Textbook Problem

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Zinc hydroxide is amphoteric (Section 16.10). Use equilibrium constants to show that, given sufficient OH⁻, Zn(OH)₂ can dissolve in NaOH.

Interpretation Introduction

Interpretation:

Using the equilibrium constant values, that $\text{Zn}{\left(\text{OH}\right)}_2$ is soluble in $\text{NaOH}$ has to be showed.

Concept introduction:

Some metal ions when present in an aqueous solution containing anions or neutral species called Lewis base or ligands having a tendency to donate electron pairs to metal ions then complex ion formation will take place.

Example of metal ions that form complex ions includes,${\text{Cd}}^{2+}$,${\text{Fe}}^{2+}$,${\text{Zn}}^{2+}$,${\text{Ni}}^{2+}$ etc.

Example of Lewis bases includes,${\text{NH}}_3$,${\text{OH}}^{-}$ etc.

The complex ion remains in equilibrium with the metal ion and the ligand called complex ion formation equilibrium and the equilibrium constant is called as formation constant $K_{\text{f}}$.

A larger value of $K_{\text{f}}$ implies that the complex ion formed is more stable. $K_{\text{f}}$ is the measure of the strength of the interaction between the metal ions and the Lewis base to form the complex ion.

For example for general complex ion formation reaction,

  $\text{xM}+\text{yL}\rightleftharpoons \left[{\text{M}}_{\text{x}}{\text{L}}_{\text{y}}\right]$

$K_{\text{f}}$ can be given as

  $K_{\text{f}}=\frac{\left[{\text{M}}_{\text{x}}{\text{L}}_{\text{y}}\right]}{{\left[\text{M}\right]}^{\text{x}}{\left[\text{L}\right]}^{\text{y}}}$

 Here,

• $\left[{\text{M}}_{\text{x}}{\text{L}}_{\text{y}}\right]$ is the equilibrium concentration of complex ion.
• $\left[\text{M}\right]$ is the equilibrium concentration of metal ion.
• $\left[\text{L}\right]$ is the equilibrium concentration of the ligand.
• $\text{x}$ and $\text{y}$ are the coefficients of metal ion and ligand respectively.

Complex ions are stable and thus formation of these increase the solubility of the salt containing the metal ions same as in complex ions. Effect of complex ion formation on solubility of salt can be explained as below,

$\text{AgBr}$ when dissolved in water does not dissolve completely and dissociates as follows,

    $\text{AgBr}(s)\stackrel{K_{\text{sp}}}{\rightleftharpoons }A{g}^{+}(aq)+{\text{Br}}^{-}(aq)$ (1)

For this reaction $K_{\text{sp}}$ expression is,

$\text{Ksp}=\left[{\text{Ag}}^{+}\right]\left[{\text{Br}}^{-}\right]$

$A{g}^{+}$ ions are capable of forming complex with ligand so when aqueous ammonia solution (strong ligand) is added to the saturated solution of $\text{AgBr}$, $A{g}^{+}$ ions present in the solution form complex with ${\text{NH}}_3$.

    ${\text{Ag}}^{+}\left(\text{aq}\right)+2{\text{NH}}_3\left(\text{aq}\right)\stackrel{K_{\text{f}}}{\rightleftharpoons }{\left[\text{Ag}{\left({\text{NH}}_3\right)}_2\right]}^{+}\left(\text{aq}\right)$ (2)

The expression for formation constant $K_{\text{f}}$ is given as,

$K_{\text{f}}=\frac{\left[{\left[\text{Ag}{\left({\text{NH}}_3\right)}_2\right]}^{+}\right]}{\left[{\text{Ag}}^{+}\right]{\left[{\text{NH}}_3\right]}^2}$

Complex formation leads to the decrease in the concentration of the $A{g}^{+}$ ions in the solution, as a result, according to Le Chatelier’s principle the equilibrium in equation (1) move in the forward direction producing more of the $A{g}^{+}$ ions and the solubility of the slightly soluble salt $\text{AgBr}$ increases.

Adding two equilibrium equation (1) and (2) new overall equilibrium constant can be defined.

Net chemical equation:

  ${\text{Ag}}^{+}\left(\text{aq}\right)+2{\text{NH}}_3\left(\text{aq}\right)\stackrel{K_{\text{net}}}{\rightleftharpoons }{\left[\text{Ag}{\left({\text{NH}}_3\right)}_2\right]}^{+}\left(\text{aq}\right)+{\text{Br}}^{-}(aq)$ (3)

Net equilibrium constant can be given as,

  $K_{\text{net}}=\left(K_{\text{sp}}\right)\left(K_{\text{f}}\right)$ (4)

Explanation of Solution

The value of net equilibrium constant, $K_{\text{net}}$ for the reaction of dissolution of  $\text{Zn}{\left(\text{OH}\right)}_2$ is calculated below.

Given:

Refer to the Appendix J and K in the textbook for the value of $K_{\text{sp}}$ and $K_{\text{f}}$ respectively.

The value of solubility product constant, $K_{\text{sp}}$ for $\text{Zn}{\left(\text{OH}\right)}_2$ is $3.0\times {10}^{-17}$ .

The value of formation constant, $K_{\text{f}}$ for ${\left[\text{Zn}{\left(\text{OH}\right)}_4\right]}^{2-}$ is $4.6\times {10}^{17}$.

$\text{Zn}{\left(\text{OH}\right)}_2$ dissociates as follows in water,

  $\text{Zn}{\left(\text{OH}\right)}_2\left(\text{s}\right)\stackrel{K_{\text{sp}}}{\rightleftharpoons }{\text{Zn}}^{2+}\left(\text{aq}\right)+2{\text{OH}}^{-}\left(\text{aq}\right)$ (5)

${\text{Zn}}^{2+}$ ions present in solution undergo complex formation. The reaction for complex formation is given as,

  ${\text{Zn}}^{2+}\left(\text{aq}\right)+4{\text{OH}}^{-1}\left(\text{aq}\right)\stackrel{K_{\text{f}}}{\rightleftharpoons }{\left[\text{Zn}{\left(\text{OH}\right)}_4\right]}^{2-}\left(\text{aq}\right)$ (6)

Add equation (5) and (6) to find the net chemical equation