Chapter F, Problem G.5E

(a)

Interpretation Introduction

Interpretation:

Volume of solution that is prepared using $\text{2}\text{.111}\text{g}$ of sodium carbonate in $\text{250}\text{.0}\text{mL}$ solution that student need to transfer in order to obtain $\text{2}\text{.15}\text{mmol}{\text{Na}}^{\text{+}}$ has to be given.

Concept Introduction:

Molarity of the solution is defined as the number of moles of solute that is dissolved in the volume of solution in liters.  Unit of molarity is $\text{mol}\cdot {\text{L}}^{-1}$.  The equation for molarity is given as follows;

    $\text{Molarity (M)}=\frac{\text{Amount of solute in moles}}{\text{Volume of solution in liters}}$

Answer to Problem G.5E

Volume of sodium carbonate solution is $13.44\text{mL}$.

Explanation of Solution

Mass of sodium carbonate dissolved is given as $\text{2}\text{.111}\text{g}$.  Molar mass of sodium carbonate is $106\text{g}\cdot {\text{mol}}^{-1}$.  Formula of sodium carbonate is ${\text{Na}}_{\text{2}}{\text{CO}}_{\text{3}}$.  Number of moles of sodium carbonate in $\text{2}\text{.111}\text{g}$ is calculated as shown below;

    $\begin{array}{c}n=\frac{m}{M}\\ =\frac{2.111\text{g}}{106\text{g}\cdot {\text{mol}}^{-1}}\\ =0.020\text{mol}\end{array}$

Volume of solution is given as $\text{250}\text{mL}$.  Number of moles of sodium carbonate is calculated as $\text{0}\text{.020}\text{mol}$.  Therefore, Molarity of the sodium carbonate solution is calculated as shown below;

    $\begin{array}{c}\text{Molarity (M)}=\frac{\text{0}\text{.020}\text{mol}}{0.250\text{L}}\\ =0.08\text{M}\end{array}$

Therefore, the molarity of the sodium carbonate solution is $0.08\text{M}$.

Required solution has $\text{2}\text{.15}\text{mmol}$ of ${\text{Na}}^{\text{+}}$ ion.  One mole of sodium carbonate gives two moles of ${\text{Na}}^{\text{+}}$.  Therefore, the number of moles of sodium carbonate that has to be taken in order to obtain $\text{2}\text{.15}\text{mmol}$ of ${\text{Na}}^{\text{+}}$ ion is shown below;

    $\begin{array}{c}{n}_{({\text{Na}}_{}{\text{CO}}_{\text{3}})}=\frac{2.15\times {10}^{-3}\text{mol}}{2}\\ =1.075\times {10}^{-3}\text{mol}\end{array}$

Molarity of the sodium carbonate solution is calculated as $\text{0}\text{.08}\text{M}$.  Therefore, the volume of sodium carbonate solution that is required to obtain $\text{2}\text{.15}\text{mmol}$ of ${\text{Na}}^{\text{+}}$ is calculated as shown below;

    $\begin{array}{c}\text{Volume}\text{of}\text{solution}(V)=\frac{n_{({\text{Na}}_{\text{2}}{\text{CO}}_{\text{3}})}}{\text{Molarity}}\\ =\frac{1.075\times {10}^{-3}\text{mol}}{0.08\text{M}}\\ =13.44\text{mL}\end{array}$

Therefore, the volume of sodium carbonate solution is $13.44\text{mL}$.

(b)

Interpretation Introduction

Interpretation:

Volume of solution that is prepared using $\text{2}\text{.111}\text{g}$ of sodium carbonate in $\text{250}\text{.0}\text{mL}$ solution, that student need to transfer in order to obtain $\text{4}\text{.98}\text{mmol}{\text{CO}}_3^{2-}$ has to be given.

Concept Introduction:

Refer part (a).

Answer to Problem G.5E

Volume of sodium carbonate solution is $62.25\text{mL}$.

Explanation of Solution

Mass of sodium carbonate dissolved is given as $\text{2}\text{.111}\text{g}$.  Molar mass of sodium carbonate is $106\text{g}\cdot {\text{mol}}^{-1}$.  Formula of sodium carbonate is ${\text{Na}}_{\text{2}}{\text{CO}}_{\text{3}}$.  Number of moles of sodium carbonate in $\text{2}\text{.111}\text{g}$ is calculated as shown below;

    $\begin{array}{c}n=\frac{m}{M}\\ =\frac{2.111\text{g}}{106\text{g}\cdot {\text{mol}}^{-1}}\\ =0.020\text{mol}\end{array}$

Volume of solution is given as $\text{250}\text{mL}$.  Number of moles of sodium carbonate is calculated as $\text{0}\text{.020}\text{mol}$.  Therefore, Molarity of the sodium carbonate solution is calculated as shown below;

    $\begin{array}{c}\text{Molarity (M)}=\frac{\text{0}\text{.020}\text{mol}}{0.250\text{L}}\\ =0.08\text{M}\end{array}$

Therefore, the molarity of the sodium carbonate solution is $0.08\text{M}$.

Required solution has $\text{4}\text{.98}\text{mmol}$ of ${\text{CO}}_3^{2-}$ ion.  One mole of sodium carbonate gives one mole of ${\text{CO}}_3^{2-}$.  Therefore, the number of moles of sodium carbonate that has to be taken in order to obtain $\text{4}\text{.98}\text{mmol}$ of ${\text{CO}}_3^{2-}$ ion is shown below;

    $\begin{array}{c}{n}_{({\text{Na}}_{}{\text{CO}}_{\text{3}})}=\frac{4.98\times {10}^{-3}\text{mol}}{1}\\ =4.98\times {10}^{-3}\text{mol}\end{array}$

Molarity of the sodium carbonate solution is calculated as $\text{0}\text{.08}\text{M}$.  Therefore, the volume of sodium carbonate solution that is required to obtain $\text{4}\text{.98}\text{mmol}$ of ${\text{CO}}_3^{2-}$ is calculated as shown below;

    $\begin{array}{c}\text{Volume}\text{of}\text{solution}(V)=\frac{n_{({\text{Na}}_{\text{2}}{\text{CO}}_{\text{3}})}}{\text{Molarity}}\\ =\frac{4.98\times {10}^{-3}\text{mol}}{0.08\text{M}}\\ =62.25\text{mL}\end{array}$

Therefore, the volume of sodium carbonate solution is $62.25\text{mL}$.

(c)

Interpretation Introduction

Interpretation:

Volume of solution that is prepared using $\text{2}\text{.111}\text{g}$ of sodium carbonate in $\text{250}\text{.0}\text{mL}$ solution, that student need to transfer in order to obtain $\text{50}\text{.0}\text{mg}$ of ${\text{Na}}_{\text{2}}{\text{CO}}_{\text{3}}$ has to be given.

Concept Introduction:

Refer part (a).

Answer to Problem G.5E

Volume of sodium carbonate solution is $5.89\text{mL}$.

Explanation of Solution

Mass of sodium carbonate dissolved is given as $\text{2}\text{.111}\text{g}$.  Molar mass of sodium carbonate is $106\text{g}\cdot {\text{mol}}^{-1}$.  Formula of sodium carbonate is ${\text{Na}}_{\text{2}}{\text{CO}}_{\text{3}}$.  Number of moles of sodium carbonate in $\text{2}\text{.111}\text{g}$ is calculated as shown below;

    $\begin{array}{c}n=\frac{m}{M}\\ =\frac{2.111\text{g}}{106\text{g}\cdot {\text{mol}}^{-1}}\\ =0.020\text{mol}\end{array}$

Volume of solution is given as $\text{250}\text{mL}$.  Number of moles of sodium carbonate is calculated as $\text{0}\text{.020}\text{mol}$.  Therefore, Molarity of the sodium carbonate solution is calculated as shown below;

    $\begin{array}{c}\text{Molarity (M)}=\frac{\text{0}\text{.020}\text{mol}}{0.250\text{L}}\\ =0.08\text{M}\end{array}$

Therefore, the molarity of the sodium carbonate solution is $0.08\text{M}$.

Mass of sodium carbonate that has to be present in the solution is given as $\text{50}\text{.0}\text{mg}$.  Molar mass of sodium carbonate is $106\text{g}\cdot {\text{mol}}^{-1}$.  Number of moles of sodium carbonate in $\text{50}\text{.0}\text{mg}$ is calculated as shown below;

    $\begin{array}{c}n=\frac{m}{\text{M}}\\ =\frac{50\text{mg}}{106\text{g}\cdot {\text{mol}}^{-1}}\\ =0.471\times {10}^{-3}\text{mol}\end{array}$

The volume of solution that is needed to obtain $\text{50}\text{.0}\text{mg}$ of sodium carbonate is given as follows;

    $\begin{array}{c}\text{Volume}\text{of}\text{solution}(V)=\frac{n_{({\text{Na}}_{\text{2}}{\text{CO}}_{\text{3}})}}{\text{Molarity}}\\ =\frac{0.471\times {10}^{-3}\text{mol}}{0.08\text{M}}\\ =5.89\text{mL}\end{array}$

Volume of sodium carbonate solution is $5.89\text{mL}$.