Chapter 9, Problem 9.167QP

A 22.02-mL solution containing 1.615 g Mg(NO₃)₂ is mixed with a 28.64-mL solution containing 1.073 g NaOH. Calculate the concentrations of the ions remaining in solution after the reaction is complete. Assume volumes are additive.

Interpretation Introduction

Interpretation:

The concentrations of ions present in mixture of $\text{NaOH}$ and ${\text{Mg(NO}}_{\text{3}}{\text{)}}_{\text{2}}$ solutions should be calculated.

Concept introduction:

Precipitation reaction:

If precipitate is formed, when two solutions are mixed together is known as precipitation reaction.

The mass of the precipitate is depends on the reactant masses.

Molarity:

The concentration of the solutions is given by the term of molarity and it is given by ratio between numbers of moles of solute present in litter of solution.

$\text{Molarity}\text{=}\frac{\text{No}\text{.mole}}{\text{volume}\text{(L)}}$

Mole:

The mole of compound is given by the ratio between taken mass of the compound and molar mass of the compound.

$\text{Mole}\text{=}\frac{\text{Massof}\text{the}\text{compound}}{\text{Molar}\text{mass}\text{of}\text{the}\text{compound}}$

Mole:

The mole of the solute is calculated by multiplication of concentration of solution and volume of the solution and it is,

$\text{Mole}\text{=}\text{Concentration}\text{(M)×}\text{volume}\text{(L)}$

Answer to Problem 9.167QP

The concentrations of ions present in mixture of $\text{NaOH}$ and ${\text{Mg(NO}}_{\text{3}}{\text{)}}_{\text{2}}$ solutions are,

Concentrations of ${\text{Na}}^{\text{+}}$ = $\text{0}\text{.5295}\text{M}$

Concentrations of ${\text{OH}}^{\text{-}}$ = $\text{0}\text{.09968}\text{M}$

Concentrations of ${\text{NO}}_{\text{3}}{}^{\text{+}}$ = $\text{0}\text{.4298}\text{M}$

Concentrations of ${\text{Mg}}^{\text{2+}}$ = $0$

Record the given data.

Volume of ${\text{Mg(NO}}_{\text{3}}{\text{)}}_{\text{2}}$ solution = $\text{22}\text{.02}\text{mL}$

Volume of $\text{NaOH}$ solution = $\text{28}\text{.64 mL}$

Mass of ${\text{Mg(NO}}_{\text{3}}{\text{)}}_{\text{2}}$ in solution = $\text{1}\text{.615 g}$

Mass of $\text{NaOH}$ in solution = 1.073 g

Explanation of Solution

The given volume and masses of $\text{NaOH}$ and ${\text{Mg(NO}}_{\text{3}}{\text{)}}_{\text{2}}$ their solutions are recorded as shown above.

To calculate the mole of $\text{NaOH}$ and ${\text{Mg(NO}}_{\text{3}}{\text{)}}_{\text{2}}$ in their solutions

The reaction of ${\text{Mg(NO}}_{\text{3}}{\text{)}}_{\text{2}}$ with $\text{NaOH}$ is,

${\text{Mg(NO}}_{\text{3}}{\text{)}}_{\text{2}}{}_{\text{(aq)}}{\text{+2NaOH}}_{\text{(aq)}}\to {\text{Mg(OH)}}_{\text{2}}{}_{\text{(s)}}\downarrow {\text{+Na}}_{\text{2}}{{\text{(NO}}_{\text{3}}\text{)}}_{\text{2}}{}_{\text{(aq)}}$

Molar mass of ${\text{Mg(NO}}_{\text{3}}{\text{)}}_{\text{2}}$= $\text{148}\text{.33 g}$

$\begin{array}{l}{\text{Mg(NO}}_{\text{3}}{\text{)}}_{\text{2}}\text{mole}\text{=}\text{1}\text{.615g}\text{×}\frac{\text{1mole}{\text{Mg(NO}}_{\text{3}}{\text{)}}_{\text{2}}}{\text{148}\text{.33}\text{g}{\text{Mg(NO}}_{\text{3}}{\text{)}}_{\text{2}}}\\ \text{=}\text{0}\text{.010888}\text{mole}{\text{Mg(NO}}_{\text{3}}{\text{)}}_{\text{2}}\end{array}$

Molar mass of $\text{NaOH}$= $\text{39}\text{.998 g}$

$\begin{array}{l}\text{NaOH}\text{mole}\text{=}\text{1}\text{.073}\text{gNaOH×}\frac{\text{1moleNaOH}}{\text{39}\text{.998}\text{g}\text{NaOH}}\\ \text{=}\text{0}\text{.026826}\text{mole}\text{NaOH}\end{array}$

• The mass of substance taken is divided by molar mass of substance to give the mole of the substance present in taken mass of substance.
• The masses and molar masses of $\text{NaOH}$ and ${\text{Mg(NO}}_{\text{3}}{\text{)}}_{\text{2}}$ are plugged in above equations to give mole of $\text{NaOH}$ and ${\text{Mg(NO}}_{\text{3}}{\text{)}}_{\text{2}}$
• The mole of ${\text{Mg(NO}}_{\text{3}}{\text{)}}_{\text{2}}$ is $\text{0}\text{.010888}\text{moles}$
• The mole of $\text{NaOH}$ is $\text{0}\text{.026826}\text{moles}$

To calculate the moles of ions in present the solution mixture of $\text{NaOH}$ and ${\text{Mg(NO}}_{\text{3}}{\text{)}}_{\text{2}}$

Concentration of ${\text{Na}}^{\text{+}}$ is,

$\begin{array}{l}{\text{[Na}}^{\text{+}}\text{]}\text{=}\text{0}\text{.026826}\text{mole}\text{NaOH×}\frac{\text{1}\text{mole}{\text{Na}}^{\text{+}}}{\text{1}\text{mol}\text{NaOH}}\text{×}\frac{\text{1}}{\text{0}\text{.05066}\text{L}}\\ \text{=}\text{0}\text{.5295}\text{M}\end{array}$

Concentration of ${\text{NO}}_{\text{3}}{}^{\text{-}}$ is,

$\begin{array}{l}{\text{[NO}}_{\text{3}}{}^{\text{-}}\text{]}\text{=}\text{0}\text{.010888}\text{mole}{\text{Mg(NO}}_{\text{3}}{\text{)}}_{\text{2}}\text{×}\frac{\text{2}\text{mole}{\text{NO}}_{\text{3}}{}^{\text{-}}}{\text{1}{\text{moleMg(NO}}_{\text{3}}{\text{)}}_{\text{2}}}\text{×}\frac{\text{1}}{\text{0}\text{.05066}\text{L}}\\ \text{=}\text{0}\text{.4298}\text{M}\end{array}$

Concentration of ${\text{OH}}^{\text{-}}$ reacted with ${\text{Mg}}^{\text{2+}}$ is,

$\begin{array}{l}{\text{[OH}}^{\text{-}}\text{]}\text{=}\text{0}\text{.010888}\text{mloe}{\text{Mg}}^{\text{2+}}\text{×}\frac{\text{2}\text{mole}{\text{OH}}^{\text{-}}}{\text{1}\text{mole}{\text{Mg}}^{\text{2+}}}\\ \text{=}\text{0}\text{.021776}\text{mole}{\text{OH}}^{\text{-}}\end{array}$

Concentration of excess ${\text{OH}}^{\text{-}}$ is,

$\begin{array}{l}{\text{[OH}}^{\text{-}}\text{]}\text{excess=}\text{0}\text{.26826}\text{mole}\text{-}\text{0}\text{.021776}\text{mole}\\ \text{=}\text{0}\text{.005050}\text{mole}\\ \text{=}\frac{\text{0}\text{.005050}\text{mole}}{\text{0}\text{.05066}\text{L}}\\ \text{=}\text{0}\text{.09968}\text{M}\end{array}$

Concentration of excess ${\text{Mg}}^{\text{2+}}$ is,

${\text{[Mg}}^{\text{2+}}\text{]}\text{=}\text{0}\text{M}$

• From the balanced equation the one mole of ${\text{Mg(NO}}_{\text{3}}{\text{)}}_{\text{2}}$ is required two mole of $\text{NaOH}$ to form a ${\text{Mg(OH)}}_{\text{2}}$ precipitate.  Therefore ${\text{Mg(NO}}_{\text{3}}{\text{)}}_{\text{2}}$ is a limiting reagent.
• The Concentration of excess ${\text{OH}}^{\text{-}}$ is calculated by subtract the calculated concentration of ${\text{OH}}^{\text{-}}$ reacted with ${\text{Mg}}^{\text{2+}}$ from total concentration of ${\text{OH}}^{\text{-}}$.
• Moles of compounds and required number of moles of ions are plugged in the equation to give concentrations of ions present in the given solution mixture.
• ${\text{Mg}}^{\text{2+}}$ ions are completely react give a precipitate.  So, the concentration of ${\text{Mg}}^{\text{2+}}$ present in the mixture of solution is zero.

Conclusion

The concentrations of ions present in mixture of $\text{NaOH}$ and ${\text{Mg(NO}}_{\text{3}}{\text{)}}_{\text{2}}$ solutions were calculated.