]^-}(aq) \qquad K_{\mathrm{f}} = 1.0\times 10^{16}$

Solution

Begin by writing out an ICE table.

 Cu+(aq) 2CN–(aq) $$\rightleftharpoons$$ CuCN2–(aq) I 0 0 0.100 C +x +2x –x E x 2x 0.100 - x

Write out the equilibrium expression and solve for x.

\begin{align*}\require{color} \dfrac{[\mathrm{CuCN_2^-}]}{[\mathrm{Cu^{+}}][\mathrm{CN^-}]^2} &= 1.0\times 10^{16}\\ \dfrac{0.100\textcolor{red}{-x}}{x(2x)^2} &= 1.0\times 10^{16}\\ \dfrac{0.100}{4x^3} &= 1.0\times 10^{16}\\ x &= 1.36\times 10^{-6}\\[2ex] [\mathrm{Cu^+}] &= 1.36\times 10^{-6}~M\\ [\mathrm{CN^{-}}] &= 2.71\times 10^{-6}~M \end{align*}

Using the molar mass of Cu+ (63.546 g mol–1), determine the mass concentration of Cu+ dissolved.

\begin{align*} \rho_{\mathrm{Cu^+}} &= \dfrac{1.36\times 10^{-6}~\mathrm{mol~Cu^{+}}}{\mathrm{L}} \left ( \dfrac{63.546~\mathrm{g}}{\mathrm{mol~Cu^+}} \right ) = \dfrac{8.64\times 10^{-5}~\mathrm{g}}{\mathrm{L}} \end{align*}

Use the total volume of solution to determine the mass of Cu+ dissolved.

$m_{\mathrm{Cu^+}} =\left (8.64\times 10^{-5}~\mathrm{g~L^{-1}} \right ) \left ( 10.0~\mathrm{L} \right ) = 8.64\times 10^{-4}~\mathrm{g~dissolved}$

Previous
Next