Chapter 19, Problem 103CP

(a)

Interpretation Introduction

Interpretation: The molar solubility of AgBr in pure water needs to be determined, if the ${\text{K}}_{\text{sp}}$ for AgBr is $5.0\times {10}^{-13}$

Concept Introduction:

A metal complex can be show as ${\text{[ML}}_{\text{n}}\text{]}$ . Here M indicates the metal atom or ion whereas L indicates the ligands which are present in ‘n’ numbers. The coordination number defined the number of coordinate bonds between metal and ligand.

Usually transition metal ions and post transition metal ions forms coordinate complexes with different ligands.

Answer to Problem 103CP

  $\text{s = 7}{\text{.1×10}}^{\text{-7}}\text{M}$

Explanation of Solution

Solid $\text{AgBr}$ soluble in water to form;

  ${\text{AgBr}}_{\text{(s)}}^{}\stackrel{}{\to }{\text{Ag}}_{\text{(aq)}}^{\text{+}}{\text{+ Br}}_{\text{(aq)}}^{\text{-}}{\text{ K}}_{\text{sp}}\text{ = }5.0\times {10}^{-13}$

The solubility product of AgBr can be written as:

  $\begin{array}{l}{\text{ K}}_{\text{sp}}\text{ = }[{\text{Ag}}_{\text{(aq)}}^{\text{+}}][{\text{Br}}_{\text{(aq)}}^{\text{-}}]\\ {\text{K}}_{\text{sp}}\text{ = }[\text{s}][\text{s }]=s^2\end{array}$

Calculate solubility:

  $\begin{array}{l}{\text{K}}_{\text{sp}}{\text{ = [s][s ]=s}}^{\text{2}}\text{= 5}{\text{.0×10}}^{\text{-13}}\\ \text{s= }\sqrt{\text{5}{\text{.0×10}}^{\text{-13}}}\\ \text{s = 7}{\text{.1×10}}^{\text{-7}}\text{M}\end{array}$

(b)

Interpretation Introduction

Interpretation: The molar solubility of $\text{AgBr}$ in 3.0 M ${\text{NH}}_{\text{3}}$ if the ${\text{K}}_f$ for ${\text{Ag(NH}}_{\text{3}}{)}_2^{+}$ is $1.7\times {10}^7$ needs to be determined.

Concept Introduction:

A metal complex can be show as ${\text{[ML}}_{\text{n}}\text{]}$ . Here M indicates the metal atom or ion whereas L indicates the ligands which are present in ‘n’ numbers. The coordination number defined the number of coordinate bonds between metal and ligand.

Usually transition metal ions and post transition metal ions forms coordinate complexes with different ligands.

Answer to Problem 103CP

  $\text{s = 8}{\text{.7×10}}^{\text{-3}}\text{M}$

Explanation of Solution

Add both equation:

  ${\text{AgBr}}_{\text{(s)}}^{}\stackrel{}{\to }{\text{Ag}}_{\text{(aq)}}^{\text{+}}{\text{+ Br}}_{\text{(aq)}}^{\text{-}}{\text{ K}}_{\text{sp}}\text{ = }5.0\times {10}^{-13}$

  ${\text{Ag}}_{\text{(aq)}}^{+}{\text{+ 2 NH}}_{\text{3(aq)}}\stackrel{}{\to }{\text{Ag(NH}}_{\text{3}}{\text{)}}_{\text{2(aq)}}^{\text{+}}{\text{ K}}_{\text{f}}\text{ =1}{\text{.7×10}}^{\text{7}}$

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  ${\text{AgBr}}_{\text{(s)}}^{}{\text{+ 2 NH}}_{\text{3(aq)}}\stackrel{}{\to }{\text{ Br}}_{\text{(aq)}}^{\text{-}}{\text{ + Ag(NH}}_{\text{3}}{\text{)}}_{\text{2(aq)}}^{\text{+}}{\text{ K = K}}_{\text{sp}}\times {\text{K}}_{\text{f}}\text{= }5.0\times {10}^{-13}\times \text{1}{\text{.7×10}}^{\text{7}}=8.5\times {10}^{-6}$

  $\begin{array}{l}{\text{ AgBr}}_{\text{(s)}}^{}{\text{+ 2 NH}}_{\text{3(aq)}}\stackrel{}{\to }{\text{ Br}}_{\text{(aq)}}^{\text{-}}{\text{ + Ag(NH}}_{\text{3}}{\text{)}}_{\text{2(aq)}}^{\text{+}}\\ \text{Initial 3}\text{.0M 0 0}\\ \text{Change - 2s s s}\\ \text{Eq}\text{. 3}\text{.0-2s s s}\end{array}$

Calculate ‘s’:

  $\begin{array}{l}{\text{AgBr}}_{\text{(s)}}^{}{\text{+ 2 NH}}_{\text{3(aq)}}\stackrel{}{\to }\\ \text{K = 8}{\text{.5×10}}^{\text{-6}}\text{=}\frac{{\text{[Br}}_{\text{(aq)}}^{\text{-}}{\text{] [Ag(NH}}_{\text{3}}{\text{)}}_{\text{2(aq)}}^{\text{+}}\text{] }}{{{\text{[NH}}_{\text{3(aq)}}\text{]}}^{\text{2}}}\\ \text{ 8}{\text{.5×10}}^{\text{-6}}\text{=}\frac{\text{[s] [s] }}{{\text{[3}\text{.0-2s]}}^{\text{2}}}\\ \text{ 8}{\text{.5×10}}^{\text{-6}}\text{=}\frac{\text{[s] [s] }}{{\text{[3}\text{.0]}}^{\text{2}}}\\ \text{s = 8}{\text{.7×10}}^{\text{-3}}\text{M}\end{array}$

(c)

Interpretation Introduction

Interpretation: The calculated solubility values needs to be compared.

  $\text{AgBr}$ in 3.0 M ${\text{NH}}_{\text{3}}$ = $\text{s = 8}{\text{.7×10}}^{\text{-3}}\text{M}$

  $\text{AgBr}$ in pure water = $\text{s = 7}{\text{.1×10}}^{\text{-7}}\text{M}$

Concept Introduction:

A metal complex can be show as ${\text{[ML}}_{\text{n}}\text{]}$ . Here M indicates the metal atom or ion whereas L indicates the ligands which are present in ‘n’ numbers. The coordination number defined the number of coordinate bonds between metal and ligand.

Usually transition metal ions and post transition metal ions forms coordinate complexes with different ligands.

Answer to Problem 103CP

The presence of ${\text{NH}}_{\text{3}}$ , the solubility of $\text{AgBr}$ increases. This is due to formation of complex ion ${\text{Ag(NH}}_{\text{3}}{\text{)}}_{\text{2(aq)}}^{\text{+}}$ .

Explanation of Solution

Calculated solubility’s:

  $\text{AgBr}$ in 3.0 M ${\text{NH}}_{\text{3}}$ = $\text{s = 8}{\text{.7×10}}^{\text{-3}}\text{M}$

  $\text{AgBr}$ in pure water = $\text{s = 7}{\text{.1×10}}^{\text{-7}}\text{M}$

The presence of ${\text{NH}}_{\text{3}}$ , the solubility of $\text{AgBr}$ increases. This is due to formation of complex ion ${\text{Ag(NH}}_{\text{3}}{\text{)}}_{\text{2(aq)}}^{\text{+}}$ which removes ${\text{Ag}}^{\text{+}}$ from equilibrium and shifted the equilibrium $\text{AgBr}$ towards right. It makes $\text{AgBr}$ more soluble.

(d)

Interpretation Introduction

Interpretation: The mass of $\text{AgBr}$ that will dissolve in 250.0 mL of 3.0 M ${\text{NH}}_{\text{3}}$ needs to be determined.

Concept Introduction:

A metal complex can be show as ${\text{[ML}}_{\text{n}}\text{]}$ . Here M indicates the metal atom or ion whereas L indicates the ligands which are present in ‘n’ numbers. The coordination number defined the number of coordinate bonds between metal and ligand.

Usually transition metal ions and post transition metal ions forms coordinate complexes with different ligands.

Answer to Problem 103CP

  $\text{ 0}\text{.41g AgBr}$

Explanation of Solution

Solubility of $\text{AgBr}$ in 3.0 M ${\text{NH}}_{\text{3}}$ = $\text{s = 8}{\text{.7×10}}^{\text{-3}}\text{M}$

Volume = 250.0 mL = 250.0 x 10-3 L

Molar mass of $\text{AgBr}$ = 187.8 g/mol

Calculate mass of $\text{AgBr}$ :

  $\text{Mass of AgBr = 250}{\text{.0×10}}^{\text{-3}}\text{L×}\frac{\text{8}{\text{.7×10}}^{\text{-3}}\text{molAgBr}}{\text{1L}}\text{×}\frac{\text{187}\text{.8 g AgBr}}{\text{1 mole AgBr}}\text{= 0}\text{.41g AgBr}$

(e)

Interpretation Introduction

Interpretation: The effect of addition of ${\text{HNO}}_{\text{3}}$ on the solubility of $\text{AgBr}$ in pure water and 3.0 M ${\text{NH}}_{\text{3}}$ needs to be determined.

Concept Introduction:

A metal complex can be show as ${\text{[ML}}_{\text{n}}\text{]}$ . Here M indicates the metal atom or ion whereas L indicates the ligands which are present in ‘n’ numbers. The coordination number defined the number of coordinate bonds between metal and ligand.

Usually transition metal ions and post transition metal ions forms coordinate complexes with different ligands.

Answer to Problem 103CP

By the addition of ${\text{HNO}}_{\text{3}}$ , there will be no effect on the solubility of $\text{AgBr}$ in pure water.

The addition of ${\text{HNO}}_{\text{3}}$ in $\text{AgBr}$ in 3.0 M ${\text{NH}}_{\text{3}}$ reduces the solubility of $\text{AgBr}$ .

Explanation of Solution

By the addition of ${\text{HNO}}_{\text{3}}$ , there will be no effect on the solubility of $\text{AgBr}$ in pure water. As ${\text{HNO}}_{\text{3}}$ will not react with $\text{AgBr}$ because ${\text{Br}}^{\text{-}}$ is the weak conjugate base of the strong acid $\text{HBr}$ .

The addition of ${\text{HNO}}_{\text{3}}$ in $\text{AgBr}$ in 3.0 M ${\text{NH}}_{\text{3}}$ reduces the solubility of $\text{AgBr}$ . Since ${\text{NH}}_{\text{3}}$ is a weak base therefore addition of hydrogen ion will react with ${\text{NH}}_{\text{3}}$ and forms ${\text{NH}}_4^{+}$ ion. It removes the ${\text{NH}}_{\text{3}}$ from solution and form very less amount of complex so it decreases the solubility of $\text{AgBr}$ .