Chapter 6, Problem 6.AE

(a)

Interpretation Introduction

Interpretation:

For given equilibria,

The equilibrium constant (K) for given reaction has to be calculated.

Concept introduction:

Equilibrium constant (K): In the equilibrium reaction, the ratio of concentration of the reactant and concentration of the product.  If value of K is small than 1 the reaction should be move to the left and k is more than 1 the reaction should be move to right.

Answer to Problem 6.AE

The equilibrium constant (K) for given reaction is $\text{= 3}{\text{.6×10}}^{\text{-7}}$

Explanation of Solution

Given

$\begin{array}{l}{\text{Ag}}^{\text{+}}\text{+ Cl}\stackrel{}{\rightleftharpoons }{\text{ AgCl}}_{\text{(aq)}}K\text{ =2}{\text{.0×10}}^{\text{3}}\\ \end{array}$

$\begin{array}{l}{\text{AgCl}}_{\text{(s)}}\stackrel{}{\rightleftharpoons }{\text{ Ag}}^{\text{+}}{\text{+ Cl}}^{\text{-}}K\text{ = 1}{\text{.8×10}}^{-10}\\ \end{array}$

By adding these two equilibria and multiply these equilibrium constant we can get equilibrium constant and given chemical equation.

$\underset{\_}{\begin{array}{l}\underset{\_}{\begin{array}{l}{\text{Ag}}^{\text{+}}\text{+ Cl}\stackrel{}{\rightleftharpoons }{\text{ AgCl}}_{\text{(aq)}}{\text{ K}}_{\text{1}}\text{ =2}{\text{.0×10}}^{\text{3}}\\ {\text{AgCl}}_{\text{(s)}}\stackrel{}{\rightleftharpoons }{\text{ Ag}}^{\text{+}}{\text{+ Cl}}^{\text{-}}{\text{ K}}_{\text{2}}\text{ = 1}{\text{.8×10}}^{\text{-10}}\end{array}}\\ {\text{AgCl}}_{\text{(s)}}\stackrel{}{\rightleftharpoons }{\text{AgCl}}_{\text{(aq)}}{\text{ K}}_{\text{3}}\text{=3}{\text{.6×10}}^{\text{-7}}\end{array}}$

(b)

Interpretation Introduction

Interpretation:

For given equilibria,

The concentration of aqueous $\text{AgCl}$ has to be calculated. ${\text{AgCl}}_{\text{(aq)}}$ is equilibrium with excess undissolved ${\text{AgCl}}_{\text{(s)}}$.

Concept introduction:

Equilibrium constant (K): In the equilibrium reaction, the ratio of concentration of the reactant and concentration of the product.  If value of K is small than 1 the reaction should be move to the left and k is more than 1 the reaction should be move to right.

When reaction attains equilibrium, the association between concentration of reactant and product are defined by equilibrium constant.

Answer to Problem 6.AE

The concentration of aqueous $\text{AgCl}$ is $\text{= 3}{\text{.6×10}}^{\text{-7}}\text{M}$

Explanation of Solution

Given

$\begin{array}{l}{\text{Ag}}^{\text{+}}\text{+ Cl}\stackrel{}{\rightleftharpoons }{\text{ AgCl}}_{\text{(aq)}}K\text{ =2}{\text{.0×10}}^{\text{3}}\\ \end{array}$

$\begin{array}{l}{\text{AgCl}}_{\text{(s)}}\stackrel{}{\rightleftharpoons }{\text{ Ag}}^{\text{+}}{\text{+ Cl}}^{\text{-}}K\text{ = 1}{\text{.8×10}}^{-10}\\ \end{array}$

By adding these two equilibria and multiply these equilibrium constant we can get equilibrium constant and given chemical equation.

$\underset{\_}{\begin{array}{l}\underset{\_}{\begin{array}{l}{\text{Ag}}^{\text{+}}\text{+ Cl}\stackrel{}{\rightleftharpoons }{\text{ AgCl}}_{\text{(aq)}}{\text{ K}}_{\text{1}}\text{ =2}{\text{.0×10}}^{\text{3}}\\ {\text{AgCl}}_{\text{(s)}}\stackrel{}{\rightleftharpoons }{\text{ Ag}}^{\text{+}}{\text{+ Cl}}^{\text{-}}{\text{ K}}_{\text{2}}\text{ = 1}{\text{.8×10}}^{\text{-10}}\end{array}}\\ {\text{AgCl}}_{\text{(s)}}\stackrel{}{\rightleftharpoons }{\text{AgCl}}_{\text{(aq)}}{\text{ K}}_{\text{3}}\text{=3}{\text{.6×10}}^{\text{-7}}\end{array}}$

When reaction attains equilibrium, the association between concentration of reactant and product are defined by equilibrium constant. Hence, the concentration will be the same as equilibrium constant.

Thus, the concentration of aqueous $\text{AgCl}$ is found to be $\text{3}{\text{.6×10}}^{\text{-7}}\text{M}$.

(c)

Interpretation Introduction

Interpretation:

For given reaction equilibrium constant has to be calculated.

Concept introduction:

Equilibrium constant (K): In the equilibrium reaction, the ratio of concentration of the reactant and concentration of the product.  If value of K is small than 1 the reaction should be move to the left and k is more than 1 the reaction should be move to right.

Answer to Problem 6.AE

The equilibrium constant (K) for given reaction is $\text{= 3}{\text{.0×10}}^4$

Explanation of Solution

Given

$\begin{array}{l}{\text{AgCl}}_{\text{2}}^{\text{-}}\stackrel{}{\rightleftharpoons }{\text{AgCl}}_{\text{(aq)}}{\text{+Cl}}^{\text{-}}\text{K=9}{\text{.3×10}}^{\text{1}}\\ \\ {\text{Ag}}^{\text{+}}\text{+ Cl}\stackrel{}{\rightleftharpoons }{\text{ AgCl}}_{\text{(s)}}\text{K = 1}{\text{.8×10}}^{\text{-10}}\\ \\ {\text{AgCl}}_{\text{(aq)}}\stackrel{}{\rightleftharpoons }{\text{AgCl}}_{\text{(s)}}{\text{+Cl}}^{\text{-}}\text{K=2}{\text{.0×10}}^{\text{3}}\\ \end{array}$

By adding these two equilibria and multiply these equilibrium constant we can get equilibrium constant and given chemical equation.

If the reaction is reversed, K = $\frac{\text{1}}{\text{K}}$

$\begin{array}{l}\underset{\_}{\begin{array}{l}{\text{AgCl}}_{\text{2}}^{\text{-}}\stackrel{}{\rightleftharpoons }{\text{AgCl}}_{\text{(aq)}}{\text{+Cl}}^{\text{-}}1/{\text{K}}_{\text{1}}\text{=9}{\text{.3×10}}^{\text{1}}\\ {\text{Ag}}^{\text{+}}\text{+ Cl}\stackrel{}{\rightleftharpoons }{\text{ AgCl}}_{\text{(s)}}1/{\text{K}}_{\text{2}}\text{= 1}{\text{.8×10}}^{\text{-10}}\\ {\text{AgCl}}_{\text{(aq)}}\stackrel{}{\rightleftharpoons }{\text{AgCl}}_{\text{(s)}}{\text{+Cl}}^{\text{-}}1/{\text{K}}_{\text{3}}\text{=2}{\text{.0×10}}^{\text{3}}\\ \end{array}}\\ {\text{AgCl}}_{\text{2}}^{\text{-}}\stackrel{}{\rightleftharpoons }{\text{AgCl}}_{\text{(s)}}{\text{+Cl}}^{\text{-}}{\text{K}}_{\text{3}}\text{=3}{\text{.0×10}}^4\end{array}$