Chapter 15, Problem 14ALQ

Interpretation Introduction

(a)

Interpretation:

The picture of the solution made by mixing solution A and B together after the precipitation reaction takes place is to be drawn.

Concept Introduction:

The molarity of a solution is the molar concentration of the solution; it measures the number of moles of solute dissolved in one liter of the solution. The formula for molarity is given as,

$\text{M}=\frac{\text{n}}{\text{V}}$

Where,

• $\text{n}$ represents the number of moles of the solute.
• $\text{V}$ represents the volume of the solution.

The limiting reagent of a reaction is that reactant of the reaction that controls the amount of product formed. The amount of a limiting reagent is lower than its required amount to complete the reaction.

Answer to Problem 14ALQ

The picture of the solution made by mixing solution A and B together after the precipitation reaction takes place is represented as,

Figure 1.

Explanation of Solution

The volume of $2.00\text{M}$

$\text{copper}\left(\text{II}\right)\text{ nitrate}$ solution is $2.00\text{L}$.

The volume of $3.00\text{M}$ potassium hydroxide solution is $2.00\text{L}$.

The molarity of a solution is given as,

$\text{M}=\frac{\text{n}}{\text{V}}$    (1)

Where,

• $\text{n}$ represents the number of moles of the solute.
• $\text{V}$ represents the volume of the solution.

Rearrange the above equation for the value of $\text{n}$.

$\text{n}=\text{M}\text{V}$    (2)

Substitute the value of molarity and volume of $\text{copper}\left(\text{II}\right)\text{ nitrate}$ solution in equation (2).

$\begin{array}{c}\text{n}=\left(2.00\text{M}\right)\left(\frac{1\text{mol}\cdot {\text{L}}^{-1}}{1\text{M}}\right)\left(2.00\text{L}\right)\\ =4.00\text{mol}\end{array}$

The number of moles of $\text{copper}\left(\text{II}\right)\text{ nitrate}$ present in the reaction mixture is $4.00\text{mol}$.

Substitute the value of molarity and volume of potassium hydroxide solution in equation (2).

$\begin{array}{c}\text{n}=\left(3.00\text{M}\right)\left(\frac{1\text{mol}\cdot {\text{L}}^{-1}}{1\text{M}}\right)\left(2.00\text{L}\right)\\ =6.00\text{mol}\end{array}$

The number of moles of potassium hydroxide present in the reaction mixture is $6.00\text{mol}$.

The balance chemical reaction between $\text{copper}\left(\text{II}\right)\text{ nitrate}$ and potassium hydroxide is represented as,

$\text{Cu}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}\left(\text{a}\text{q}\right)+2\text{KOH}\left(\text{a}\text{q}\right)\to \text{Cu}{\left(\text{OH}\right)}_{\text{2}}\left(\text{s}\right)+2{\text{KNO}}_{\text{3}}\left(\text{a}\text{q}\right)$

The complete ionic reaction between $\text{copper}\left(\text{II}\right)\text{ nitrate}$ and potassium hydroxide is represented as,

$\left(\begin{array}{l}{\text{Cu}}^{2+}\left(\text{a}\text{q}\right)+2{\text{NO}}_3^{-}\left(\text{a}\text{q}\right)\\ +2{\text{K}}^{+}\left(\text{a}\text{q}\right)+2{\text{OH}}^{-}\left(\text{a}\text{q}\right)\end{array}\right)\to \text{Cu}{\left(\text{OH}\right)}_{\text{2}}\left(\text{s}\right)+2{\text{K}}^{+}\left(\text{a}\text{q}\right)+2{\text{NO}}_3^{-}\left(\text{a}\text{q}\right)$

One moles of $\text{Cu}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}$ reacts with two moles of $\text{KOH}$ to produce one mole of $\text{Cu}{\left(\text{OH}\right)}_{\text{2}}$ and two moles of ${\text{KNO}}_{\text{3}}$. Therefore, the relation between the number of moles of $\text{Cu}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}$ and $\text{KOH}$ reacts is given as,

${\text{n}}_{\text{Cu}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}}=\frac{{\text{n}}_{\text{KOH}}}{2}$

Where,

• ${\text{n}}_{\text{KOH}}$ represents the number of moles of $\text{KOH}$.
• ${\text{n}}_{\text{Cu}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}}$ represents the number of moles of $\text{Cu}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}$.

Substitute the value of number of moles of $\text{KOH}$ in the above equation.

$\begin{array}{c}{\text{n}}_{\text{Cu}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}}=\frac{6.00\text{mol}}{2}\\ =3.00\text{mol}\end{array}$

The number of moles of $\text{Cu}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}$ required to react with $6.00\text{mol}$ of $\text{KOH}$ is $3.00\text{mol}$ but the number of moles of $\text{Cu}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}$ present in the reaction mixture is $4.00\text{mol}$. Therefore, some amount of $\text{Cu}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}$ will remain unreacted in the reaction mixture. The ions that will remain in the solution even after the completion of the reaction are ${\text{K}}^{+}$, ${\text{Cu}}^{2+}$ and ${\text{NO}}_3^{-}$. The picture of the solution made by mixing solution A and B together after the precipitation reaction takes place is represented as,

Figure 1.

Interpretation Introduction

(b)

Interpretation:

The concentration of all ions left in solution after completion of the reaction and mass of the solid formed is to be calculated.

Concept Introduction:

The molarity of a solution is the molar concentration of the solution; it measures the number of moles of solute dissolved in one liter of the solution. The formula for molarity is given as,

$\text{M}=\frac{\text{n}}{\text{V}}$

Where,

• $\text{n}$ represents the number of moles of the solute.
• $\text{V}$ represents the volume of the solution.

Answer to Problem 14ALQ

The concentration of ${\text{Cu}}^{2+}$, ${\text{NO}}_3^{-}$ and ${\text{K}}^{+}$ ions left in solution after completion of the reaction between given solution of $\text{copper}\left(\text{II}\right)\text{ nitrate}$ and potassium hydroxide is $0.25\text{M}$, $2.00\text{M}$ and $1.50\text{M}$ respectively. The mass of the precipitate formed after reaction between given solution of $\text{copper}\left(\text{II}\right)\text{ nitrate}$ and potassium hydroxide is $292.68\text{g}$.

Explanation of Solution

The complete ionic reaction between $\text{copper}\left(\text{II}\right)\text{ nitrate}$ and potassium hydroxide is represented as,

$\left(\begin{array}{l}{\text{Cu}}^{2+}\left(\text{a}\text{q}\right)+2{\text{NO}}_3^{-}\left(\text{a}\text{q}\right)\\ +2{\text{K}}^{+}\left(\text{a}\text{q}\right)+2{\text{OH}}^{-}\left(\text{a}\text{q}\right)\end{array}\right)\to \text{Cu}{\left(\text{OH}\right)}_{\text{2}}\left(\text{s}\right)+2{\text{K}}^{+}\left(\text{a}\text{q}\right)+2{\text{NO}}_3^{-}\left(\text{a}\text{q}\right)$

The initial number of moles of $\text{copper}\left(\text{II}\right)\text{ nitrate}$ is $4.00\text{mol}$.

The number of mole of $\text{copper}\left(\text{II}\right)\text{ nitrate}$ reacted is $3.00\text{mol}$.

The number of moles of $\text{copper}\left(\text{II}\right)\text{ nitrate}$ remains unreacted in the solution is given as,

$\text{n}={\text{n}}_1-{\text{n}}_2$

Where,

• ${\text{n}}_1$ represents the initial number of moles of $\text{copper}\left(\text{II}\right)\text{ nitrate}$.
• ${\text{n}}_2$ represents the number of moles of $\text{copper}\left(\text{II}\right)\text{ nitrate}$ reacted.

Substitute the value of ${\text{n}}_1$ and ${\text{n}}_2$ in the above equation.

$\begin{array}{c}\text{n}=4.00\text{mol}-3.00\text{mol}\\ =1.00\text{mol}\end{array}$

One mole of $\text{Cu}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}$ dissociates in one mole of ${\text{Cu}}^{2+}$ ion and two moles of ${\text{NO}}_3^{-}$ ion. The number of moles of ${\text{Cu}}^{2+}$ remains in the solution is $1.00\text{mol}$.

The nitrate ions are not precipitated in the reaction therefore, the number of moles of nitrate ion present in the reaction before the reaction started and after the reaction completed will remain same. The relation between the number of moles of $\text{Cu}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}$ and ${\text{NO}}_3^{-}$ ion is given as,

${\text{n}}_{{\text{NO}}_3^{-}}=2{\text{n}}_{\text{Cu}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}}$

Where,

• ${\text{n}}_{{\text{NO}}_3^{-}}$ represents the number of moles of ${\text{NO}}_3^{-}$.
• ${\text{n}}_{\text{Cu}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}}$ represents the number of moles of $\text{Cu}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}$.

Substitute the value of ${\text{n}}_{\text{Cu}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}}$ in the above equation.

$\begin{array}{c}{\text{n}}_{{\text{NO}}_3^{-}}=\left(2\right)\left(4.00\text{mol}\right)\\ =8.00\text{mol}\end{array}$

The number of moles of ${\text{NO}}_3^{-}$ ions present in the solution after completion of the reaction is $8.00\text{mol}$.

The ${\text{K}}^{+}$ ions are also not precipitated in the reaction. Therefore, the number of moles of ${\text{K}}^{+}$ present in the reaction mixture is $6.00\text{mol}$.

The volume of $2.00\text{M}$

$\text{copper}\left(\text{II}\right)\text{ nitrate}$ solution is $2.00\text{L}$.

The volume of $3.00\text{M}$ potassium hydroxide solution is $2.00\text{L}$.

The total volume of the reaction mixture is given as,

$\text{V}={\text{V}}_1+{\text{V}}_2$

Where,

• ${\text{V}}_1$ represents the volume of $\text{copper}\left(\text{II}\right)\text{ nitrate}$ solution.
• ${\text{V}}_2$ represents the volume of potassium hydroxide solution.

Substitute the value of ${\text{V}}_1$ and ${\text{V}}_2$ in the above equation.

$\begin{array}{c}\text{V}=2.00\text{L}+2.00\text{L}\\ =4.00\text{L}\end{array}$

The molarity of a solution is given as,

$\text{M}=\frac{\text{n}}{\text{V}}$    (1)

Where,

• $\text{n}$ represents the number of moles of the solute.
• $\text{V}$ represents the volume of the solution.

Substitute the value of number of moles of ${\text{Cu}}^{2+}$ ion and total volume of the reaction mixture in the equation (1).

$\begin{array}{c}\text{M}=\left(\frac{1.00\text{mol}}{4.00\text{L}}\right)\left(\frac{1\text{M}}{1\text{mol}\cdot {\text{L}}^{-1}}\right)\\ =0.25\text{M}\end{array}$

Therefore, the concentration of ${\text{Cu}}^{2+}$ ions left in solution after completion of the reaction between given solution of $\text{copper}\left(\text{II}\right)\text{ nitrate}$ and potassium hydroxide is $0.25\text{M}$.

Substitute the value of number of moles of nitrate ion and total volume of the reaction mixture in the equation (1).

$\begin{array}{c}\text{M}=\left(\frac{8.00\text{mol}}{4.00\text{L}}\right)\left(\frac{1\text{M}}{1\text{mol}\cdot {\text{L}}^{-1}}\right)\\ =2.00\text{M}\end{array}$

Therefore, the concentration of nitrate ions left in solution after completion of the reaction between given solution of $\text{copper}\left(\text{II}\right)\text{ nitrate}$ and potassium hydroxide is $2.00\text{M}$.

Substitute the value of number of moles of ${\text{K}}^{+}$ ion and total volume of the reaction mixture in the equation (1).

$\begin{array}{c}\text{M}=\left(\frac{6.00\text{mol}}{4.00\text{L}}\right)\left(\frac{1\text{M}}{1\text{mol}\cdot {\text{L}}^{-1}}\right)\\ =1.50\text{M}\end{array}$

Therefore, the concentration of ${\text{K}}^{+}$ ions left in solution after completion of the reaction between given solution of $\text{copper}\left(\text{II}\right)\text{ nitrate}$ and potassium hydroxide is $1.50\text{M}$.

The limiting agent of the reaction is $\text{KOH}$. The amount of $\text{KOH}$ available will decide the amount of $\text{Cu}{\left(\text{OH}\right)}_{\text{2}}$ produce.

The relation between the amount of number of moles of $\text{KOH}$ reacted and number of moles of $\text{Cu}{\left(\text{OH}\right)}_{\text{2}}$ produced is given as,

${\text{n}}_{\text{Cu}{\left(\text{OH}\right)}_{\text{2}}}=\frac{{\text{n}}_{\text{KOH}}}{2}$

Where,

• ${\text{n}}_{\text{Cu}{\left(\text{OH}\right)}_{\text{2}}}$ represents the number of moles of $\text{Cu}{\left(\text{OH}\right)}_{\text{2}}$.
• ${\text{n}}_{\text{KOH}}$ represents the number of moles of $\text{KOH}$.

Substitute the value of the available number of moles of $\text{KOH}$ in the above equation.

$\begin{array}{c}{\text{n}}_{\text{Mg}{\left(\text{OH}\right)}_{\text{2}}}=\frac{6.00\text{mol}}{2}\\ =3.00\text{mol}\end{array}$

The number of $\text{Cu}{\left(\text{OH}\right)}_{\text{2}}$ produced on mixing given solutions of $\text{copper}\left(\text{II}\right)\text{ nitrate}$ and potassium hydroxide is $3.00\text{mol}$.

The molar mass of $\text{Cu}{\left(\text{OH}\right)}_{\text{2}}$ is $97.56\text{g}\cdot {\text{mol}}^{-1}$.

The relation between the number of moles and mass of a substance is given as,

$\text{m}=\text{n}{\text{M}}_{\text{m}}$

Where,

• $\text{n}$ represents the number of the substance.
• ${\text{M}}_{\text{m}}$ represents the molar mass of the substance.

Substitute the value of $\text{n}$ and ${\text{M}}_{\text{m}}$ in the above equation for the mass of $\text{Cu}{\left(\text{OH}\right)}_{\text{2}}$ produced.

$\begin{array}{c}\text{m}=\left(3.00\text{mol}\right)\left(97.56\text{g}\cdot {\text{mol}}^{-1}\right)\\ =292.68\text{g}\end{array}$

Therefore, the mass of the precipitate formed after reaction between given solution of $\text{copper}\left(\text{II}\right)\text{ nitrate}$ and potassium hydroxide is $292.68\text{g}$.