Chapter 20, Problem 113QRT

(a)

Interpretation Introduction

Interpretation:

Moles of ions that is present when $\text{1}\text{.00}\text{mol}$ of ${\text{[Pt(en)Cl}}_{\text{2}}\text{]}$ is dissolved in water has to be determined.

Explanation of Solution

Given complex has formula of ${\text{[Pt(en)Cl}}_{\text{2}}\text{]}$.

The complex ${\text{[Pt(en)Cl}}_{\text{2}}\text{]}$ do not dissociate into ions in water.  This is because there are no counter ions present in it.  The complex dissolves in water but does not dissociate in water.  The equation can be given as shown below.

    ${\text{[Pt(en)Cl}}_{\text{2}}\text{]}(s)\to {\text{[Pt(en)Cl}}_{\text{2}}\text{](aq)}$

Therefore, there are no ions in the solution of ${\text{[Pt(en)Cl}}_{\text{2}}\text{]}$.

(b)

Interpretation Introduction

Interpretation:

Moles of ions that is present when $\text{1}\text{.00}\text{mol}$ of ${\text{Na[Cr(en)}}_{\text{2}}{{\text{(SO}}_{\text{4}}\text{)}}_{\text{2}}\text{]}$ is dissolved in water has to be determined.

Explanation of Solution

Given complex has formula of ${\text{Na[Cr(en)}}_{\text{2}}{{\text{(SO}}_{\text{4}}\text{)}}_{\text{2}}\text{]}$.

The complex ${\text{Na[Cr(en)}}_{\text{2}}{{\text{(SO}}_{\text{4}}\text{)}}_{\text{2}}\text{]}$ contains a complex anion and a counter cation.  Therefore, dissociation takes place in water.  The equation can be written as shown below.

    ${\text{Na[Cr(en)}}_{\text{2}}{{\text{(SO}}_{\text{4}}\text{)}}_{\text{2}}\text{]}(s)\to {{\text{[Cr(en)}}_{\text{2}}{{\text{(SO}}_{\text{4}}\text{)}}_{\text{2}}\text{]}}^{-}(aq){\text{+Na}}^{\text{+}}(aq)$

Therefore, there are two ions in the solution of ${\text{Na[Cr(en)}}_{\text{2}}{{\text{(SO}}_{\text{4}}\text{)}}_{\text{2}}\text{]}$.

(c)

Interpretation Introduction

Interpretation:

Moles of ions that is present when $\text{1}\text{.00}\text{mol}$ of ${\text{K}}_{\text{3}}{\text{[Au(CN)}}_{\text{4}}\text{]}$ is dissolved in water has to be determined.

Explanation of Solution

Given complex has formula of ${\text{K}}_{\text{3}}{\text{[Au(CN)}}_{\text{4}}\text{]}$.

The complex ${\text{K}}_{\text{3}}{\text{[Au(CN)}}_{\text{4}}\text{]}$ contains a complex anion and a counter cation.  Therefore, dissociation takes place in water.  The equation can be written as shown below.

    ${\text{K}}_{\text{3}}{\text{[Au(CN)}}_{\text{4}}\text{]}(s)\to {{\text{[Au(CN)}}_{\text{4}}\text{]}}^{3-}(aq){\text{+3K}}^{\text{+}}(aq)$

Therefore, there are four ions in the solution of ${\text{K}}_{\text{3}}{\text{[Au(CN)}}_{\text{4}}\text{]}$.

(d)

Interpretation Introduction

Interpretation:

Moles of ions that is present when $\text{1}\text{.00}\text{mol}$ of ${\text{[Ni(H}}_{\text{2}}{\text{O)}}_{\text{2}}{{\text{(NH}}_{\text{3}}\text{)}}_{\text{4}}{\text{]Cl}}_{\text{2}}$ is dissolved in water has to be determined.

Explanation of Solution

Given complex has formula of ${\text{[Ni(H}}_{\text{2}}{\text{O)}}_{\text{2}}{{\text{(NH}}_{\text{3}}\text{)}}_{\text{4}}{\text{]Cl}}_{\text{2}}$.

The complex ${\text{[Ni(H}}_{\text{2}}{\text{O)}}_{\text{2}}{{\text{(NH}}_{\text{3}}\text{)}}_{\text{4}}{\text{]Cl}}_{\text{2}}$ contains a complex cation and a counter anion.  Therefore, dissociation takes place in water.  The equation can be written as shown below.

    ${\text{[Ni(H}}_{\text{2}}{\text{O)}}_{\text{2}}{{\text{(NH}}_{\text{3}}\text{)}}_{\text{4}}{\text{]Cl}}_{\text{2}}(s)\to {{\text{[Ni(H}}_{\text{2}}{\text{O)}}_{\text{2}}{{\text{(NH}}_{\text{3}}\text{)}}_{\text{4}}\text{]}}^{2+}(aq){\text{+2Cl}}^{-}(aq)$

Therefore, there are three ions in the solution of ${\text{[Ni(H}}_{\text{2}}{\text{O)}}_{\text{2}}{{\text{(NH}}_{\text{3}}\text{)}}_{\text{4}}{\text{]Cl}}_{\text{2}}$.