Chapter 19, Problem 19.148P

(a)

Interpretation Introduction

Interpretation:

${\text{[Ag}}^{\text{+}}\text{]}$ present in the solution is $\frac{(1.8\times {10}^{-10})}{\left[{\text{Cl}}^{\text{-}}\right]}$ and $\left[{\text{AgCl}}_2^{-}\right]$ present in the solution is $(3.2\times {10}^{-5})\times \left[{\text{Cl}}^{\text{-}}\right]$ have to be shown.

Concept introduction:

Solubility:

Solubility is defined as the maximum amount of the solute that can be dissolved in the solvent at equilibrium.

Solubility product constant:

Solubility product constant is defined for equilibrium between solids and its respective ions in the solution. Generally, solubility product refers only to insoluble or slightly soluble ionic substances that make equilibrium in water.

It is defined as the product of concentration of ions of a sparingly soluble salt in its saturated solution at ${25}^{\text{o}}\text{C}$. The concentration terms are raised to the coefficients of their respective ions. It is denoted by ${\text{K}}_{\text{sp}}$.

This value indicates the degree of dissociation of a compound in water. More the value of ${\text{K}}_{\text{sp}}$ more is the solubility of the compound.

Considering an equilibrium of salt ${\text{A}}_{\text{x}}{\text{B}}_{\text{y}}$ with its respective ions in solution and thus ${\text{K}}_{\text{sp}}$ is calculated.

  ${\text{A}}_{\text{x}}{\text{B}}_{\text{y}}\stackrel{}{\rightleftarrows }{\text{xA}}^{\text{y+}}+{\text{yB}}^{\text{x}}{}^{\text{-}}$

  ${\text{K}}_{\text{sp}}={\left[{\text{xA}}^{\text{y+}}\right]}^{\text{x}}{\left[{\text{yB}}^{\text{x-}}\right]}^{\text{y}}$

Formation constant:

A stability constant or formation constant is an equilibrium constant for the formation of a complex ion in the solution and it measures the strength of interaction between the reactants that forms the complex.

Answer to Problem 19.148P

It is proved that $\left[{\text{AgCl}}_2^{-}\right]=(3.2\times {10}^{-5})\times \left[{\text{Cl}}^{\text{-}}\right]$ and ${\text{[Ag}}^{\text{+}}\text{]}=\frac{(1.8\times {10}^{-10})}{\left[{\text{Cl}}^{\text{-}}\right]}$.

Explanation of Solution

Given that the equations are,

  ${\text{[Ag}}^{\text{+}}\text{]}\text{}\text{(aq)}\text{+}\left[{\text{Cl}}^{\text{-}}\right]\text{(aq)}\stackrel{}{\rightleftarrows }\left[\text{AgCl}\right]\text{(s)}({\text{K}}_{\text{sp}}\text{=}\text{1}{\text{.8×10}}^{\text{-10}})\mathrm{...}(1)$

  ${\text{[Ag}}^{\text{+}}\text{]}\text{}\text{(aq)}\text{+}2\left[{\text{Cl}}^{\text{-}}\right]\text{(aq)}\stackrel{}{\rightleftarrows }\left[{\text{AgCl}}_2^{-}\right]\text{(aq)}({\text{K}}_{\text{f}}\text{=}\text{1}{\text{.8×10}}^{\text{5}})\mathrm{...}(2)$

  ${\text{K}}_{\text{sp}}={\text{[Ag}}^{\text{+}}\text{]}\text{}\times \left[{\text{Cl}}^{\text{-}}\right]$

  $\therefore {\text{[Ag}}^{\text{+}}\text{]}=\frac{(1.8\times {10}^{-10})}{\left[{\text{Cl}}^{\text{-}}\right]}$ [As ${\text{K}}_{\text{sp}}\text{=}\text{1}{\text{.8×10}}^{\text{-10}}$ ]

Subtracting the above equation (1) from equation (2),

  $\left[\text{AgCl}\right]\text{(s)}\text{+}\left[{\text{Cl}}^{\text{-}}\right]\text{(aq)}\stackrel{}{\rightleftarrows }\left[{\text{AgCl}}_2^{-}\right]\text{(aq)}$

  ${\text{K}}_{\text{f}}=\frac{\left[{\text{AgCl}}_2^{-}\right]\text{(aq)}}{{\text{[Ag}}^{\text{+}}\text{]}\text{}\text{(aq)}\times {\left[{\text{Cl}}^{\text{-}}\right]}^2\text{(aq)}}\mathrm{...}(3)$

  ${\text{[Ag}}^{\text{+}}\text{]}=\frac{{\text{K}}_{\text{sp}}}{\left[{\text{Cl}}^{\text{-}}\right]}$

Substituting the value of ${\text{[Ag}}^{\text{+}}\text{]}$ in the equation (3),

  $\left[{\text{AgCl}}_2^{-}\right]={\text{K}}_{\text{sp}}\times {\text{K}}_{\text{f}}\times \left[{\text{Cl}}^{\text{-}}\right]$

  $\left[{\text{AgCl}}_2^{-}\right]=(1.8\times {10}^{-10})\times (1.8\times {10}^5)\times \left[{\text{Cl}}^{\text{-}}\right]$

  $\therefore$$\left[{\text{AgCl}}_2^{-}\right]=(3.2\times {10}^{-5})\times \left[{\text{Cl}}^{\text{-}}\right]$

Therefore it is proved that $\left[{\text{AgCl}}_2^{-}\right]=\left(3.2\times {10}^{-5}\right)\times \left[{\text{Cl}}^{\text{-}}\right]$ and ${\text{[Ag}}^{\text{+}}\text{]}=\frac{(1.8\times {10}^{-10})}{\left[{\text{Cl}}^{\text{-}}\right]}$.

(b)

Interpretation Introduction

Interpretation:

Value of $\left[{\text{Cl}}^{\text{-}}\right]$ has to be found for the given ${\text{[Ag}}^{\text{+}}\text{]}=\left[{\text{AgCl}}_2^{-}\right]$.

Concept introduction:

Solubility:

Solubility is defined as the maximum amount of the solute that can be dissolved in the solvent at equilibrium.

Solubility product constant:

Solubility product constant is defined for equilibrium between solids and its respective ions in the solution. Generally, solubility product refers only to insoluble or slightly soluble ionic substances that make equilibrium in water.

It is defined as the product of concentration of ions of a sparingly soluble salt in its saturated solution at ${25}^{\text{o}}\text{C}$. The concentration terms are raised to the coefficients of their respective ions. It is denoted by ${\text{K}}_{\text{sp}}$.

This value indicates the degree of dissociation of a compound in water. More the value of ${\text{K}}_{\text{sp}}$ more is the solubility of the compound.

Considering an equilibrium of salt ${\text{A}}_{\text{x}}{\text{B}}_{\text{y}}$ with its respective ions in solution and thus ${\text{K}}_{\text{sp}}$ is calculated.

  ${\text{A}}_{\text{x}}{\text{B}}_{\text{y}}\stackrel{}{\rightleftarrows }{\text{xA}}^{\text{y+}}+{\text{yB}}^{\text{x}}{}^{\text{-}}$

  ${\text{K}}_{\text{sp}}={\left[{\text{xA}}^{\text{y+}}\right]}^{\text{x}}{\left[{\text{yB}}^{\text{x-}}\right]}^{\text{y}}$

Formation constant:

A stability constant or formation constant is an equilibrium constant for the formation of a complex ion in the solution and it measures the strength of interaction between the reactants that forms the complex.

Answer to Problem 19.148P

The value of $\left[{\text{Cl}}^{\text{-}}\right]$ is equals to $\text{2}{\text{.37×10}}^{\text{-3}}\text{M}$

Explanation of Solution

Given that,

  ${\text{[Ag}}^{\text{+}}\text{]}\text{}\text{(aq)}\text{+}\left[{\text{Cl}}^{\text{-}}\right]\text{(aq)}\stackrel{}{\rightleftarrows }\left[\text{AgCl}\right]\text{(s)}({\text{K}}_{\text{sp}}\text{=}\text{1}{\text{.8×10}}^{\text{-10}})\mathrm{...}(1)$

  ${\text{[Ag}}^{\text{+}}\text{]}\text{}\text{(aq)}\text{+}2\left[{\text{Cl}}^{\text{-}}\right]\text{(aq)}\stackrel{}{\rightleftarrows }\left[{\text{AgCl}}_2^{-}\right]\text{(aq)}({\text{K}}_{\text{f}}\text{=}\text{1}{\text{.8×10}}^{\text{5}})\mathrm{...}(2)$

Now the condition is given that ${\text{[Ag}}^{\text{+}}\text{]}=\left[{\text{AgCl}}_2^{-}\right]$.

  ${\text{K}}_{\text{sp}}={\text{[Ag}}^{\text{+}}\text{]}\text{}\times \left[{\text{Cl}}^{\text{-}}\right]$

  $\therefore {\text{[Ag}}^{\text{+}}\text{]}=\frac{(1.8\times {10}^{-10})}{\left[{\text{Cl}}^{\text{-}}\right]}$ [as ${\text{K}}_{\text{sp}}\text{=}\text{1}{\text{.8×10}}^{\text{-10}}$ ]

From the formation constant,

  ${\text{K}}_{\text{f}}=\frac{\left[{\text{AgCl}}_2^{-}\right]\text{(aq)}}{{\text{[Ag}}^{\text{+}}\text{]}\text{}\text{(aq)}\times {\left[{\text{Cl}}^{\text{-}}\right]}^2\text{(aq)}}\mathrm{...}(3)$

  ${\text{[Ag}}^{\text{+}}\text{]}=\frac{{\text{K}}_{\text{sp}}}{\left[{\text{Cl}}^{\text{-}}\right]}$

Substituting the value of ${\text{[Ag}}^{\text{+}}\text{]}$ in the equation (3),

  $\left[{\text{AgCl}}_2^{-}\right]={\text{K}}_{\text{sp}}\times {\text{K}}_{\text{f}}\times \left[{\text{Cl}}^{\text{-}}\right]$

  $\left[{\text{AgCl}}_2^{-}\right]=(1.8\times {10}^{-10})\times (1.8\times {10}^5)\times \left[{\text{Cl}}^{\text{-}}\right]$

  $\therefore$$\left[{\text{AgCl}}_2^{-}\right]=(3.2\times {10}^{-5})\times \left[{\text{Cl}}^{\text{-}}\right]$

Now according to given condition,

  ${\text{[Ag}}^{\text{+}}\text{]}=\left[{\text{AgCl}}_2^{-}\right]$

  $\frac{(1.8\times {10}^{-10})}{\left[{\text{Cl}}^{\text{-}}\right]}=(3.2\times {10}^{-5})\times \left[{\text{Cl}}^{\text{-}}\right]$

  $\therefore$${\left[{\text{Cl}}^{\text{-}}\right]}^2=\frac{(1.8\times {10}^{-10})}{(3.2\times {10}^{-5})}$

   $\therefore$$\left[{\text{Cl}}^{\text{-}}\right]\text{=}\text{2}{\text{.37×10}}^{\text{-3}}\text{M}$

Therefore $\left[{\text{Cl}}^{\text{-}}\right]$ is equals to $\text{2}{\text{.37×10}}^{\text{-3}}\text{M}$.

(c)

Interpretation Introduction

Interpretation:

Plot of solubility of $\text{AgCl}$ versus ${\text{Cl}}^{\text{-}}$ has to be explained.

Concept introduction:

Solubility:

Solubility is defined as the maximum amount of the solute that can be dissolved in the solvent at equilibrium.

 Solubility product constant:

Solubility product constant is defined for equilibrium between solids and its respective ions in the solution. Generally, solubility product refers only to insoluble or slightly soluble ionic substances that make equilibrium in water.

It is defined as the product of concentration of ions of a sparingly soluble salt in its saturated solution at ${25}^{\text{o}}\text{C}$. The concentration terms are raised to the coefficients of their respective ions. It is denoted by ${\text{K}}_{\text{sp}}$.

This value indicates the degree of dissociation of a compound in water. More the value of ${\text{K}}_{\text{sp}}$ more is the solubility of the compound.

Considering an equilibrium of salt ${\text{A}}_{\text{x}}{\text{B}}_{\text{y}}$ with its respective ions in solution and thus ${\text{K}}_{\text{sp}}$ is calculated.

  ${\text{A}}_{\text{x}}{\text{B}}_{\text{y}}\stackrel{}{\rightleftarrows }{\text{xA}}^{\text{y+}}+{\text{yB}}^{\text{x}}{}^{\text{-}}$

  ${\text{K}}_{\text{sp}}={\left[{\text{xA}}^{\text{y+}}\right]}^{\text{x}}{\left[{\text{yB}}^{\text{x-}}\right]}^{\text{y}}$

Le Chatelier’s principle:

When a system in equilibrium then is subjected to any external disturbance like change of pressure, volume, temperature etc… Then the system acts in a way to prevent that change. This is called Le-Chatelier’s principle.

Explanation of Solution

Given that,

  ${\text{[Ag}}^{\text{+}}\text{]}\text{}\text{(aq)}\text{+}\left[{\text{Cl}}^{\text{-}}\right]\text{(aq)}\stackrel{}{\rightleftarrows }\left[\text{AgCl}\right]\text{(s)}\left({\text{K}}_{\text{sp}}\text{=}\text{1}{\text{.8×10}}^{\text{-10}}\right)\mathrm{...}(1)$

  ${\text{K}}_{\text{sp}}={\text{[Ag}}^{\text{+}}\text{]}\text{}\times \left[{\text{Cl}}^{\text{-}}\right]$

When the concentration of ${\text{Cl}}^{\text{-}}$ is less than the formation of $\text{AgCl}$ is also less. Then ${\text{[Ag}}^{\text{+}}\text{]}$ is more in the solution.

As the $\left[{\text{Cl}}^{\text{-}}\right]$ starts to increase then the formation of  $\text{AgCl}$ also increases and it continues till all the ${\text{Ag}}^{\text{+}}$ has not been consumed. As the concentration of $\text{AgCl}$ is increased then solubility of $\text{AgCl}$ decreases to maintain the equilibrium. Hence equilibrium will shift to left side (According to Le Chatelier’s Principle).

After that if $\left[{\text{Cl}}^{\text{-}}\right]$ is increased then ${\text{AgCl}}_2^{-}$ will form and its rate of formation increases with the increase in concentration of ${\text{Cl}}^{\text{-}}$. As ${\text{AgCl}}_2^{-}$ forms hence concentration of $\text{AgCl}$ decreases and as a result to maintain equilibrium solubility of $\text{AgCl}$ increases and equilibrium shifts to right.

The graph is given below,

Figure.1

(d)

Interpretation Introduction

Interpretation:

The solubility of $\text{AgCl}$ has to be found at the $\left[{\text{Cl}}^{\text{-}}\right]$ calculated at part (b) which is the minimum solubility of $\text{AgCl}$ at that $\left[{\text{Cl}}^{\text{-}}\right]$.

Concept introduction:

Solubility:

Solubility is defined as the maximum amount of the solute that can be dissolved in the solvent at equilibrium.

 Solubility product constant:

Solubility product constant is defined for equilibrium between solids and its respective ions in the solution. Generally, solubility product refers only to insoluble or slightly soluble ionic substances that make equilibrium in water.

It is defined as the product of concentration of ions of a sparingly soluble salt in its saturated solution at ${25}^{\text{o}}\text{C}$. The concentration terms are raised to the coefficients of their respective ions. It is denoted by ${\text{K}}_{\text{sp}}$.

This value indicates the degree of dissociation of a compound in water. More the value of ${\text{K}}_{\text{sp}}$ more is the solubility of the compound.

Considering equilibrium of salt ${\text{A}}_{\text{x}}{\text{B}}_{\text{y}}$ with its respective ions in solution and thus the ${\text{K}}_{\text{sp}}$ is calculated.

  ${\text{A}}_{\text{x}}{\text{B}}_{\text{y}}\stackrel{}{\rightleftarrows }{\text{xA}}^{\text{y+}}+{\text{yB}}^{\text{x}}{}^{\text{-}}$

  ${\text{K}}_{\text{sp}}={\left[{\text{xA}}^{\text{y+}}\right]}^{\text{x}}{\left[{\text{yB}}^{\text{x-}}\right]}^{\text{y}}$

Formation constant:

A stability constant or formation constant is an equilibrium constant for the formation of a complex ion in the solution and it measures the strength of interaction between the reactants that forms the complex.

Answer to Problem 19.148P

The solubility of $\text{AgCl}$ at $\left[{\text{Cl}}^{\text{-}}\right]\text{=}\text{2}{\text{.37×10}}^{\text{-3}}\text{M}$ is $1.52{\text{×10}}^{\text{-7}}\text{M}$.

Explanation of Solution

The condition given ${\text{[Ag}}^{\text{+}}\text{]}=\left[{\text{AgCl}}_2^{-}\right]$.

From that it can be concluded that the value of $\left[{\text{Cl}}^{\text{-}}\right]\text{=}\text{2}{\text{.37×10}}^{\text{-3}}\text{M}$.

The condition is that solubility of $\text{AgCl}={\text{[Ag}}^{\text{+}}\text{]+}\left[{\text{AgCl}}_2^{-}\right]$.

It is already found that,

  ${\text{[Ag}}^{\text{+}}\text{]}=\frac{(1.8\times {10}^{-10})}{\left[{\text{Cl}}^{\text{-}}\right]}$

  $\left[{\text{AgCl}}_2^{-}\right]=(3.2\times {10}^{-5})\times \left[{\text{Cl}}^{\text{-}}\right]$

  $\begin{array}{l}\text{Hence, solubility of }\text{AgCl}\text{=}\frac{\text{(1}{\text{.8×10}}^{\text{-10}}\text{)}}{\left[{\text{Cl}}^{\text{-}}\right]}\text{+}\text{(3}{\text{.2×10}}^{\text{-5}}\text{)×}\left[{\text{Cl}}^{\text{-}}\right]\\ \\ \text{=}\frac{\text{(1}{\text{.8×10}}^{\text{-10}}\text{)}}{\text{2}{\text{.37×10}}^{\text{-3}}}\text{+}\text{(3}{\text{.2×10}}^{\text{-5}}\text{)×2}{\text{.37×10}}^{\text{-3}}\\ \\ \text{=}\text{1}{\text{.52×10}}^{\text{-7}}\text{M}\end{array}$

Hence the solubility of $\text{AgCl}$ at $\left[{\text{Cl}}^{\text{-}}\right]\text{=}\text{2}{\text{.37×10}}^{\text{-3}}\text{M}$ is $1.52{\text{×10}}^{\text{-7}}\text{M}$.