Chapter 4, Problem 127AP

Interpretation Introduction

Interpretation:

The concentration of ions remaining in the solution after completion of the given reaction is to be calculated, assume volumes are additive.

Concept introduction:

The number of moles of a given compound that take part in the given reaction depends upon the concentration of the limiting reagent.

The reagent that is present in lesser amount in the given reaction is known as limiting reagent.

The conversion of volume from $\text{mL}$ to liter is done as,

$1\text{mL}={10}^{-3}\text{L}$

The number of moles of compound in the given solution is calculated as:

$\text{Moles of compound }=\frac{\text{mass of compund }}{\text{molar mass of compound}}$

Answer to Problem 127AP

Solution:

The concentration of ${\text{Na}}^{+},{\text{NO}}_3^{-},{\text{OH}}^{-}$, and magnesium ions is $0.5295\text{M},0.4298\text{M},0.09968\text{M}$, and $0.0\text{M}$, respectively.

Explanation of Solution

Given information: The reaction of $\text{Mg}{\left({\text{NO}}_3\right)}_2$ solution with $\text{NaOH}$ solution.

The mass of $\text{Mg}{\left({\text{NO}}_3\right)}_2$ in the given reaction is $1.615\text{g}$.

The volume of $\text{Mg}{\left({\text{NO}}_3\right)}_2$ is $22.02\text{mL}$.

The mass of $\text{NaOH}$ in the given reaction is $1.073\text{g}$.

The volume of $\text{NaOH}$ solution is $28.64\text{mL}$.

The balanced chemical equation for the given reaction is,

$\text{Mg}{\left({\text{NO}}_3\right)}_2\left(aq\right)+2\text{NaOH}\left(aq\right)\to \text{Mg}{\left(\text{OH}\right)}_2\left(s\right)+2{\text{NaNO}}_3\left(aq\right)$

The conversion of volume from $\text{mL}$ to liter is done as,

$1\text{mL}={10}^{-3}\text{L}$

Hence, the conversion of the volume of $22.02\text{mL}$ to liters is,

$22.02\text{mL}=22.02\times {10}^{-3}\text{L}$

Similarly, the conversion of the volume of $28.64\text{mL}$ to liters is,

$28.64\text{mL}=28.64\times {10}^{-3}\text{L}$

The total volume of the final solution is calculated by adding the volume of $\text{Mg}{\left({\text{NO}}_3\right)}_2$ solution and the volume of $\text{NaOH}$ solution.

$\begin{array}{c}\text{Volume}\text{of}\text{the final}\text{solution}=\left(22.02\times {10}^{-3}\text{L}+28.64\times {10}^{-3}\text{L}\right)\\ =0.05066\text{L}\end{array}$

The number of moles of a compound in the given reaction is calculated by the formula,

$\text{Moles of compound }=\frac{\text{mass of compund }}{\text{molar mass of compound}}$ …… (1)

Substitute the values of mass and molar mass of $\text{Mg}{\left({\text{NO}}_3\right)}_2$ in the equation (1),

$\begin{array}{c}\text{Moles}\text{of}\text{Mg}{\left({\text{NO}}_3\right)}_2=\left(\frac{1.615\text{g}}{148.33\text{g/mol}}\right)\\ =0.010888\text{mol}\end{array}$

Substitute the values of mass and molar mass of $\text{NaOH}$ in the equation (1),

$\begin{array}{c}\text{Moles}\text{of}\text{NaOH}=\left(\frac{1.073\text{g}}{39.998\text{g/mol}}\right)\\ =0.026826\text{mol}\end{array}$

The number of moles of hydroxide ions that take part in the given reaction is calculated by the formula,

$\text{Moles}\text{of}{\text{OH}}^{\text{-}}\text{ions}\text{reacted}=2\times \text{Moles}\text{of}\text{Mg}{\left({\text{NO}}_3\right)}_2$

Substitute the value of number of moles of $\text{Mg}{\left({\text{NO}}_3\right)}_2$ in the above equation,

$\begin{array}{c}\text{Moles}\text{of}{\text{OH}}^{\text{-}}\text{ions}\text{reacted}=\left(2\times 0.010888\text{mol}\right)\\ =0.021776\text{mol}\end{array}$

Hence, the number of moles of hydroxide ions are present in excess, and $\text{Mg}{\left({\text{NO}}_3\right)}_2$ is a limiting reagent, which fully consumed in the given reaction. As all magnesium ions present in the solution get precipitated after completion of the reaction, therefore, the molarity of ${\text{Mg}}^{2+}$ ions in the solution after the completion of reaction is $0.0\text{M}$.

The number of moles of hydroxide ions remaining in the solution after completion of reaction is,

$\begin{array}{c}\text{Moles}\text{of}{\text{OH}}^{\text{-}}\text{remaining}=\left(0.026826\text{mol}-0.021776\text{mol}\right)\\ =0.00505\text{mol}\end{array}$

The molarity of the ions present in the given solution is calculated by the formula,

$\text{Molarity of solute (M)}=\frac{\text{moles of solute (n)}}{\text{volume of solution (V)}}$ …… (2)

Here, $\text{M}$ is the molarity of final solution, $\text{n}$ is the number of moles of magnesium nitrate in the final solution and $\text{V}$ is the volume of final solution in liters.

Substitute the values of $\text{n}$ and $\text{V}$ for ${\text{OH}}^{+}$ ions in the equation (2),

$\begin{array}{c}\text{M}=\left(\frac{0.00505\text{mol}}{0.05066\text{L}}\right)\\ =0.09968\text{M}\end{array}$

Substitute the values of $\text{n}$ and $\text{V}$ for ${\text{Na}}^{+}$ ions in the equation (2),

$\begin{array}{c}\text{M}=\left(\frac{0.026826\text{mol}}{0.05066\text{L}}\right)\\ =0.5295\text{M}\end{array}$

Substitute the values of $\text{n}$ and $\text{V}$ for ${\text{NO}}_3^{-}$ ions in the equation (2),

$\begin{array}{c}\text{M}=\left(\frac{2\times 0.010888\text{mol}}{0.05066\text{L}}\right)\\ =0.4298\text{M}\end{array}$

Conclusion

The concentration of ${\text{Na}}^{+},{\text{NO}}_3^{-},{\text{OH}}^{-}$, and magnesium ions in the solution after the completion of reaction is $0.5295\text{M},0.4298\text{M},0.09968\text{M}$, and $0.0\text{M}$, respectively.