Chapter 17, Problem 100GQ

(a)

Interpretation Introduction

Interpretation:

The value of the equilibrium constant $K_{\text{net}}$ has to be calculated for the given reaction.

  $\text{AgBr}\left(\text{s}\right)+2{\text{S}}_2{\text{O}}_3{}^{2-}\left(\text{aq}\right)\stackrel{K_{\text{net}}}{\rightleftharpoons }{\left[\text{Ag}{\left({\text{S}}_2{\text{O}}_3\right)}_2\right]}^{3-}\left(\text{aq}\right)+{\text{Br}}^{-}\left(\text{aq}\right)$

Concept introduction:

Some metal ions when present in an aqueous solution containing anions or neutral species called Lewis base or ligands having a tendency to donate electron pairs to metal ions then complex ion formation will take place.

Example of metal ions that form complex ions includes,${\text{Cd}}^{2+}$,${\text{Fe}}^{2+}$,${\text{Zn}}^{2+}$,${\text{Ni}}^{2+}$ etc.

Example of Lewis bases includes,${\text{NH}}_3$,${\text{OH}}^{-}$ etc.

The complex ion remains in equilibrium with the metal ion and the ligand called complex ion formation equilibrium and the equilibrium constant is called as formation constant $K_{\text{f}}$.

A larger value of $K_{\text{f}}$ implies that the complex ion formed is more stable. $K_{\text{f}}$ is the measure of the strength of the interaction between the metal ions and the Lewis base to form the complex ion.

For example for general complex ion formation reaction,

$\text{xM}+\text{yL}\rightleftharpoons \left[{\text{M}}_{\text{x}}{\text{L}}_{\text{y}}\right]$

$K_{\text{f}}$ can be given as

$K_{\text{f}}=\frac{\left[{\text{M}}_{\text{x}}{\text{L}}_{\text{y}}\right]}{{\left[\text{M}\right]}^{\text{x}}{\left[\text{L}\right]}^{\text{y}}}$

 Here,

• $\left[{\text{M}}_{\text{x}}{\text{L}}_{\text{y}}\right]$ is the equilibrium concentration of complex ion.
• $\left[\text{M}\right]$ is the equilibrium concentration of metal ion.
• $\left[\text{L}\right]$ is the equilibrium concentration of the ligand.
• $\text{x}$ and $\text{y}$ are the coefficients of metal ion and ligand respectively.

Complex ions are stable and thus formation of these increase the solubility of the salt containing the metal ions same as in complex ions. Effect of complex ion formation on the solubility of salt can be explained as below,

$\text{AgBr}$ when dissolved in water does not dissolve completely and dissociates as follows,

  $\text{AgBr}(s)\stackrel{K_{\text{sp}}}{\rightleftharpoons }A{g}^{+}(aq)+{\text{Br}}^{-}(aq)$ (1)

For this reaction $K_{\text{sp}}$ expression is,

$\text{Ksp}=\left[{\text{Ag}}^{+}\right]\left[{\text{Br}}^{-}\right]$

$A{g}^{+}$ ions are capable of forming complex with ligand so when aqueous ammonia solution (strong ligand) is added to the saturated solution of $\text{AgBr}$, $A{g}^{+}$ ions present in the solution form complex with ${\text{NH}}_3$.

  ${\text{Ag}}^{+}\left(\text{aq}\right)+2{\text{NH}}_3\left(\text{aq}\right)\stackrel{K_{\text{f}}}{\rightleftharpoons }{\left[\text{Ag}{\left({\text{NH}}_3\right)}_2\right]}^{+}\left(\text{aq}\right)$ (2)

The expression for formation constant $K_{\text{f}}$ is given as,

$K_{\text{f}}=\frac{\left[{\left[\text{Ag}{\left({\text{NH}}_3\right)}_2\right]}^{+}\right]}{\left[{\text{Ag}}^{+}\right]{\left[{\text{NH}}_3\right]}^2}$

Complex formation leads to the decrease in the concentration of the $A{g}^{+}$ ions in the solution, as a result, according to Le Chatelier’s principle the equilibrium in equation (1) move in the forward direction producing more of the $A{g}^{+}$ ions and the solubility of the slightly soluble salt $\text{AgBr}$ increases.

Adding two equilibrium equation (1) and (2) new overall equilibrium constant can be defined.

Net chemical equation:

  ${\text{Ag}}^{+}\left(\text{aq}\right)+2{\text{NH}}_3\left(\text{aq}\right)\stackrel{K_{\text{net}}}{\rightleftharpoons }{\left[\text{Ag}{\left({\text{NH}}_3\right)}_2\right]}^{+}\left(\text{aq}\right)+{\text{Br}}^{-}(aq)$ (3)

Net equilibrium constant can be given as,

$K_{\text{net}}=\left(K_{\text{sp}}\right)\left(K_{\text{f}}\right)$ (4)

Solubility product constant $K_{\text{sp}}$ is an equilibrium constant and is defined as the product of the equilibrium concentration of the ions of the salt raised to the power of their coefficients in the balanced chemical equation.

For example, general salt ${\text{A}}_x{\text{B}}_y$ when dissolved in water dissociates as,

${\text{A}}_x{\text{B}}_y\left(\text{s}\right)\rightleftharpoons x{\text{A}}^{y+}\left(\text{aq}\right)+y{\text{B}}^{-x}\left(\text{aq}\right)$

The expression for $K_{\text{sp}}$ of a salt is,

$K_{\text{sp}}={\left[{\text{A}}^{y+}\right]}^x{\left[{\text{B}}^{-x}\right]}^y$

Explanation of Solution

The value of equilibrium constant, $K_{\text{net}}$ for the given reaction is calculated below.

Given:

Refer to the Appendix J in the textbook for the value of $K_{\text{sp}}$ .

The value of solubility product constant, $K_{\text{sp}}$ for $\text{AgBr}$ is $5.4\times {10}^{-13}$ .

The value of formation constant, $K_f$ for ${\left[\text{Ag}{\left({\text{S}}_2{\text{O}}_3\right)}_2\right]}^{3-}$ is $2.9\times {10}^{13}$ .

$\text{AgBr}$ dissociates as follows in water,

  $\text{AgBr}\left(\text{s}\right)\stackrel{K_{\text{sp}}}{\rightleftharpoons }{\text{Ag}}^{+}\left(\text{aq}\right)+{\text{Br}}^{-}\left(\text{aq}\right)$ (5)

${\text{Ag}}^{+}$ ions present in solution react with the thio ions added to the solution to form ${\left[\text{Ag}{\left({\text{S}}_2{\text{O}}_3\right)}_2\right]}^{3-}$.

  ${\text{Ag}}^{+}\left(\text{aq}\right)+2{\text{S}}_2{\text{O}}_3{}^{2-}\left(\text{aq}\right)\stackrel{K_{\text{f}}}{\rightleftharpoons }{\left[\text{Ag}{\left({\text{S}}_2{\text{O}}_3\right)}_2\right]}^{3-}\left(\text{aq}\right)$ (6)

Net chemical equation is,

  $\text{AgBr}\left(\text{s}\right)+2{\text{S}}_2{\text{O}}_3{}^{2-}\left(\text{aq}\right)\stackrel{K_{\text{net}}}{\rightleftharpoons }{\left[\text{Ag}{\left({\text{S}}_2{\text{O}}_3\right)}_2\right]}^{3-}\left(\text{aq}\right)+{\text{Br}}^{-}\left(\text{aq}\right)$                   (7)

The net equilibrium constant, $K_{\text{net}}$ for equation (7) is given as,

$K_{\text{net}}=K_{\text{sp}}\times K_{\text{f}}$

Substitute $5.4\times {10}^{-13}$ for $K_{\text{sp}}$ and $2.9\times {10}^{13}$ for $K_{\text{f}}$.

$\begin{array}{c}{K}_{\text{net}}=\left(5.4\times {10}^{-13}\right)\left(2.9\times {10}^{13}\right)\\ =1.56\times 10^1\end{array}$

The equilibrium constant, $K_{\text{net}}$ for the given reaction is $1.56\times 10^1$.

(b)

Interpretation Introduction

Interpretation:

For the given reaction,

  $\text{AgBr}\left(\text{s}\right)+2{\text{S}}_2{\text{O}}_3{}^{2-}\left(\text{aq}\right)\stackrel{K_{\text{net}}}{\rightleftharpoons }{\left[\text{Ag}{\left({\text{S}}_2{\text{O}}_3\right)}_2\right]}^{3-}\left(\text{aq}\right)+{\text{Br}}^{-}\left(\text{aq}\right)$

The mass of ${\text{Na}}_2{\text{S}}_2{\text{O}}_3$ that must be added to dissolve $1.00\text{}\text{ g}$ of $\text{AgBr}$ suspended in $1.00\text{ L}$ of water has to be calculated.

Concept introduction:

Some metal ions when present in an aqueous solution containing anions or neutral species called Lewis base or ligands having a tendency to donate electron pairs to metal ions then complex ion formation will take place.

Example of metal ions that form complex ions includes,${\text{Cd}}^{2+}$,${\text{Fe}}^{2+}$,${\text{Zn}}^{2+}$,${\text{Ni}}^{2+}$ etc.

Example of Lewis bases includes,${\text{NH}}_3$,${\text{OH}}^{-}$ etc.

The complex ion remains in equilibrium with the metal ion and the ligand called complex ion formation equilibrium and the equilibrium constant is called as formation constant $K_{\text{f}}$.

A larger value of $K_{\text{f}}$ implies that the complex ion formed is more stable. $K_{\text{f}}$ is the measure of the strength of the interaction between the metal ions and the Lewis base to form the complex ion.

For example for general complex ion formation reaction,

$\text{xM}+\text{yL}\rightleftharpoons \left[{\text{M}}_{\text{x}}{\text{L}}_{\text{y}}\right]$

$K_{\text{f}}$ can be given as

$K_{\text{f}}=\frac{\left[{\text{M}}_{\text{x}}{\text{L}}_{\text{y}}\right]}{{\left[\text{M}\right]}^{\text{x}}{\left[\text{L}\right]}^{\text{y}}}$

 Here,

• $\left[{\text{M}}_{\text{x}}{\text{L}}_{\text{y}}\right]$ is the equilibrium concentration of complex ion.
• $\left[\text{M}\right]$ is the equilibrium concentration of metal ion.
• $\left[\text{L}\right]$ is the equilibrium concentration of the ligand.
• $\text{x}$ and $\text{y}$ are the coefficients of metal ion and ligand respectively.

Complex ions are stable and thus formation of these increase the solubility of the salt containing the metal ions same as in complex ions. Effect of complex ion formation on the solubility of salt can be explained as below,

$\text{AgBr}$ when dissolved in water does not dissolve completely and dissociates as follows,

  $\text{AgBr}(s)\stackrel{K_{\text{sp}}}{\rightleftharpoons }A{g}^{+}(aq)+{\text{Br}}^{-}(aq)$ (1)

For this reaction $K_{\text{sp}}$ expression is,

$\text{Ksp}=\left[{\text{Ag}}^{+}\right]\left[{\text{Br}}^{-}\right]$

$A{g}^{+}$ ions are capable of forming complex with ligand so when aqueous ammonia solution (strong ligand) is added to the saturated solution of $\text{AgBr}$, $A{g}^{+}$ ions present in the solution form complex with ${\text{NH}}_3$.

  ${\text{Ag}}^{+}\left(\text{aq}\right)+2{\text{NH}}_3\left(\text{aq}\right)\stackrel{K_{\text{f}}}{\rightleftharpoons }{\left[\text{Ag}{\left({\text{NH}}_3\right)}_2\right]}^{+}\left(\text{aq}\right)$ (2)

The expression for formation constant $K_{\text{f}}$ is given as,

$K_{\text{f}}=\frac{\left[{\left[\text{Ag}{\left({\text{NH}}_3\right)}_2\right]}^{+}\right]}{\left[{\text{Ag}}^{+}\right]{\left[{\text{NH}}_3\right]}^2}$

Complex formation leads to the decrease in the concentration of the $A{g}^{+}$ ions in the solution, as a result, according to Le Chatelier’s principle the equilibrium in equation (1) move in the forward direction producing more of the $A{g}^{+}$ ions and the solubility of the slightly soluble salt $\text{AgBr}$ increases.

Adding two equilibrium equation (1) and (2) new overall equilibrium constant can be defined.

Net chemical equation:

  ${\text{Ag}}^{+}\left(\text{aq}\right)+2{\text{NH}}_3\left(\text{aq}\right)\stackrel{K_{\text{net}}}{\rightleftharpoons }{\left[\text{Ag}{\left({\text{NH}}_3\right)}_2\right]}^{+}\left(\text{aq}\right)+{\text{Br}}^{-}(aq)$ (3)

Net equilibrium constant can be given as,

$K_{\text{net}}=\left(K_{\text{sp}}\right)\left(K_{\text{f}}\right)$ (4)

Solubility product constant $K_{\text{sp}}$ is an equilibrium constant and is defined as the product of the equilibrium concentration of the ions of the salt raised to the power of their coefficients in the balanced chemical equation.

For example, general salt ${\text{A}}_x{\text{B}}_y$ when dissolved in water dissociates as,

${\text{A}}_x{\text{B}}_y\left(\text{s}\right)\rightleftharpoons x{\text{A}}^{y+}\left(\text{aq}\right)+y{\text{B}}^{-x}\left(\text{aq}\right)$

The expression for $K_{\text{sp}}$ of a salt is,

$K_{\text{sp}}={\left[{\text{A}}^{y+}\right]}^x{\left[{\text{B}}^{-x}\right]}^y$

The mass of ${\text{Na}}_2{\text{S}}_2{\text{O}}_3$ that must be added to dissolve $1.00\text{}\text{ g}$ of $\text{AgBr}$ is calculated below.

Explanation of Solution

Given:

Mass of $\text{AgBr}$ is $1.00\text{ L}$.

Volume of solution is $1.00\text{ L}$.

The equilibrium constant, $K_{\text{net}}$ for the given reaction is $1.56\times {10}^1$.

Molar mass of ${\text{Na}}_2{\text{S}}_2{\text{O}}_3$ is $158\text{ g}\cdot {\text{mol}}^{-1}$.

In water $\text{AgBr}$ dissociates as,

  $\text{AgBr}\left(\text{s}\right)\rightleftharpoons {\text{Ag}}^{+}\left(\text{aq}\right)+{\text{Br}}^{-}\left(\text{aq}\right)$

Concentration of $\text{AgBr}$ is,

$\begin{array}{c}\left[\text{AgBr}\right]=\frac{1\text{}\text{ g}}{\left(188\text{ g}\cdot {\text{mol}}^{-1}\right)\left(1\text{ L}\right)}\\ =5.33\times {10}^{-3}\text{ mol}\cdot {\text{L}}^{-1}\end{array}$

If all the $\text{AgBr}$ dissolves and all ${\text{Ag}}^{+}$ ions produce undergoes complex formation with thio ions then the concentration of ${\text{Br}}^{-}$ and ${\left[\text{Ag}{\left({\text{S}}_2{\text{O}}_3\right)}_2\right]}^{3-}$ is equal to $5.33\times {10}^{-3}\text{ mol}\cdot {\text{L}}^{-1}$.

Each ${\text{Ag}}^{+}$ ions requires $2$ moles of ${\text{S}}_2{\text{O}}_3{}^{2-}$ ions therefore concentration of ${\text{S}}_2{\text{O}}_3{}^{2-}$ is,

$\begin{array}{c}\left[{\text{S}}_2{\text{O}}_3{}^{2-}\right]=2\left(5.33\times {10}^{-3}\text{ mol}\cdot {\text{L}}^{-1}\right)\\ =0.01064\text{mol}\cdot {\text{L}}^{-1}\end{array}$

Additional concentration of ${\text{S}}_2{\text{O}}_3{}^{2-}$ is required for maintaining equilibrium. From equation (7),

$K_{\text{net}}=\frac{\left[{\left[\text{Ag}{\left({\text{S}}_2{\text{O}}_3\right)}_2\right]}^{3-}\right]\left[{\text{Br}}^{-1}\right]}{{\left[{\text{S}}_2{\text{O}}_3{}^{2-}\right]}^2}$

Substitute the values.

$\begin{array}{c}\left[{\text{S}}_2{\text{O}}_3{}^{2-}\right]=\sqrt{\frac{{\left(5.32\times {10}^{-3}\right)}^2}{\left(1.56\times {10}^1\right)}}\\ =1.34\times {10}^{-3}\text{ mol}\cdot {\text{L}}^{-1}\end{array}$

Total moles of ${\text{S}}_2{\text{O}}_3{}^{2-}$ required to dissolve $1.00\text{}\text{ g}$ of $\text{AgBr}$ per liter of water is,

$\begin{array}{c}\text{n}=\left(0.01064\text{mol}\right)+\left(1.34\times {10}^{-3}\text{ mol}\right)\\ =0.0119\text{}\text{ mol}\end{array}$

Mass of ${\text{Na}}_2{\text{S}}_2{\text{O}}_3$ is,

$\begin{array}{c}\text{m}=\left(0.0119\text{}\text{ mol}\right)\left(158\text{ g}\cdot {\text{mol}}^{-1}\right)\\ =1.89g\end{array}$

The mass of ${\text{Na}}_2{\text{S}}_2{\text{O}}_3$ that must be added to dissolve $1.00\text{}\text{ g}$ of $\text{AgBr}$ suspended in $1.00\text{ L}$ of water is $1.89\text{}\text{}g$.