Chapter 4, Problem 4.150P

(a)

Interpretation Introduction

Interpretation:

The balanced molecular, total ionic, and net ionic equations for the reaction that occurs when the solutions are mixed are to be determined.

Concept introduction:

There are three types of equations that are utilized to represent an ionic reaction:

1. Molecular equation

2. Total ionic equation

3. Net ionic equation

The molecular equation represents the reactants and products of the ionic reaction in undissociated form. In total ionic reaction, all the dissociated ions that are present in the reaction mixture are represented and in net ionic reaction, the useful ions that participate in the reaction are represented.

Precipitation reaction involves the reaction of two soluble ionic compounds to form an insoluble product. The insoluble product is known as a precipitate.

Answer to Problem 4.150P

The balanced molecular equation of the reaction is as follows:

${\text{Na}}_{\text{2}}{\text{CO}}_{\text{3}}\left(aq\right)+{\text{CaCl}}_2\left(aq\right)\to {\text{CaCO}}_{\text{3}}\left(s\right)+2\text{NaCl}\left(aq\right)$

The total ionic equation for the given reaction is as follows:

$2{\text{Na}}^{+}\left(aq\right)+{\text{CO}}_3^{2-}\left(aq\right)+{\text{Ca}}^{2+}\left(aq\right)+2{\text{Cl}}^{-}\left(aq\right)\to {\text{CaCO}}_{\text{3}}\left(s\right)+2{\text{Na}}^{+}\left(aq\right)+2{\text{Cl}}^{-}\left(aq\right)$

The net ionic equation for the given reaction is as follows:

${\text{CO}}_3^{2-}\left(aq\right)+{\text{Ca}}^{2+}\left(aq\right)\to {\text{CaCO}}_{\text{3}}\left(s\right)$

Explanation of Solution

The first beaker consists of sodium carbonate $\left({\text{Na}}_{\text{2}}{\text{CO}}_{\text{3}}\right)$ and the second beaker consists of calcium chloride $\left({\text{CaCl}}_2\right)$. The balanced molecular equation of the reaction is as follows:

${\text{Na}}_{\text{2}}{\text{CO}}_{\text{3}}\left(aq\right)+{\text{CaCl}}_2\left(aq\right)\to {\text{CaCO}}_{\text{3}}\left(s\right)+2\text{NaCl}\left(aq\right)$

The total ionic equation for the given reaction is as follows:

$2{\text{Na}}^{+}\left(aq\right)+{\text{CO}}_3^{2-}\left(aq\right)+{\text{Ca}}^{2+}\left(aq\right)+2{\text{Cl}}^{-}\left(aq\right)\to {\text{CaCO}}_{\text{3}}\left(s\right)+2{\text{Na}}^{+}\left(aq\right)+2{\text{Cl}}^{-}\left(aq\right)$

${\text{Na}}^{+}$ and ${\text{Cl}}^{-}$ are the spectator ions that are present in the reaction mixture. Spectator ions are not present in the net ionic equation.

The net ionic equation for the given reaction is as follows:

${\text{CO}}_3^{2-}\left(aq\right)+{\text{Ca}}^{2+}\left(aq\right)\to {\text{CaCO}}_{\text{3}}\left(s\right)$

Conclusion

${\text{CaCO}}_{\text{3}}$ is the solid insoluble substance that is formed in the reaction. ${\text{Na}}^{+}$ and ${\text{Cl}}^{-}$ are the spectator ions that are present in the reaction mixture. These ions are present in the total ionic equation but are absent in the net ionic equation.

(b)

Interpretation Introduction

Interpretation:

Mass of precipitate formed when the yield is $100\%$ is to be calculated.

Concept introduction:

Precipitation reaction involves the reaction of two soluble ionic compounds to form an insoluble product. The insoluble product is known as a precipitate.

The reason for the precipitation reaction to occur is the formation of a product that is insoluble in nature. The insoluble product is formed when the electrostatic attraction between the ions is greater as compared to the attraction between ions and water molecule. The product will remain intact and precipitate out from the solution.

Answer to Problem 4.150P

Mass of precipitate formed when the yield is $100\%$ is $10\text{g}$.

Explanation of Solution

${\text{CaCO}}_{\text{3}}$ dissociates to give one ${\text{Ca}}^{2+}$ and one ${\text{CO}}_3^{2-}$ ion. There are two ${\text{Ca}}^{2+}$ ions and three ${\text{CO}}_3^{2-}$ ions in the solution so ${\text{Ca}}^{2+}$ is the limiting reagent.

The formula to calculate the mass of ${\text{CaCO}}_{\text{3}}$ is as follows:

${\text{Mass of CaCO}}_{\text{3}}=\left[\begin{array}{l}\left(\text{number}\text{}{\text{of Ca}}^{2+}\text{sphere}\right)\left(\frac{\text{mole of}1{\text{Ca}}^{\text{2+}}\text{sphere}}{\text{1}\text{sphere}}\right)\\ \left(\frac{1\text{mol}{\text{CaCO}}_{\text{3}}}{\text{1}\text{mol}{\text{Ca}}^{\text{2+}}}\right)\left({\text{molar mass of CaCO}}_{\text{3}}\right)\end{array}\right]$ (1)

Substitute $2\text{sphere}$ for a number of ${\text{Ca}}^{2+}$ sphere, $0.050\text{mol}$ for a mole of 1${\text{Ca}}^{2+}$ sphere and $100.09\text{g}$ for molar mass of ${\text{CaCO}}_{\text{3}}$ in the equation (1).

$\begin{array}{c}{\text{Mass of CaCO}}_{\text{3}}=\left[\left(2\text{sphere}\right)\left(\frac{0.050\text{mol}}{\text{1}\text{sphere}}\right)\left(\frac{1\text{mol}{\text{CaCO}}_{\text{3}}}{\text{1}\text{mol}{\text{Ca}}^{\text{2+}}}\right)\left(100.09\text{g}\right)\right]\\ =10.009\text{g}\\ \approx 10\text{g}\end{array}$

Conclusion

Mass of precipitate formed when the yield is $100\%$ is $10\text{g}$.

(c)

Interpretation Introduction

Interpretation:

The concentration of each ion in solution after reaction is to be calculated.

Concept introduction:

Precipitation reaction involves the reaction of two soluble ionic compounds to form an insoluble product. The insoluble product is known as a precipitate.

The reason for the precipitation reaction to occur is the formation of a product that is insoluble in nature. The insoluble product is formed when the electrostatic attraction between the ions is greater as compared to the attraction between ions and water molecule. The product will remain intact and precipitate out from the solution.

Answer to Problem 4.150P

The concentration of ${\text{Na}}^{+}$ is $0.60M$, ${\text{Cl}}^{-}$ is $0.40M$, ${\text{CO}}_3^{2-}$ is $0.10M$ and ${\text{Ca}}^{2+}$ is 0 after the reaction.

Explanation of Solution

The formula to calculate moles of ${\text{Na}}^{+}$ is as follows:

${\text{Moles of Na}}^{+}=\left[\left(\text{number}\text{}{\text{of Na}}^{+}\text{sphere}\right)\left(\frac{\text{mole of}1{\text{Na}}^{+}\text{sphere}}{\text{1}\text{sphere}}\right)\right]$ (2)

Substitute 6 spheres for number of ${\text{Na}}^{+}$ sphere and $0.050\text{mol}$ for mole of 1${\text{Na}}^{+}$ sphere in the equation (2).

$\begin{array}{c}{\text{Moles of Na}}^{+}=\left[\left(\text{6 sphere}\right)\left(\frac{0.050\text{mol}}{\text{1}\text{sphere}}\right)\right]\\ =0.30\text{mol}\end{array}$

The formula to calculate moles of ${\text{CO}}_3^{2-}$ is as follows:

${\text{Moles of CO}}_3^{2-}=\left[\left(\text{number}\text{}{\text{of CO}}_3^{2-}\text{sphere}\right)\left(\frac{\text{mole of}1{\text{CO}}_3^{2-}\text{sphere}}{\text{1}\text{sphere}}\right)\right]$ (3)

Substitute 3 spheres for number of ${\text{CO}}_3^{2-}$ sphere and $0.050\text{mol}$ for mole of 1${\text{CO}}_3^{2-}$ sphere in the equation (3).

$\begin{array}{c}{\text{Moles of CO}}_3^{2-}=\left[\left(\text{3 sphere}\right)\left(\frac{0.050\text{mol}}{\text{1}\text{sphere}}\right)\right]\\ =0.15\text{mol}\end{array}$

The formula to calculate moles of ${\text{Ca}}^{2+}$ is as follows:

${\text{Moles of Ca}}^{2+}=\left[\left(\text{number}\text{}{\text{of Ca}}^{2+}\text{sphere}\right)\left(\frac{\text{mole of}1{\text{Ca}}^{2+}\text{sphere}}{\text{1}\text{sphere}}\right)\right]$ (4)

Substitute 2 spheres for number of ${\text{Ca}}^{2+}$ sphere and $0.050\text{mol}$ for mole of 1${\text{Ca}}^{2+}$ sphere in the equation (4).

$\begin{array}{c}{\text{Moles of Ca}}^{2+}=\left[\left(\text{2 sphere}\right)\left(\frac{0.050\text{mol}}{\text{1}\text{sphere}}\right)\right]\\ =0.10\text{mol}\end{array}$

The formula to calculate moles of ${\text{Cl}}^{-}$ is as follows:

${\text{Moles of Cl}}^{-}=\left[\left(\text{number}\text{}{\text{of Cl}}^{-}\text{sphere}\right)\left(\frac{\text{mole of}1{\text{Cl}}^{-}\text{sphere}}{\text{1}\text{sphere}}\right)\right]$ (5)

Substitute 4 spheres for a number of ${\text{Cl}}^{-}$ sphere and $0.050\text{mol}$ for mole of 1${\text{Cl}}^{-}$ sphere in the equation (5).

$\begin{array}{c}{\text{Moles of Cl}}^{-}=\left[\left(\text{4 sphere}\right)\left(\frac{0.050\text{mol}}{\text{1}\text{sphere}}\right)\right]\\ =0.20\text{mol}\end{array}$

After the reaction, the moles of ${\text{Na}}^{+}$ and ${\text{Cl}}^{-}$ remain the same. The moles of ${\text{Ca}}^{2+}$ becomes zero as it is a limiting reagent. $0.10\text{mol}$ of ${\text{Ca}}^{2+}$ reacts with $0.10\text{mol}$ of ${\text{CO}}_3^{2-}$.

The formula to calculate moles of ${\text{CO}}_3^{2-}$ left is as follows:

$\text{Moles of}{\text{CO}}_3^{2-}\text{ left}=\text{Total moles of}{\text{CO}}_3^{2-}-{\text{moles of CO}}_3^{2-}\text{reacted}$ (6)

Substitute $0.15\text{mol}$ for total moles of ${\text{CO}}_3^{2-}$ and $0.10\text{mol}$ for mole of ${\text{CO}}_3^{2-}$ reacted in the equation (6).

$\begin{array}{c}\text{Moles of}{\text{CO}}_3^{2-}\text{ left}=0.15\text{mol}-0.10\text{mol}\\ =0.050\text{mol}\end{array}$

The total volume of solution is the sum of the volume of both the beaker that is $500\text{mL}$.

The formula to calculate the molarity of ${\text{Na}}^{+}$ is as follows:

${\text{Molarity of Na}}^{+}=\left(\frac{{\text{moles of Na}}^{+}}{\text{volume of solution}}\right)$ (7)

Substitute $0.30\text{mol}$ for moles of ${\text{Na}}^{+}$ and $500\text{mL}$ for the volume of solution in the equation (7).

$\begin{array}{c}{\text{Molarity of Na}}^{+}=\left(\frac{0.30\text{mol}}{500\text{mL}}\right)\left(\frac{1000\text{mL}}{1\text{L}}\right)\\ =0.60M\end{array}$

The formula to calculate the molarity of ${\text{Cl}}^{-}$ is as follows:

${\text{Molarity of Cl}}^{-}=\left(\frac{{\text{moles of Cl}}^{-}}{\text{volume of solution}}\right)$ (8)

Substitute $0.20\text{mol}$ for moles of ${\text{Cl}}^{-}$ and $500\text{mL}$ for volume of solution in the equation (8).

$\begin{array}{c}{\text{Molarity of Cl}}^{-}=\left(\frac{0.20\text{mol}}{500\text{mL}}\right)\left(\frac{1000\text{mL}}{1\text{L}}\right)\\ =0.40M\end{array}$

The formula to calculate molarity of ${\text{CO}}_3^{2-}$ is as follows:

${\text{Molarity of CO}}_3^{2-}=\left(\frac{{\text{moles of CO}}_3^{2-}}{\text{volume of solution}}\right)$ (9)

Substitute $0.050\text{mol}$ for moles of ${\text{CO}}_3^{2-}$ and $500\text{mL}$ for volume of solution in the equation (9).

$\begin{array}{c}{\text{Molarity of CO}}_3^{2-}=\left(\frac{0.050\text{mol}}{500\text{mL}}\right)\left(\frac{1000\text{mL}}{1\text{L}}\right)\\ =0.10M\end{array}$

Conclusion

The concentration of ${\text{Na}}^{+}$ is $0.60M$, ${\text{Cl}}^{-}$ is $0.40M$, ${\text{CO}}_3^{2-}$ is $0.10M$ and ${\text{Ca}}^{2+}$ is 0 after the reaction.