Chapter 16, Problem 41E

Interpretation Introduction

(a)

Interpretation:

The direction of the equilibrium with the increase in the concentration of $\left[{\text{Cu}}^{\text{2+}}\right]$ is to be stated.

Concept introduction:

In a chemical reaction, when the rate of both forward and reverse reaction are equal. Then the chemical reaction is in equilibrium. In other words, their concentration does not vary with the time change. They are reversible in nature.

It is represented as $\rightleftarrows$ between the chemical reaction. It is denoted as $K_{eq}$. It is known as equilibrium constant. Mathematically, it is represented as the ratio of product concentration to reactant concentration raised to their stoichiometric coefficients as shown below.

$K_{eq}=\frac{{\left[\text{Product}\text{concentration}\right]}^{\text{a}}}{{\left[\text{Reactant}\text{concentration}\right]}^{\text{b}}}$

Also, the concentration of liquid and solid is not considered in the equilibrium constant expression.

Answer to Problem 41E

The equilibrium shifts to left with the increase in the concentration of $\left[{\text{Cu}}^{\text{2+}}\right]$.

Explanation of Solution

According to the Le Chatelier’s principle, the change in concentration, volume, pressure and temperature affect the equilibrium of the reaction.

The reaction is given below.

$\text{Cu}{\left(\text{OH}\right)}_{\text{2}}\left(s\right)\rightleftarrows {\text{Cu}}^{\text{2+}}\left(aq\right)+{\text{2OH}}^{-}\left(aq\right)$…(1)

According to the definition of the equilibrium constant,. The $\left[\text{Cu}{\left(\text{OH}\right)}_{\text{2}}\right]$ molecule is present in solid state so its concentration is not considered. The equilibrium constant is written as shown below.

$K_{eq}=\left[{\text{Cu}}^{\text{2+}}\right]{\left[{\text{OH}}^{-}\right]}^2$…(2)

In this case, $\left[{\text{Cu}}^{\text{2+}}\right]$ ions are present on the product side or right side. When its concentration is increased, it increases the equilibrium constant. Therefore, in order to maintain the equilibrium, the reaction will shift towards left.

Conclusion

The equilibrium shifts to left if the concentration of $\left[{\text{Cu}}^{\text{2+}}\right]$ is increased.

Interpretation Introduction

(b)

Interpretation:

The direction of the equilibrium with the increase in the concentration of $\left[{\text{OH}}^{-}\right]$ is to be stated.

Concept introduction:

In a chemical reaction, when the rate of both forward and reverse reaction are equal. Then the chemical reaction is in equilibrium. In other words, their concentration does not vary with the time change. They are reversible in nature.

It is represented as $\rightleftarrows$ between the chemical reaction. It is denoted as $K_{eq}$. It is known as equilibrium constant. Mathematically, it is represented as the ratio of product concentration to reactant concentration raised to their stoichiometric coefficients as shown below.

$K_{eq}=\frac{{\left[\text{Product}\text{concentration}\right]}^{\text{a}}}{{\left[\text{Reactant}\text{concentration}\right]}^{\text{b}}}$

Also, the concentration of liquid and solid is not considered in the equilibrium constant expression.

Answer to Problem 41E

The equilibrium shifts to left with the increase in the concentration of $\left[{\text{OH}}^{-}\right]$.

Explanation of Solution

According to the Le Chatelier’s principle, the change in concentration, volume, pressure and temperature affect the equilibrium of the reaction.

The reaction is given below.

$\text{Cu}{\left(\text{OH}\right)}_{\text{2}}\left(s\right)\rightleftarrows {\text{Cu}}^{\text{2+}}\left(aq\right)+{\text{2OH}}^{-}\left(aq\right)$…(1)

According to the definition of the equilibrium constant,. The $\left[\text{Cu}{\left(\text{OH}\right)}_{\text{2}}\right]$ molecule is present in solid state so its concentration is not considered. The equilibrium constant is written as shown below.

$K_{eq}=\left[{\text{Cu}}^{\text{2+}}\right]{\left[{\text{OH}}^{-}\right]}^2$…(2)

In this case, $\left[{\text{OH}}^{-}\right]$ ions are present on the product side or right side. When its concentration is increased, it increases the equilibrium constant. Therefore, in order to maintain the equilibrium, the reaction will shift towards left.

Conclusion

The equilibrium shifts to left if the concentration of $\left[{\text{OH}}^{-}\right]$ is increased.

Interpretation Introduction

(c)

Interpretation:

The direction of the equilibrium with the decrease of the concentration of $\left[{\text{Cu}}^{\text{2+}}\right]$ is to be stated.

Concept introduction:

In a chemical reaction, when the rate of both forward and reverse reaction are equal. Then the chemical reaction is in equilibrium. In other words, their concentration does not vary with the time change. They are reversible in nature.

It is represented as $\rightleftarrows$ between the chemical reaction. It is denoted as $K_{eq}$. It is known as equilibrium constant. Mathematically, it is represented as the ratio of product concentration to reactant concentration raised to their stoichiometric coefficients as shown below.

$K_{eq}=\frac{{\left[\text{Product}\text{concentration}\right]}^{\text{a}}}{{\left[\text{Reactant}\text{concentration}\right]}^{\text{b}}}$

Also, the concentration of liquid and solid is not considered in the equilibrium constant expression.

Answer to Problem 41E

The equilibrium shifts to right with the decrease in the concentration of $\left[{\text{Cu}}^{\text{2+}}\right]$.

Explanation of Solution

According to the Le Chatelier’s principle, the change in concentration, volume, pressure and temperature affect the equilibrium of the reaction.

The reaction is given below.

$\text{Cu}{\left(\text{OH}\right)}_{\text{2}}\left(s\right)\rightleftarrows {\text{Cu}}^{\text{2+}}\left(aq\right)+{\text{2OH}}^{-}\left(aq\right)$…(1)

According to the definition of the equilibrium constant $K_{\text{eq}}$. The $\left[\text{Cu}{\left(\text{OH}\right)}_{\text{2}}\right]$ molecule is present in solid state so its concentration is not considered. The equilibrium constant is written as shown below.

$K_{eq}=\left[{\text{Cu}}^{\text{2+}}\right]{\left[{\text{OH}}^{-}\right]}^2$…(2)

In this case, $\left[{\text{Cu}}^{\text{2+}}\right]$ ions are present on the product side or right side. When its concentration is decreased, it decreases the equilibrium constant. Therefore, in order to maintain the equilibrium, the reaction will shift towards right.

Conclusion

The equilibrium shifts to the right if the concentration of $\left[{\text{Cu}}^{\text{2+}}\right]$ is decreased.

Interpretation Introduction

(d)

Interpretation:

The direction of the equilibrium with the decrease in the concentration of $\left[{\text{OH}}^{-}\right]$ is to be stated.

Concept introduction:

In a chemical reaction, when the rate of both forward and reverse reaction are equal. Then the chemical reaction is in equilibrium. In other words, their concentration does not vary with the time change. They are reversible in nature.

It is represented as $\rightleftarrows$ between the chemical reaction. It is denoted as $K_{eq}$. It is known as equilibrium constant. Mathematically, it is represented as the ratio of product concentration to reactant concentration raised to their stoichiometric coefficients as shown below.

$K_{eq}=\frac{{\left[\text{Product}\text{concentration}\right]}^{\text{a}}}{{\left[\text{Reactant}\text{concentration}\right]}^{\text{b}}}$

Also, the concentration of liquid and solid is not considered in the equilibrium constant expression.

Answer to Problem 41E

The equilibrium shifts to right with the decrease in the concentration of $\left[{\text{OH}}^{-}\right]$.

Explanation of Solution

According to the Le Chatelier’s principle, the change in concentration, volume, pressure and temperature affect the equilibrium of the reaction.

The reaction is given below.

$\text{Cu}{\left(\text{OH}\right)}_{\text{2}}\left(s\right)\rightleftarrows {\text{Cu}}^{\text{2+}}\left(aq\right)+{\text{2OH}}^{-}\left(aq\right)$…(1)

According to the definition of the equilibrium constant $K_{\text{eq}}$. The $\left[\text{Cu}{\left(\text{OH}\right)}_{\text{2}}\right]$ molecule is present in solid state so its concentration is not considered. The equilibrium constant is written as shown below.

$K_{eq}=\left[{\text{Cu}}^{\text{2+}}\right]{\left[{\text{OH}}^{-}\right]}^2$…(2)

In this case, $\left[{\text{OH}}^{-}\right]$ ions are present on the product side or right side. When its concentration is decreased, it decreases the equilibrium constant. Therefore, in order to maintain the equilibrium, the reaction will shift towards right.

Conclusion

The equilibrium shifts to the right if the concentration of $\left[{\text{OH}}^{-}\right]$ is decreased.

Interpretation Introduction

(e)

Interpretation:

The direction of the equilibrium with the addition of solid $\text{Cu}\left({\text{OH}}_{\text{2}}\right)$ is to be stated.

Concept introduction:

In a chemical reaction, when the rate of both forward and reverse reaction are equal. Then the chemical reaction is in equilibrium. In other words, their concentration does not vary with the time change. They are reversible in nature.

It is represented as $\rightleftarrows$ between the chemical reaction. It is denoted as $K_{eq}$. It is known as equilibrium constant. Mathematically, it is represented as the ratio of product concentration to reactant concentration raised to their stoichiometric coefficients as shown below.

$K_{eq}=\frac{{\left[\text{Product}\text{concentration}\right]}^{\text{a}}}{{\left[\text{Reactant}\text{concentration}\right]}^{\text{b}}}$

Also, the concentration of liquid and solid is not considered in the equilibrium constant expression.

Answer to Problem 41E

It does not affect the equilibrium direction if solid $\text{Cu}\left({\text{OH}}_{\text{2}}\right)$ is added.

Explanation of Solution

According to the Le Chatelier’s principle, the change in concentration, volume, pressure and temperature affect the equilibrium of the reaction.

The reaction is given below.

$\text{Cu}{\left(\text{OH}\right)}_{\text{2}}\left(s\right)\rightleftarrows {\text{Cu}}^{\text{2+}}\left(aq\right)+{\text{2OH}}^{-}\left(aq\right)$…(1)

According to the definition of the equilibrium constant $K_{eq}$. The $\left[\text{Cu}{\left(\text{OH}\right)}_{\text{2}}\right]$ molecule is present in solid state so its concentration is not considered. The equilibrium constant is written as shown below.

$K_{eq}=\left[{\text{Cu}}^{\text{2+}}\right]{\left[{\text{OH}}^{-}\right]}^2$…(2)

Since $\text{Cu}{\left(\text{OH}\right)}_{\text{2}}$ is not present in the equilibrium constant equation. Therefore the addition of solid $\text{Cu}{\left(\text{OH}\right)}_{\text{2}}$ does not cause any change in the value of the equilibrium constant value or no change in the direction of equilibrium.

Conclusion

If solid $\text{Cu}\left({\text{OH}}_{\text{2}}\right)$ is added then it does not cause any change in the direction of the equilibrium.

Interpretation Introduction

(f)

Interpretation:

The direction of the equilibrium with the addition of solid $\text{NaOH}$ is to be stated.

Concept introduction:

In a chemical reaction, when the rate of both forward and reverse reaction are equal. Then the chemical reaction is in equilibrium. In other words, their concentration does not vary with the time change. They are reversible in nature.

It is represented as $\rightleftarrows$ between the chemical reaction. It is denoted as $K_{eq}$. It is known as equilibrium constant. Mathematically, it is represented as the ratio of product concentration to reactant concentration raised to their stoichiometric coefficients as shown below.

$K_{eq}=\frac{{\left[\text{Product}\text{concentration}\right]}^{\text{a}}}{{\left[\text{Reactant}\text{concentration}\right]}^{\text{b}}}$

Also, the concentration of liquid and solid is not considered in the equilibrium constant expression.

Answer to Problem 41E

If solid $\text{NaOH}$ is added it shifts the equilibrium towards left.

Explanation of Solution

According to the Le Chatelier’s principle, the change in concentration, volume, pressure and temperature affect the equilibrium of the reaction.

The reaction is given below.

$\text{Cu}{\left(\text{OH}\right)}_{\text{2}}\left(s\right)\rightleftarrows {\text{Cu}}^{\text{2+}}\left(aq\right)+{\text{2OH}}^{-}\left(aq\right)$…(1)

According to the definition of the equilibrium constant $K_{eq}$. The $\left[\text{Cu}{\left(\text{OH}\right)}_{\text{2}}\right]$ molecule is present in solid state so its concentration is not considered. The equilibrium constant is written as shown below.

$K_{eq}=\left[{\text{Cu}}^{\text{2+}}\right]{\left[{\text{OH}}^{-}\right]}^2$…(2)

The solid $\left[\text{NaOH}\right]$ is basically deliquescent in nature so its phase is aqueous. It also contains $\left[{\text{OH}}^{-}\right]$ ions. Threfore, the addition of solid $\left[\text{NaOH}\right]$ increases the equilibrium constant value which shift the equilibrium toward left.

Conclusion

The direction of equilibrium is left if solid $\text{NaOH}$ is added.

Interpretation Introduction

(g)

Interpretation:

The direction of the equilibrium with the addition of solid $\text{NaCl}$ is to be stated.

Concept introduction:

In a chemical reaction, when the rate of both forward and reverse reaction are equal. Then the chemical reaction is in equilibrium. In other words, their concentration does not vary with the time change. They are reversible in nature.

It is represented as $\rightleftarrows$ between the chemical reaction. It is denoted as $K_{eq}$. It is known as equilibrium constant. Mathematically, it is represented as the ratio of product concentration to reactant concentration raised to their stoichiometric coefficients as shown below.

$K_{eq}=\frac{{\left[\text{Product}\text{concentration}\right]}^{\text{a}}}{{\left[\text{Reactant}\text{concentration}\right]}^{\text{b}}}$

Also, the concentration of liquid and solid is not considered in the equilibrium constant expression.

Answer to Problem 41E

It does not affect the equilibrium direction if solid $\text{NaCl}$ is added.

Explanation of Solution

According to the Le Chatelier’s principle, the change in concentration, volume, pressure and temperature affect the equilibrium of the reaction.

The reaction is given below.

$\text{Cu}{\left(\text{OH}\right)}_{\text{2}}\left(s\right)\rightleftarrows {\text{Cu}}^{\text{2+}}\left(aq\right)+{\text{2OH}}^{-}\left(aq\right)$…(1)

According to the definition of the equilibrium constant ${\text{K}}_{\text{eq}}$. The $\left[\text{Cu}{\left(\text{OH}\right)}_{\text{2}}\right]$ molecule is present in solid state so its concentration is not considered. The equilibrium constant is written as shown below.

$K_{eq}=\left[{\text{Cu}}^{\text{2+}}\right]{\left[{\text{OH}}^{-}\right]}^2$…(2)

From equation (2) there is no effect on the equilibrium. When the solid $\text{NaCl}$ is added.

Conclusion

It does not cause any change in the equilibrium direction if solid $\text{NaCl}$ is added.

Interpretation Introduction

(h)

Interpretation:

The direction of the equilibrium with the decrease in its $\text{pH }$ is to be stated.

Concept introduction:

In a chemical reaction, when the rate of both forward and reverse reaction are equal. Then the chemical reaction is in equilibrium. In other words, their concentration does not vary with the time change. They are reversible in nature.

It is represented as $\rightleftarrows$ between the chemical reaction. It is denoted as $K_{eq}$. It is known as equilibrium constant. Mathematically, it is represented as the ratio of product concentration to reactant concentration raised to their stoichiometric coefficients as shown below.

$K_{eq}=\frac{{\left[\text{Product}\text{concentration}\right]}^{\text{a}}}{{\left[\text{Reactant}\text{concentration}\right]}^{\text{b}}}$

Also, the concentration of liquid and solid is not considered in the equilibrium constant expression.

Answer to Problem 41E

It shifts the direction of equilibrium to the left if $\text{pH}$ is decreased.

Explanation of Solution

According to the Le Chatelier’s principle, the change in concentration, volume, pressure and temperature affect the equilibrium of the reaction.

The reaction is given below.

$\text{Cu}{\left(\text{OH}\right)}_{\text{2}}\left(s\right)\rightleftarrows {\text{Cu}}^{\text{2+}}\left(aq\right)+{\text{2OH}}^{-}\left(aq\right)$…(1)

According to the definition of the equilibrium constant $K_{eq}$. The $\left[\text{Cu}{\left(\text{OH}\right)}_{\text{2}}\right]$ molecule is present in solid state so its concentration is not considered. The equilibrium constant is written as shown below.

$K_{eq}=\left[{\text{Cu}}^{\text{2+}}\right]{\left[{\text{OH}}^{-}\right]}^2$…(2)

When $\text{pH}$ is decreased it means the number of $\left[{\text{OH}}^{-}\right]$ ions decreases. The $\left[{\text{OH}}^{-}\right]$ ions are present on the right side. When the $\left[{\text{OH}}^{-}\right]$ ions decrease, it decreases the equilibrium constant value. Therefore, in order to maintain the equilibrium, the reaction will shift towards right.

Conclusion

The direction of equilibrium is left if $\text{pH}$ is decreased.