Chapter 4, Problem 4.34P

(a)

Interpretation Introduction

Interpretation:

The molarity of each ion is to be calculated.

Concept introduction:

Molarity $\left(M\right)$ is one of the concentration terms that determine the number of moles of solute present in per litre of solution. Unit of molarity is $\text{mol}/\text{L}$.

The expression to calculate the molarity of solution when moles of solute and volume of solution are given is as follows:

  $\text{Molarity}=\frac{\text{moles of solute}\left(\text{mol}\right)}{\text{volume of solution}\left(\text{L}\right)}$        (1)

Density is defined as mass per unit volume. Mass and volume are physical quantities and the units of mass and volume are fundamental units. Density is the ratio of mass to the volume. The unit of volume is derived from the units of mass and volume. The SI unit of density is ${\text{kg/m}}^{\text{3}}$. The formula to calculate density is,

  $\text{Density}=\frac{\text{Mass}}{\text{Volume}}$        (2)

Answer to Problem 4.34P

The molarity of ${\text{Na}}^{+}$, ${\text{Cl}}^{-}$, ${\text{Mg}}^{2+}$, ${\text{SO}}_4^{2-}$, ${\text{Ca}}^{2+}$, ${\text{K}}^{+}$, ${\text{HCO}}_3^{-}$ and ${\text{Br}}^{-}$ ions are $0.470M$, $0.553M$, $0.0544M$, $0.0285M$, $0.0111M$, $0.0144M$, $0.00383M$ and $0.00098M$ respectively.

Explanation of Solution

Rearrange the equation (2) to calculate the volume.

  $\text{Volume}=\frac{\text{Mass}}{\text{Density}}$        (3)

Substitute $1.025{\text{g/cm}}^3$ for density and $1\text{ kg}$ for mass in the equation (3) to calculate the volume of seawater.

  $\begin{array}{c}\text{Volume}\text{of sea water}=\left(\frac{1\text{ kg}}{1.025{\text{g/cm}}^3}\right)\left(\frac{1000\text{g}}{\text{1}\text{kg}}\right)\left(\frac{1\text{mL}}{\text{1}{\text{cm}}^3}\right)\left(\frac{1\text{L}}{1000\text{mL}}\right)\\ =0.97560976\text{L}\end{array}$

The formula to calculate the moles of $\text{NaCl}$ is as follows:

  $\text{Moles of NaCl}=\left[\frac{\text{given mass}\text{of NaCl}}{\text{molecular mass of NaCl}}\right]$        (4)

Substitute $26.5\text{g}$ for the mass of $\text{NaCl}$ and $58.44\text{ g/mol}$ for the molar mass of $\text{NaCl}$ in the equation (4).

  $\begin{array}{c}\text{Moles of NaCl}=\left[\frac{26.5\text{g}}{58.44\text{ g/mol}}\right]\\ =0.4534565\text{mol}\end{array}$

One mole of $\text{NaCl}$ dissociates to form one mole of ${\text{Na}}^{+}$ and one mole of ${\text{Cl}}^{-}$. Therefore, $0.4534565\text{mol}$ of $\text{NaCl}$ dissociates to form $0.4534565\text{mol}$ of ${\text{Na}}^{+}$ and $0.4534565\text{mol}$ of ${\text{Cl}}^{-}$.

The formula to calculate the moles of ${\text{MgCl}}_2$ is as follows:

  ${\text{Moles of MgCl}}_2=\left[\frac{\text{given mass}{\text{of MgCl}}_2}{{\text{molecular mass of MgCl}}_2}\right]$        (5)

Substitute $2.40\text{g}$ for the mass of ${\text{MgCl}}_2$ and $95.21\text{ g/mol}$ for the molar mass of ${\text{MgCl}}_2$ in the equation (5).

  $\begin{array}{c}{\text{Moles of MgCl}}_2=\left[\frac{2.40\text{g}}{95.21\text{ g/mol}}\right]\\ =0.025207\text{mol}\end{array}$

One mole of ${\text{MgCl}}_2$ dissociates to form one mole of ${\text{Mg}}^{2+}$ and two moles of ${\text{Cl}}^{-}$. Therefore, $0.025207\text{mol}$ of ${\text{MgCl}}_2$ dissociates to form $0.025207\text{mol}$ of ${\text{Mg}}^{2+}$ and $0.050415\text{mol}$ of ${\text{Cl}}^{-}$.

The formula to calculate the moles of ${\text{MgSO}}_{\text{4}}$ is as follows:

  ${\text{Moles of MgSO}}_{\text{4}}=\left[\frac{\text{given mass}{\text{of MgSO}}_{\text{4}}}{{\text{molecular mass of MgSO}}_{\text{4}}}\right]$        (6)

Substitute $3.35\text{g}$ for the mass of ${\text{MgSO}}_{\text{4}}$ and $120.37\text{ g/mol}$ for the molar mass of ${\text{MgSO}}_{\text{4}}$ in the equation (6).

  $\begin{array}{c}{\text{Moles of MgSO}}_{\text{4}}=\left[\frac{3.35\text{g}}{120.37\text{ g/mol}}\right]\\ =0.0278308\text{mol}\end{array}$

One mole of ${\text{MgSO}}_{\text{4}}$ dissociates to form one mole of ${\text{Mg}}^{2+}$ and one mole of ${\text{SO}}_4^{2-}$. Therefore, $0.0278308\text{mol}$ of ${\text{MgSO}}_{\text{4}}$ dissociates to form $0.0278308\text{mol}$ of ${\text{Mg}}^{2+}$ and $0.0278308\text{mol}$ of ${\text{SO}}_4^{2-}$.

The formula to calculate the moles of ${\text{CaCl}}_2$ is as follows:

  ${\text{Moles of CaCl}}_2=\left[\frac{\text{given mass}{\text{of CaCl}}_2}{{\text{molecular mass of CaCl}}_2}\right]$        (7)

Substitute $1.20\text{g}$ for the mass of ${\text{CaCl}}_2$ and $110.98\text{ g/mol}$ for the molar mass of ${\text{CaCl}}_2$ in the equation (7).

  $\begin{array}{c}{\text{Moles of CaCl}}_2=\left[\frac{1.20\text{g}}{110.98\text{ g/mol}}\right]\\ =0.0108128\text{mol}\end{array}$

One mole of ${\text{CaCl}}_2$ dissociates to form one mole of ${\text{Ca}}^{2+}$ and two moles of ${\text{Cl}}^{-}$. Therefore, $0.0108128\text{mol}$ of ${\text{CaCl}}_2$ dissociates to form $0.0108128\text{mol}$ of ${\text{Ca}}^{2+}$ and $0.0216255\text{mol}$ of ${\text{Cl}}^{-}$.

The formula to calculate the moles of $\text{KCl}$ is as follows:

  $\text{Moles of KCl}=\left[\frac{\text{given mass}\text{of KCl}}{\text{molecular mass of KCl}}\right]$        (8)

Substitute $1.05\text{g}$ for the mass of $\text{KCl}$ and $74.55\text{ g/mol}$ for the molar mass of $\text{KCl}$ in the equation (4).

  $\begin{array}{c}\text{Moles of KCl}=\left[\frac{1.05\text{g}}{74.55\text{ g/mol}}\right]\\ =0.0140845\text{mol}\end{array}$

One mole of $\text{KCl}$ dissociates to form one mole of ${\text{K}}^{+}$ and one mole of ${\text{Cl}}^{-}$. Therefore, $0.0140845\text{mol}$ of $\text{KCl}$ dissociates to form $0.0140845\text{mol}$ of ${\text{K}}^{+}$ and $0.0140845\text{mol}$ of ${\text{Cl}}^{-}$.

The formula to calculate the moles of ${\text{NaHCO}}_3$ is as follows:

  ${\text{Moles of NaHCO}}_3=\left[\frac{\text{given mass}{\text{of NaHCO}}_3}{{\text{molecular mass of NaHCO}}_3}\right]$        (9)

Substitute $0.315\text{g}$ for the mass of ${\text{NaHCO}}_3$ and $84.01\text{ g/mol}$ for the molar mass of ${\text{NaHCO}}_3$ in the equation (9).

  $\begin{array}{c}{\text{Moles of NaHCO}}_3=\left[\frac{0.315\text{g}}{84.01\text{ g/mol}}\right]\\ =0.00374955\text{mol}\end{array}$

One mole of ${\text{NaHCO}}_3$ dissociates to form one mole of ${\text{Na}}^{+}$ and one mole of ${\text{HCO}}_3^{-}$. Therefore, $0.00374955\text{mol}$ of ${\text{NaHCO}}_3$ dissociates to form $0.00374955\text{mol}$ of ${\text{Na}}^{+}$ and $0.00374955\text{mol}$ of ${\text{HCO}}_3^{-}$.

The formula to calculate the moles of $\text{NaBr}$ is as follows:

  $\text{Moles of NaBr}=\left[\frac{\text{given mass}\text{of NaBr}}{\text{molecular mass of NaBr}}\right]$        (10)

Substitute $0.098\text{g}$ for the mass of $\text{NaBr}$ and $102.98\text{ g/mol}$ for the molar mass of $\text{NaBr}$ in the equation (10).

  $\begin{array}{c}\text{Moles of NaBr}=\left[\frac{0.098\text{g}}{102.98\text{ g/mol}}\right]\\ =0.0009524735\text{mol}\end{array}$

One mole of $\text{NaBr}$ dissociates to form one mole of ${\text{Na}}^{+}$ and one mole of ${\text{Br}}^{-}$. Therefore, $0.0009524735\text{mol}$ of $\text{NaBr}$ dissociates to form $0.0009524735\text{mol}$ of ${\text{Na}}^{+}$ and $0.0009524735\text{mol}$ of ${\text{Br}}^{-}$.

The ${\text{Cl}}^{-}$ ions come from $\text{NaCl}$, ${\text{MgCl}}_2$, ${\text{CaCl}}_2$ and $\text{KCl}$. Therefore the total moles of ${\text{Cl}}^{-}$ ions are calculated as follows:

  $\begin{array}{c}\text{Total moles of}{\text{Cl}}^{-}\text{ions}={\left({\text{Cl}}^{-}\right)}_{\text{NaCl}}+{\left({\text{Cl}}^{-}\right)}_{{\text{MgCl}}_2}+{\left({\text{Cl}}^{-}\right)}_{{\text{CaCl}}_2}+{\left({\text{Cl}}^{-}\right)}_{\text{KCl}}\\ =\left(\begin{array}{c}0.4534565\text{mol}+0.050415\text{mol}+0.0216255\text{mol}\\ +0.0140845\text{mol}\end{array}\right)\\ =0.5395815\text{mol}\end{array}$

The ${\text{Na}}^{+}$ ions come from $\text{NaCl}$, ${\text{NaHCO}}_3$ and $\text{NaBr}$. Therefore the total moles of ${\text{Na}}^{+}$ ions are calculated as follows:

  $\begin{array}{c}\text{Total moles of}{\text{Na}}^{+}\text{ions}={\left({\text{Na}}^{+}\right)}_{\text{NaCl}}+{\left({\text{Na}}^{+}\right)}_{{\text{NaHCO}}_3}+{\left({\text{Na}}^{+}\right)}_{\text{NaBr}}\\ =\left(0.4534565\text{mol}+0.00374955\text{mol}+0.0009524735\text{mol}\right)\\ =0.458158523\text{mol}\end{array}$

The ${\text{Mg}}^{2+}$ ions come from ${\text{MgCl}}_2$ and ${\text{MgSO}}_{\text{4}}$. Therefore the total moles of ${\text{Mg}}^{2+}$ ions are calculated as follows:

  $\begin{array}{c}\text{Total moles of}{\text{Mg}}^{2+}\text{ions}={\left({\text{Mg}}^{2+}\right)}_{{\text{MgCl}}_2}+{\left({\text{Mg}}^{2+}\right)}_{{\text{MgSO}}_{\text{4}}}\\ =\left(0.025207\text{mol}+0.0278308\text{mol}\right)\\ =0.0530355\text{mol}\end{array}$

The formula to calculate the molarity of an ion is as follows:

  $\text{Molarity}=\frac{\text{moles of ions}}{\text{volume of sea water}}$        (11)

Substitute $0.5395815\text{mol}$ for the moles of ion and $0.97560976\text{L}$ for the volume of seawater in the equation (11) to calculate the molarity of ${\text{Cl}}^{-}$ ions.

  $\begin{array}{c}{\left(\text{Molarity}\right)}_{{\text{Cl}}^{-}}=\frac{0.5395815\text{mol}}{0.97560976\text{L}}\\ =0.55307M\\ \approx 0.553M\end{array}$

Substitute $0.458158523\text{mol}$ for the moles of ion and $0.97560976\text{L}$ for the volume of seawater in the equation (11) to calculate the molarity of ${\text{Na}}^{+}$ ions.

  $\begin{array}{c}{\left(\text{Molarity}\right)}_{{\text{Na}}^{+}}=\frac{0.458158523\text{mol}}{0.97560976\text{L}}\\ =0.469612M\\ \approx 0.470M\end{array}$

Substitute $0.0530355\text{mol}$ for the moles of ion and $0.97560976\text{L}$ for the volume of seawater in the equation (11) to calculate the molarity of ${\text{Mg}}^{2+}$ ions.

  $\begin{array}{c}{\left(\text{Molarity}\right)}_{{\text{Mg}}^{2+}}=\frac{0.0530355\text{mol}}{0.97560976\text{L}}\\ =0.054361M\\ \approx 0.0544M\end{array}$

Substitute $0.0278308\text{mol}$ for the moles of ion and $0.97560976\text{L}$ for the volume of seawater in the equation (11) to calculate the molarity of ${\text{SO}}_4^{2-}$ ions.

  $\begin{array}{c}{\left(\text{Molarity}\right)}_{{\text{SO}}_4^{2-}}=\frac{0.0278308\text{mol}}{0.97560976\text{L}}\\ =0.028526M\\ \approx 0.0285M\end{array}$

Substitute $0.0108128\text{mol}$ for the moles of ion and $0.97560976\text{L}$ for the volume of seawater in the equation (11) to calculate the molarity of ${\text{Ca}}^{2+}$ ions.

  $\begin{array}{c}{\left(\text{Molarity}\right)}_{{\text{Ca}}^{2+}}=\frac{0.0108128\text{mol}}{0.97560976\text{L}}\\ =0.011083M\\ \approx 0.0111M\end{array}$

Substitute $0.0140845\text{mol}$ for the moles of ion and $0.97560976\text{L}$ for the volume of seawater in the equation (11) to calculate the molarity of ${\mathrm{K}}^{+}$ ions.

  $\begin{array}{c}{\left(\text{Molarity}\right)}_{{\mathrm{K}}^{+}}=\frac{0.0140845\text{mol}}{0.97560976\text{L}}\\ =0.014437M\\ \approx 0.0144M\end{array}$

Substitute $0.00374955\text{mol}$ for the moles of ion and $0.97560976\text{L}$ for the volume of sea water in the equation (11) to calculate the molarity of ${\text{HCO}}_3^{-}$ ions.

  $\begin{array}{c}{\left(\text{Molarity}\right)}_{{\text{HCO}}_3^{-}}=\frac{0.00374955\text{mol}}{0.97560976\text{L}}\\ =0.003843M\\ \approx 0.00384M\end{array}$

Substitute $0.0009524735\text{mol}$ for the moles of ion and $0.97560976\text{L}$ for the volume of seawater in the equation (11) to calculate the molarity of ${\text{Br}}^{-}$ ions.

  $\begin{array}{c}{\left(\text{Molarity}\right)}_{{\text{Br}}^{-}}=\frac{0.0009524735\text{mol}}{0.97560976\text{L}}\\ =0.0009763M\\ \approx 0.00098M.\end{array}$

Conclusion

Strong electrolytes are the substance that dissociates completely into its ions when dissolved in the solution and conducts a large amount of electricity.

(b)

Interpretation Introduction

Interpretation:

The total molarity of alkali metal ions is to be calculated.

Concept introduction:

Molarity $\left(M\right)$ is one of the concentration terms that determine the number of moles of solute present in per litre of solution. Unit of molarity is $\text{mol}/\text{L}$.

The expression to calculate the molarity of solution when moles of solute and volume of solution are given is as follows:

  $\text{Molarity}=\frac{\text{moles of solute}\left(\text{mol}\right)}{\text{volume of solution}\left(\text{L}\right)}$        (1)

Strong electrolytes are the substance that dissociates completely into its ions when dissolved in the solution and conducts a large amount of electricity.

Answer to Problem 4.34P

The total molarity of alkali metal ions in sea water is $0.484M$.

Explanation of Solution

The alkali metal ions include from ${\text{Na}}^{+}$ and ${\text{K}}^{+}$. Therefore the molarity of alkali metal ions is calculated as follows:

  $\begin{array}{c}\text{Molarity of}\text{alkali metal ions}=\left({\text{molarity of Na}}^{+}\right)+\left(\text{molarity of}{\text{K}}^{+}\right)\\ =\left(0.469612M+0.014437M\right)\\ =0.484049M\\ \approx 0.484M.\end{array}$

Conclusion

${\text{Na}}^{+}$ and ${\text{K}}^{+}$ lie in the first group and hence they are alkali metal ions.

(c)

Interpretation Introduction

Interpretation:

The total molarity of alkaline earth metal ions is to be calculated.

Concept introduction:

Molarity $\left(M\right)$ is one of the concentration terms that determine the number of moles of solute present in per litre of solution. Unit of molarity is $\text{mol}/\text{L}$.

The expression to calculate the molarity of solution when moles of solute and volume of solution are given is as follows:

  $\text{Molarity}=\frac{\text{moles of solute}\left(\text{mol}\right)}{\text{volume of solution}\left(\text{L}\right)}$        (1)

Strong electrolytes are the substance that dissociates completely into its ions when dissolved in the solution and conducts a large amount of electricity.

Answer to Problem 4.34P

The total molarity of alkaline earth metal ions in sea water is $0.065M$.

Explanation of Solution

The alkaline earth metal ions include from ${\text{Mg}}^{2+}$ and ${\text{Ca}}^{2+}$. Therefore the molarity of alkaline earth metal ions is calculated as follows:

  $\begin{array}{c}\text{Molarity of}\text{alkaline earth metal ions}=\left({\text{molarity of Mg}}^{2+}\right)+\left(\text{molarity of}{\text{Ca}}^{2+}\right)\\ =\left(0.054361M+0.011083M\right)\\ =0.065444M\\ \approx 0.065M.\end{array}$

Conclusion

${\text{Mg}}^{2+}$ and ${\text{Ca}}^{2+}$ lie in the second group and hence they are alkaline earth metal ions.

(d)

Interpretation Introduction

The total molarity of anions is to be calculated.

Concept introduction:

Molarity $\left(M\right)$ is one of the concentration terms that determine the number of moles of solute present in per litre of solution. Unit of molarity is $\text{mol}/\text{L}$.

The expression to calculate the molarity of solution when moles of solute and volume of solution are given is as follows:

  $\text{Molarity}=\frac{\text{moles of solute}\left(\text{mol}\right)}{\text{volume of solution}\left(\text{L}\right)}$        (1)

Strong electrolytes are the substance that dissociates completely into its ions when dissolved in the solution and conducts large amount of electricity.

Answer to Problem 4.34P

The total molarity of anions in seawater is $0.586M$.

Explanation of Solution

The anions include come from ${\text{Cl}}^{-}$, ${\text{SO}}_4^{2-}$, ${\text{HCO}}_3^{-}$ and ${\text{Br}}^{-}$. Therefore the total molarity of anions is calculated as follows:

  $\begin{array}{c}\text{Molarity of}\text{anions}=\left[\begin{array}{l}\left({\text{molarity of Cl}}^{-}\right)+\left({\text{molarity of SO}}_4^{2-}\right)\\ +\left({\text{molarity of HCO}}_3^{-}\right)+\left({\text{molarity of Br}}^{-}\right)\end{array}\right]\\ =0.55307M+0.028526M+0.003843M+0.0009763M\\ =0.5864153M\\ \approx 0.586M.\end{array}$

Conclusion

${\text{Cl}}^{-}$, ${\text{SO}}_4^{2-}$, ${\text{HCO}}_3^{-}$ and ${\text{Br}}^{-}$ have negative charge on them and hence, they are anions.