Chapter 2, Problem 2.128P

(a)

Interpretation Introduction

Interpretation:

The mass percentage each ion in the seawater is to be calculated.

Concept introduction:

Mass percent is defined as the mass of a component divided by the total mass of the mixture multiplied by 100.

The expression to calculate the percentage mass of ions is as follows:

  $\text{Mass }\%\text{ of ion}=\frac{\text{Mass of ion}}{\text{Mass of seawater}}$        (1)

Answer to Problem 2.128P

The mass percentage of ${\text{Cl}}^{-}$, ${\text{Na}}^{+}$, ${\text{SO}}_4^{2-}$, ${\text{Mg}}^{\text{2+}}$, ${\text{K}}^{\text{+}}$, ${\text{Ca}}^{\text{2+}}$, and ${\text{HCO}}_3^{-}$ ion in the seawater is $1.898\%$, $1.056\%$, $0.265\%$, $0.127\%$, $0.038\%$, $0.04\%$, and $0.014\%$ respectively.

Explanation of Solution

Convert the mass of seawater from $\text{kg}$ to $\text{mg}$ as follows:

  $\begin{array}{c}\text{Mass of water}=\left(1\text{ kg}\right)\left(\frac{1000\text{ g}}{1\text{ kg}}\right)\left(\frac{1000\text{ mg}}{1\text{ g}}\right)\\ =1\times {10}^6\text{ mg}\end{array}$

Substitute $18980\text{ mg}$ for the mass of ion and $1\times {10}^6\text{ mg}$ for the mass of seawater in equation (1) to calculate the mass $\%{\text{ Cl}}^{-}$ ion.

  $\begin{array}{c}\text{Mass }\%{\text{ of Cl}}^{-}\text{ ion}=\left(\frac{18980\text{ mg}}{1\times {10}^6\text{ mg}}\right)\left(100\text{ \%}\right)\\ =1.898\%\end{array}$

Substitute $10560\text{ mg}$ for the mass of ion and $1\times {10}^6\text{ mg}$ for seawater in equation (1) to calculate the mass $\%{\text{ Na}}^{+}$ ion.

  $\begin{array}{c}\text{Mass }\%{\text{ of Na}}^{+}\text{ ion}=\left(\frac{10560\text{ mg}}{1\times {10}^6\text{ mg}}\right)\left(100\text{ \%}\right)\\ =1.056\%\end{array}$

Substitute $\text{2650 mg}$ for the mass of ion and $1\times {10}^6\text{ mg}$ for seawater in equation (1) to calculate the mass $\%{\text{ SO}}_4^{2-}$ ion.

  $\begin{array}{c}\text{Mass }\%{\text{ of SO}}_4^{2-}\text{ ion}=\left(\frac{2650\text{ mg}}{1\times {10}^6\text{ mg}}\right)\left(100\text{ \%}\right)\\ =0.256\%\end{array}$

Substitute $\text{1270 mg}$ for the mass of ion and $1\times {10}^6\text{ mg}$ for seawater in equation (1) to calculate the mass $\%{\text{ Mg}}^{\text{2+}}$ ion.

  $\begin{array}{c}\text{Mass }\%{\text{ of Mg}}^{\text{2+}}\text{ ion}=\left(\frac{\text{1270 mg}}{1\times {10}^6\text{ mg}}\right)\left(100\text{ \%}\right)\\ =0.127\%\end{array}$

Substitute $\text{400 mg}$ for the mass of ion and $1\times {10}^6\text{ mg}$ for seawater in the equation (1) to calculate the mass $\%{\text{ Ca}}^{\text{2+}}$.

  $\begin{array}{c}\text{Mass }\%{\text{ of Ca}}^{\text{2+}}=\left(\frac{\text{400 mg}}{1\times {10}^6\text{ mg}}\right)\left(100\text{ \%}\right)\\ =0.04\%\end{array}$

Substitute $\text{380 mg}$ for the mass of ion and $1\times {10}^6\text{ mg}$ for seawater in equation (1) to calculate the mass $\%{\text{ K}}^{\text{+}}$ ion.

  $\begin{array}{c}\text{Mass }\%{\text{ of K}}^{\text{+}}\text{ ion}=\left(\frac{\text{380 mg}}{1\times {10}^6\text{ mg}}\right)\left(100\text{ \%}\right)\\ =0.038\%\end{array}$

Substitute $\text{140 mg}$ for the mass of ion and $1\times {10}^6\text{ mg}$ for seawater in equation (1) to calculate the mass $\%{\text{ HCO}}_3^{-}$ ion.

  $\begin{array}{c}\text{Mass }\%{\text{ of HCO}}_3^{-}\text{ ion}=\left(\frac{\text{140 mg}}{1\times {10}^6\text{ mg}}\right)\left(100\text{ \%}\right)\\ =0.014\%.\end{array}$

Conclusion

The mass percentage is the unit of concentration and is expressed as the mass of ion to the total mass of seawater.

(b)

Interpretation Introduction

Interpretation:

Percentage of the total mass of ions represented by sodium ions is to be calculated.

Concept introduction:

Mass percent is defined as the mass of a component divided by the total mass of the mixture multiplied by 100.

The expression to calculate the percentage mass of ions is as follows:

  $\text{Mass }\%\text{ of ion}=\frac{\text{Mass of ion}}{\text{Mass of seawater}}$        (1)

Answer to Problem 2.128P

Percentage of the total mass of ions represented by sodium ions is $30.72\text{ \%}$.

Explanation of Solution

The total mass of ions in $1\text{ kg}$ of seawater is equal to the sum of all mass of ions. The total mass of all the ions is $34380\text{ mg}$.

The expression to calculate the percentage of ${\text{Na}}^{+}$ present in the total mass of an ion is as follows:

  $\text{Mass }\%{\text{ of Na}}^{\text{+}}=\left(\frac{{\text{Mass of Na}}^{\text{+}}}{\text{Total mass}}\right)\left(100\text{ \%}\right)$        (2)

Substitute $10560\text{ mg}$ for the mass of ${\text{Na}}^{+}$ and $34380\text{ mg}$ for total mass in equation (2).

  $\begin{array}{c}\text{Mass }\%{\text{ of Na}}^{\text{+}}=\left(\frac{10560\text{ mg}}{34380\text{ mg}}\right)\left(100\text{ \%}\right)\\ =30.72\text{ \%}\text{.}\end{array}$

Conclusion

The mass percentage of sodium ions present in the total mass is $30.72\text{ \%}$.

(c)

Interpretation Introduction

Interpretation:

The total mass $\%$ of alkali ions and total mass $\%$ of alkaline earth metal ions is to be compared.

Concept introduction:

Mass percent is defined as the mass of a component divided by the total mass of the mixture multiplied by 100.

The expression to calculate the percentage mass of ions is as follows:

  $\text{Mass }\%\text{ of ion}=\frac{\text{Mass of ion}}{\text{Mass of seawater}}$        (1)

The alkali metal is defined as the group of elements that are found in Group I. The alkaline earth metal is defined as the group of elements that are found in Group II.

Answer to Problem 2.128P

The total mass percent for alkali metal ions is $6.6$ times greater than mass percent for alkaline earth metal ions.

Explanation of Solution

${\text{Mg}}^{2+}$ and ${\text{Ca}}^{2+}$ ions are alkaline earth metal ions. The percentage of the total mass of alkaline earth metal ions is equal to the sum of the mass percentage of ${\text{Mg}}^{2+}$ and ${\text{Ca}}^{2+}$.

  $\begin{array}{c}\text{Mass \% of alkaline earth metal ions}={\text{Mg}}^{2+}+{\text{Ca}}^{2+}\\ =0.127\text{ \%}+0.04\text{ \%}\\ =0.167\text{ \%}\end{array}$

${\text{Na}}^{+}$ and ${\text{K}}^{+}$ ions are found in the alkali metal ions. The percentage of the total mass of alkali metal ions is equal to the sum of the mass percentage of ${\text{Na}}^{+}$ and ${\text{K}}^{+}$.

  $\begin{array}{c}\text{Mass \% of alkali metal ions}={\text{Na}}^{+}+{\text{K}}^{+}\\ =1.056\text{ \%}+0.038\text{ \%}\\ =1.094\text{ \%}\end{array}$

The expression to calculate the ratio of the mass $\%$ of alkali ions to the mass $\%$ of alkaline earth metal ions is as follows:

  $\left(\begin{array}{l}\text{Ratio of mass \% of alkali }\\ \text{ions to mass \% of }\\ \text{alkaline earth metal ions}\end{array}\right)=\frac{\text{Mass \% of alkali ions}}{\text{Mass \% of alkaline earth metal ions}}$        (3)

Substitute $1.094\text{ \%}$ for the mass $\%$ of alkali ions and $0.167\text{ \%}$ for mass $\%$ alkaline earth metal ions in the equation (3).

  $\begin{array}{c}\left(\begin{array}{l}\text{Ratio of mass \% of alkali ions to }\\ \text{mass \% of alkaline earth metal ions}\end{array}\right)=\frac{1.094\text{ \%}}{0.167\text{ \%}}\\ =\mathrm{6.6.}\end{array}$

Therefore, the total mass percent for alkali metal ions is $6.6$ times greater than mass percent for alkaline earth metal ions.

Conclusion

The total mass percent for alkali metal ions is $6.6$ times greater than a mass percent for alkaline earth metal ions. This suggests that alkali metal ions are dominant in sea water.

(d)

Interpretation Introduction

Interpretation:

The cation or anion that has larger mass fraction of dissolved component is to be identified.

Concept introduction:

Mass percent is defined as the mass of a component divided by the total mass of the mixture multiplied by 100.

The expression to calculate the percentage mass of ions is as follows:

  $\text{Mass }\%\text{ of ion}=\frac{\text{Mass of ion}}{\text{Mass of seawater}}$        (1)

Ions that are formed by the loss of electrons form a positively charged ion. The positively charged ions are called cation.

Ions that are formed by the gain of electrons form a negatively charged ions. The negatively charged ions are called anion.

Answer to Problem 2.128P

The mass fraction of anion is larger than the mass fraction of cations.

Explanation of Solution

The anions that are found in the seawater are ${\text{Cl}}^{-}$, ${\text{SO}}_4^{2-}$ and ${\text{HCO}}_3^{-}$.

The total mass percent of anions in seawater is calculated as follows:

  $\text{Total mass percent}=\left(\begin{array}{l}{\text{Mass percent of Cl}}^{-}+\\ {\text{Mass percent of SO}}_4^{2-}+\\ {\text{Mass percent of HCO}}_3^{-}\end{array}\right)$        (4)

Substitute $1.898\text{ \%}$ for ${\text{Cl}}^{-}$, $0.265\text{ \%}$ for ${\text{SO}}_4^{2-}$ and $0.014\text{ \%}$ for ${\text{HCO}}_3^{-}$ in equation (4).

  $\begin{array}{c}\text{Total mass percent}=\left(1.898\text{ \%}+0.265\text{ \%}+0.014\text{ \%}\right)\\ =2.177\text{ \%}\end{array}$

The cations that are found in the sea water such as ${\text{Na}}^{+}$, ${\text{Mg}}^{2+}$, ${\text{Ca}}^{2+}$ and ${\text{K}}^{+}$.

The total mass percent of anions in seawater is calculated as follows:

  $\text{Total mass percent}=\left(\begin{array}{l}{\text{Mass percent of Na}}^{+}+\\ {\text{Mass percent of Mg}}^{2+}+\\ {\text{Mass percent of Ca}}^{2+}+\\ {\text{Mass percent of K}}^{+}\end{array}\right)$        (5)

Substitute $1.056\text{ \%}$ for ${\text{Na}}^{+}$, $0.127\text{ \%}$ for ${\text{Mg}}^{2+}$, $0.038\text{ \%}$ for ${\text{K}}^{+}$ and $0.04\text{ \%}$ for ${\text{Ca}}^{2+}$ in equation (5).

  $\begin{array}{c}\text{Total cations mass percent}=\left(1.056\text{ \%}+0.127\text{ \%}+0.04\text{ \% + 0}\text{.038\%}\right)\\ =1.26\text{ \%}\text{.}\end{array}$

The mass fraction of anion is larger than the mass fraction of cations.

Conclusion

Anions form a larger mass fraction of the dissolved components in the seawater.