Skip to main content
Chemistry LibreTexts

Metal Coordination Complexes

  • Page ID
  • \( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \) \( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)\(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\) \(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\)

    Learning Objectives

    • Explain what complex ions or metal complexes are.
    • Apply the concept of equilibrium in complex formation.
    • Calculate concentrations by applying the formation constant.
    • Derive the overall formation constant from stepwise formation constants.
    • Derive the dissociation constant from formation constant.

    Metals are Lewis acids because of their positive charge. When dissolved in water, they react with water to form hydrated compounds such as \(\ce{Na(H2O)6+}\) and \(\ce{Cu(H2O)6^2+}\). These are called metal complexes, or coordination compounds. These coordination reactions are Lewis acid-base reactions. The neutral molecules such as \(\ce{H2O}\) and \(\ce{NH3}\), and anions such as \(\ce{CN-}\), \(\ce{CH3COO-}\) are called ligands. The coordination reaction can be represented by

    \[\ce{CuSO4 + 6 H2O \rightleftharpoons} \ce{Cu(H2O)6^2+ + SO4^2-}\]

    Usually, copper sulfate solids are hydrated with 5 water molecules per CuSO_4 complex ( \(\mathrm{ {\color{Periwinkle} CuSO_4\cdot(H_2O)_5}}\)), and its color is light blue. When heated, it loses water of crystallization, and becomes \(\ce{CuSO4}\), which is colorless.

    Most people know that when ammonia is added to a \(\mathrm{ {\color{Periwinkle} Cu(H_2O)_6^{2+}}}\) solution, it turns deep blue. This is due to the formation of complexes:

    \[ \color{Periwinkle} {Cu(H_2O)_6^{2+} } + 4 NH_3 \: \rightleftharpoons {\color{Blue} Cu(H_2O)_2(NH_3)_4^{2+} + 4 H_2O}\]

    Actually, the above reaction takes place in steps. The \(\ce{H2O}\) molecules are displaced one at a time as the ammonia concentration, \(\ce{[NH3]}\), increases. As more \(\ce{NH3}\) is bonded to \(\ce{Cu^2+}\), the blue color deepens.

    Ammonia forms complexes with many metals. It forms a very strong complex with \(\ce{Ag+}\) such that \(\ce{AgCl}\) solid will dissolve in ammonia solution.

    \[\ce{AgCl_{\large{(s)}} + 2 NH3 \rightleftharpoons Ag(NH3)2+ + 2 Cl-}\]

    The silver ammonia complex is colorless, however.

    Other commonly encountered ligands are \(\ce{CN-}\), \(\ce{SCN-}\), \(\ce{Cl-}\), ethylene diamine (\(\ce{NH2CH3CH3NH2}\)), and acetate (\(\ce{CH3COO-}\)). For example,

    \(\mathrm{ {\color{Tan} Fe(H_2O)_6^{3+}} + SCN^-}\) \(\rightleftharpoons\) \(\mathrm{ {\color{Red} Fe(H_2O)_5SCN^{2+}} + H_2O}\)
    brown blood red

    Formation Constant of Complexes

    Complex Kf
    \(\ce{Ag(NH3)2}\) 1.6e7
    \(\ce{Ag(S2O3)2^3-}\) 1.7e13
    \(\ce{Al(OH)4-}\) 7.7e33
    \(\ce{AlF6^3-}\) 6.7e19
    \(\ce{Zn(EDTA)^2-}\) 3.8e16

    Formation of complexes is also a thermal dynamic phenomenon. For the equilibrium,

    \(\ce{Ag+ + 2 NH3 \rightleftharpoons Ag(NH3)2+}\),

    the formation constant is very large,

    \(K_{\ce f} = \ce{\dfrac{[Ag(NH3)2+]}{[Ag+] [NH3]^2}} = \textrm{1.6e7 M}^{-2}\)

    because \(\ce{[Ag+]}\) is very small in such an equilibrium.

    The formation constants of some other complexes are given in a table form on the right. Although only three metal ions are involved, the complexes are formed by five ligands, and the overall formation constants range from 1.6e7 to 7.7e33. The EDTA is a complicated organic molecule, \(\mathrm{( ^- OOCCH_2)_2}\ce{N-CH2-CH2-N(CH2COO- )2}\) with six sites (\(\ce{4\: O}\) and \(\ce{2\: N}\)) embracing the zinc ion in the zinc complex.

    Example 1

    Calculate \(\ce{[Ag+]}\) in a solution containing 0.10 M \(\ce{AgNO3}\) and 1.0 M \(\ce{NH3}\).

    For simplicity of formulation, let \(\mathrm{M = [Ag^+]}\). The equilibrium equation and concentration are:

    \(\begin{array}{cccccl}\ce{Ag+ &+ &2 NH3 &\rightleftharpoons &Ag(NH3)2+} &\hspace{10px}K_{\ce f} = \textrm{1.6e7 M}^{-2}\\
    x &&1.00-0.20+x &&0.10-x &

    \(\dfrac{0.10-x}{x (0.80+x)^2} = \textrm{1.6e7 M}^{-2}\)

    \(x = \textrm{9.7e-9 M}\)

    If we assume \(x \ce M = \ce{[Ag(NH3)2+]}\), the calculation will be very difficult. Try it and find out why.

    Suppose we now introduce 0.10 M \(\ce{Cl-}\) into the equilibrium; will a precipitate form? For \(\ce{AgCl}\), Ksp = 1.8e-10.

    Ans. Since 9.7e-9*0.10 = 9.7e-10 > Ksp, a precipitate will form.

    Example 2

    What is the solubility of \(\ce{AgCl}\) in a solution which contains 1.0 M \(\ce{NH3}\)? For \(\ce{Ag(NH3)2+}\), Kf = 1.6e7, and for \(\ce{AgCl}\), Ksp = 1.8e-10.

    First consider the equilibria:

    \(\ce{AgCl \rightleftharpoons Ag+ + Cl-} \hspace{15px} K_{\ce{sp}} = \textrm{1.8e-10 M}^2\)

    \(\ce{Ag+ + 2 NH3 \rightleftharpoons Ag(NH3)2+} \hspace{15px} K_{\ce f} = \textrm{1.6e7 M}^{-2}\)

    Adding the two equations together results in the equilibrium equation below.

    \(\begin{array}{cccccccl}\ce{AgCl &+ &2 NH3 &\rightleftharpoons &Ag(NH3)2+ &+ &Cl-} &\:\:\: K = K_{\ce{sp}} K_{\ce f} = \textrm{2.9e-3}\\
    &&1.0-x &&x &&x &\:\:\: \Leftarrow \textrm{equilibrium concentrations}

    where x is the molar solubility of \(\ce{AgCl}\).

    \(\dfrac{x^2}{(1.0-x)^2} = \textrm{2.9e-3}\)

    \(\dfrac{x}{(1.0-x)} = 0.054\)

    \(x = \textrm{0.051 M}\)

    Answer the following questions and review the discussion in Example 1.

    The solubility product of \(\ce{AgBr}\) is 5.0e-13 M2. Estimate the molar solubility of \(\ce{AgBr}\) in a 1.0 M \(\ce{NH3}\) solution.

    Ans. 2.8e-3 M

    Stepwise Formation Constants and Overall Constants

    As indicated earlier, the formation of a complex takes place in steps. The formation constants in these steps are called stepwise formation constants.

    For the reaction,

    \[\ce{Ag+ + NH3 \rightleftharpoons Ag(NH3)+}\]

    \[K_{\ce{\large f_{\Large 1}}} = \ce{\dfrac{[Ag(NH3)+]}{[Ag+] [NH3]}} = \textrm{2.2e3 M}\]

    And for the reaction,

    \[\ce{Ag(NH3)+ + NH3 \rightleftharpoons Ag(NH3)2+}\]

    \[K_{\ce{\large f_{\Large 2}}} = \ce{\dfrac{[Ag(NH3)2+]}{[Ag(NH3)+] [NH3]}} = \textrm{7.2e3 M}\]

    And obviously, for the overall reaction,

    \[\ce{Ag+ + 2 NH3 \rightleftharpoons Ag(NH3)2+}\]

    \[K_{\ce{\large f}} = \ce{\dfrac{[Ag(NH3)2+]}{[Ag+] [NH3]+}} = K_{\ce{\large f_{\Large 1}}} \times K_{\ce{\large f_{\Large 2}}} = \textrm{1.6e7 M}^2\]

    A generalized formula is

    \[K_{\ce{\large f}} = K_{\ce{\large f_{\Large 1}}} \times K_{\ce{\large f_{\Large 2}}} \times K_{\ce{\large f_{\Large 3}}} \times \: ...\]

    Dissociation Constants and the Formation Constants

    The reverse reaction of the complex formation is called dissociation, and the equilibrium constant is called dissociation constant Kd.

    \p\ce{Ag(NH3)2+ \rightleftharpoons Ag+ + 2 NH3}, \hspace{10px} K_{\ce d}.\]

    Obviously, we have

    \pK_{\ce d} = \dfrac{1}{K_{\ce f}}\]

    Similar to stepwise formation constants, we can also apply the concept to give a stepwise dissociation constant.

    Example 3

    Calculate \(\ce{[Ag+]}\) when equal volume of 0.10 M \(\ce{AgNO3}\) and 0.10 M \(\ce{Na2S2O3}\) solutions are mixed, giving Kf = 1.7e13 M-2 for \(\ce{Ag(S2O3)2^3-}\).

    When the solutions are mixed, \(\ce{[Ag(S2O3)2^3- ]} = \textrm{0.05 M}\). Let \(x \ce M = \ce{[Ag+]}\) at equilibrium. The dissociation equilibrium equation rather than the formation equilibrium equation is more convenient in this case.

    \ce{Ag(S2O3)2^3- &\rightleftharpoons &Ag+ &+ &2 S2O3^3-}\\
    0.05-x &&x &&2 x

    \(\dfrac{x (2 x)^2}{0.05-x} = \dfrac{1}{\textrm{1.7e13}} = \textrm{5.9e-14 M}^2\)

    Since x is very small, 0.05-x ~ 0.050, and

    \(\ce{[Ag+]} = x = \left(\dfrac{\textrm{5.9e-14}}{4} \right)^{1/3} = \textrm{2.5e-5 M}\)

    The approximation is justified.

    If you use the formation equilibrium equation and let \(\ce{[Ag(S2O3)2^3- ]} = x\), the equation is very difficult to solve.


    1. Which of the following is a complex ion?

      \(\ce{CH4}\), \(\ce{H2O}\), \(\ce{NH3}\), \(\ce{Al(OH)4-}\), \(\ce{CCl4}\), \(\ce{CO3^2-}\), \(\ce{NH4+}\)

    2. To a sample of 9.0 mL solution with \(\ce{[NH3]} = \textrm{1.1 M}\), 1.0 mL of 0.1 M \(\ce{AgNO3}\) solution is added. Calculate \(\ce{[Ag+]}\). Assume the final volume to be 10.0 mL. The Kf for \(\ce{Ag(NH3)2} = \textrm{1.6e7}\).
    3. Calculate the dissociation constant Kd for \(\ce{Ag(NH3)+}\), if Kf for \(\ce{Ag(NH3)2} = \textrm{1.6e7}\).
    4. The solubility product of \(\ce{AgBr}\) is 5.0e-13 M2, and Kf = 1.6e7 M-2 for \(\ce{Ag(NH3)2+}\). Estimate the molar solubility of \(\ce{AgBr}\) in a 1.0 M \(\ce{NH3}\) solution.
    5. The solubility product of \(\ce{AgI}\) is 8.3e-17 M2, and Kf = 1.6e7 M-2 for \(\ce{Ag(NH3)2+}\). Estimate the molar solubility of \(\ce{AgI}\) in a 1.0 M \(\ce{NH3}\) solution.
    6. The solubility product of \(\ce{AgBr}\) is 5.0e-13 M2, and Kf = 1.7e13 M-2 for \(\ce{Ag(S2O3)2^3-}\). Estimate the molar solubility of \(\ce{AgBr}\) in a 1.0 M \(\ce{Na2S2O3}\) solution.


    1. Answer \(\ce{Al(OH)4-}\)
      Explain what complex ions or metal complexes are.
    2. Answer \(\ce{[Ag+]} = \textrm{6.25e-10}\)
      Assume \(\ce{[Ag+]} = x\), then

      \ce{Ag+ &+ &2 NH3 &\rightleftharpoons &Ag(NH3)2+}, &K_{\ce f} = \textrm{1.6e7}\\
      x &&1.0 &&0.01 &\leftarrow \textrm{equilibrium concentration}

      \(\dfrac{0.01}{x 1.0^2} = \textrm{1.6e7}\); \(x =\: ?\)

      Apply the concept of equilibrium in complex formation.

    3. Answer = Kd = 6.25e-8

      Derive the dissociation constant from formation constant, and show that \(K_{\ce d} = \dfrac{1}{K_{\ce f}}\).

    4. Answer Molar solubility = 2.8e-3 M
      Review Example 2.
    5. Answer Molar solubility = 3.6e-5 M

      The solubilities for \(\ce{AgCl}\), \(\ce{AgBr}\), and \(\ce{AgI}\) in 1.0 M \(\ce{NH3}\) solution are 0.051, 0.0028, and 0.000036 M respectively.

    6. Answer The molar solubility is 0.49 M
      Use the method of Example 2, but no approximation can be made.

    Contributors and Attributions

    Metal Coordination Complexes is shared under a not declared license and was authored, remixed, and/or curated by LibreTexts.

    • Was this article helpful?