Chapter 20, Problem 114QRT

(a)

Interpretation Introduction

Interpretation:

The compound ${\text{[Pt(en)Cl}}_{\text{2}}\text{]}$ resembles ${{\text{(NH}}_{\text{2}}\text{)}}_{\text{2}}\text{CO}$, $\text{KCl}$, ${\text{K}}_{\text{2}}{\text{SO}}_{\text{4}}$, or ${\text{K}}_{\text{3}}{\text{PO}}_{\text{4}}$ in colligative properties and conductivity has to be given.

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 ${\text{[Pt(en)Cl}}_{\text{2}}\text{]}$.

Among the given compounds, urea alone dissolves in water without forming ions.

    ${{\text{(NH}}_{\text{2}}\text{)}}_{\text{2}}\text{CO}(s)\to {{\text{(NH}}_{\text{2}}\text{)}}_{\text{2}}\text{CO(aq)}$

Therefore, ${\text{[Pt(en)Cl}}_{\text{2}}\text{]}$ resembles ${{\text{(NH}}_{\text{2}}\text{)}}_{\text{2}}\text{CO}$ in colligative properties and conductivity.

(b)

Interpretation Introduction

Interpretation:

The compound ${\text{Na[Cr(en)}}_{\text{2}}{{\text{(SO}}_{\text{4}}\text{)}}_{\text{2}}\text{]}$ resembles ${{\text{(NH}}_{\text{2}}\text{)}}_{\text{2}}\text{CO}$, $\text{KCl}$, ${\text{K}}_{\text{2}}{\text{SO}}_{\text{4}}$, or ${\text{K}}_{\text{3}}{\text{PO}}_{\text{4}}$ in colligative properties and conductivity has to be given.

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 ${\text{Na[Cr(en)}}_{\text{2}}{{\text{(SO}}_{\text{4}}\text{)}}_{\text{2}}\text{]}$.

Among the given compounds, $\text{KCl}$ dissociates in water to form two ions.

    $\text{KCl}(s)\to {\text{K}}^{\text{+}}(aq)\text{+}{\text{Cl}}^{-}(aq)$

Therefore, ${\text{Na[Cr(en)}}_{\text{2}}{{\text{(SO}}_{\text{4}}\text{)}}_{\text{2}}\text{]}$ resembles $\text{KCl}$ in colligative properties and conductivity.

(c)

Interpretation Introduction

Interpretation:

The compound ${\text{K}}_{\text{3}}{\text{[Au(CN)}}_{\text{4}}\text{]}$ resembles ${{\text{(NH}}_{\text{2}}\text{)}}_{\text{2}}\text{CO}$, $\text{KCl}$, ${\text{K}}_{\text{2}}{\text{SO}}_{\text{4}}$, or ${\text{K}}_{\text{3}}{\text{PO}}_{\text{4}}$ in colligative properties and conductivity has to be given.

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 ${\text{K}}_{\text{3}}{\text{[Au(CN)}}_{\text{4}}\text{]}$.

Among the given compounds, ${\text{K}}_{\text{3}}{\text{PO}}_{\text{4}}$ dissociates in water to form four ions.

    ${\text{K}}_{\text{3}}{\text{PO}}_{\text{4}}(s)\to 3{\text{K}}^{\text{+}}(aq)\text{+}{\text{PO}}_{\text{4}}{}^{3-}(aq)$

Therefore, ${\text{K}}_{\text{3}}{\text{[Au(CN)}}_{\text{4}}\text{]}$ resembles ${\text{K}}_{\text{3}}{\text{PO}}_{\text{4}}$ in colligative properties and conductivity.

(d)

Interpretation Introduction

Interpretation:

The compound ${\text{[Ni(H}}_{\text{2}}{\text{O)}}_{\text{2}}{{\text{(NH}}_{\text{3}}\text{)}}_{\text{4}}{\text{]Cl}}_{\text{2}}$ resembles ${{\text{(NH}}_{\text{2}}\text{)}}_{\text{2}}\text{CO}$, $\text{KCl}$, ${\text{K}}_{\text{2}}{\text{SO}}_{\text{4}}$, or ${\text{K}}_{\text{3}}{\text{PO}}_{\text{4}}$ in colligative properties and conductivity has to be given.

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 ${\text{[Ni(H}}_{\text{2}}{\text{O)}}_{\text{2}}{{\text{(NH}}_{\text{3}}\text{)}}_{\text{4}}{\text{]Cl}}_{\text{2}}$.

Among the given compounds, ${\text{K}}_{\text{2}}{\text{SO}}_{\text{4}}$ dissociates in water to form three ions.

    ${\text{K}}_{\text{2}}{\text{SO}}_{\text{4}}(s)\to 2{\text{K}}^{\text{+}}(aq)\text{+}{\text{SO}}_{\text{4}}{}^{2-}(aq)$

Therefore, ${\text{[Ni(H}}_{\text{2}}{\text{O)}}_{\text{2}}{{\text{(NH}}_{\text{3}}\text{)}}_{\text{4}}{\text{]Cl}}_{\text{2}}$ resembles ${\text{K}}_{\text{2}}{\text{SO}}_{\text{4}}$ in colligative properties and conductivity.