Chapter 15, Problem 1QAP

Write a net ionic equation for the reaction between aqueous solutions of

(a) ammonia and hydrofluoric acid.

(b) perchloric acid and rubidium hydroxide.

(c) sodium sulfite and hydriodic acid.

(d) nitric acid and calcium hydroxide.

Interpretation Introduction

Interpretation: The concentration of ammonia for the given conditions of complex ions needs to be determined.

  $[{\text{Cd}}^{\text{+2}}\text{]}=[\text{Cd}{\left({\text{NH}}_{\text{3}}\right)}_{\text{4}}{}^{\text{+2}}]$

Concept Introduction: The coordination complexes and complex ions are formed by the combination of metal ion and ligands. Here, ligands are the electron donating atoms or molecules. Therefore, ligands form coordinate bond with metal ion. The stability of coordination complexes can be determined in terms of their formation constant. It is represented as K_f. It is the ratio of product and reactant at equilibrium condition of formation of coordinate complex.

Answer to Problem 1QAP

  ${\text{[NH}}_{\text{3}}\text{] =}0.014\text{ M}$

Explanation of Solution

The formation reaction for $\left[\text{Cd}{\left({\text{NH}}_{\text{3}}\right)}_{\text{4}}{}^{\text{+2}}\right]$ can be written as:

  ${\text{Cd}}^{\text{+2}}{\text{ + 4 NH}}_{\text{3}}\leftrightarrow \left[\text{Cd}{\left({\text{NH}}_{\text{3}}\right)}_{\text{4}}{}^{\text{+2}}\right]$

  $\text{Kf = }\frac{\left[\text{Cd}{\left({\text{NH}}_{\text{3}}\right)}_{\text{4}}{}^{\text{+2}}\right]}{{\text{[Cd}}^{\text{+2}}{\text{] [NH}}_{\text{3}}{\text{]}}^{\text{4}}}$

Given: $[{\text{Cd}}^{\text{+2}}\text{]}=[\text{Cd}{\left({\text{NH}}_{\text{3}}\right)}_{\text{4}}{}^{\text{+2}}]$

Plug this in Kf expression:

  $\begin{array}{l}\text{Kf = }\frac{\left[\text{Cd}{\left({\text{NH}}_{\text{3}}\right)}_{\text{4}}{}^{\text{+2}}\right]}{\text{[Cd}{\left({\text{NH}}_{\text{3}}\right)}_{\text{4}}{}^{\text{+2}}{\text{] [NH}}_{\text{3}}{\text{]}}^{\text{4}}}\\ {\text{[NH}}_{\text{3}}\text{] = }{\left[\frac{\text{1}}{\text{Kf}}\right]}^{\raisebox{1ex}{$\text{1}$}\!\left/ \!\raisebox{-1ex}{$\text{4}$}\right.}\\ {\text{[NH}}_{\text{3}}\text{] = }{\left[\frac{\text{1}}{\text{2}{\text{.8 x 10}}^{\text{7}}}\right]}^{\raisebox{1ex}{$\text{1}$}\!\left/ \!\raisebox{-1ex}{$\text{4}$}\right.}\\ {\text{[NH}}_{\text{3}}\text{] = }0.014\text{ M}\end{array}$

At ${\text{[NH}}_{\text{3}}\text{] =}0.014\text{ M}$ ; the $[{\text{Cd}}^{\text{+2}}\text{]}=[\text{Cd}{\left({\text{NH}}_{\text{3}}\right)}_{\text{4}}{}^{\text{+2}}]$ relation will be true.

Interpretation Introduction

Interpretation: The concentration of ammonia for the given conditions of complex ions needs to be determined.

  $[{\text{Co}}^{\text{+2}}\text{]}=[\text{Co}\left({\text{NH}}_{\text{3}}\right){\text{6}}^{\text{+2}}]$

Concept Introduction: The coordination complexes and complex ions are formed by the combination of metal ion and ligands. Here ligands are the electron donating atoms or molecules. Therefore, ligands form coordinate bond with metal ion. The stability of coordination complexes can be determined in terms of their formation constant. It is represented as K_f. It is the ratio of product and reactant at equilibrium condition of formation of coordinate complex.

Answer to Problem 1QAP

  ${\text{[NH}}_{\text{3}}\text{] = }0.147\text{ M}$

Explanation of Solution

The formation reaction for $\left[\text{Co}\left({\text{NH}}_{\text{3}}\right){\text{6}}^{\text{+2}}\right]$ can be written as:

  ${\text{Co}}^{\text{+2}}{\text{ + 6 NH}}_{\text{3}}\leftrightarrow \left[\text{Co}\left({\text{NH}}_{\text{3}}\right){\text{6}}^{\text{+2}}\right]$

  $\text{Kf = }\frac{\left[\text{Co}{\left({\text{NH}}_{\text{3}}\right)}_{\text{6}}{}^{\text{+2}}\right]}{{\text{[Co}}^{\text{+2}}{\text{] [NH}}_{\text{3}}{\text{]}}^{\text{6}}}$

Given: $[{\text{Co}}^{\text{+2}}\text{]}=[\text{Co}\left({\text{NH}}_{\text{3}}\right){\text{6}}^{\text{+2}}]$

Plug this in K_fexpression:

  $\begin{array}{l}\text{Kf = }\frac{\left[\text{Co}{\left({\text{NH}}_{\text{3}}\right)}_{\text{6}}{}^{\text{+2}}\right]}{{\text{[Co}}^{\text{+2}}{\text{] [NH}}_{\text{3}}{\text{]}}^{\text{6}}}\\ {\text{[NH}}_{\text{3}}\text{] = }{\left[\frac{\text{1}}{\text{Kf}}\right]}^{\raisebox{1ex}{$\text{1}$}\!\left/ \!\raisebox{-1ex}{$6$}\right.}\\ {\text{[NH}}_{\text{3}}\text{] = }{\left[\frac{\text{1}}{{\text{1 x 10}}^{\text{5}}}\right]}^{\raisebox{1ex}{$\text{1}$}\!\left/ \!\raisebox{-1ex}{$\text{6}$}\right.}\\ {\text{[NH}}_{\text{3}}\text{] = }0.147\text{ M}\end{array}$

At ${\text{[NH}}_{\text{3}}\text{] =}0.147\text{ M}$ ; the $[{\text{Co}}^{\text{+2}}\text{]}=[\text{Co}\left({\text{NH}}_{\text{3}}\right){\text{6}}^{\text{+2}}]$ relation will be true.