Chapter 13, Problem 142AP

What masses of sodium chloride, magnesium chloride, sodium sulfate, calcium chloride, potassium chloride, and sodium bicarbonate are needed to produce 1 L of artificial seawater for an aquarium? The required ionic concentrations are $\left[{\text{Na}}^{\text{+}}\right]\text{ = 2}\text{.56 }M\text{,}\left[{\text{K}}^{\text{+}}\right]\text{ = 0}\text{.0090 }M\text{, }\left[{\text{Mg}}^{\text{2+}}\right]\text{ = 0}\text{.054 }M\text{, }\left[{\text{Ca}}^{\text{2+}}\right]\text{ = 0}\text{.010 }M\text{, }\left[{\text{HCO}}_{\text{3}}\right]\text{=0}\text{.0020 }M\text{, }\left[{\text{Cl}}^{\text{-}}\right]\text{ = 2}\text{.60 }M\text{, }\left[{\text{SO}}_4^{2-}\right]\text{ =0}\text{.051 }M$.

Interpretation Introduction

Interpretation:

The masses of different salts required to prepare artificial sea water are to be calculated.

Concept introduction:

Molarity is calculated as thenumber of moles of solute dissolved in one litre of solution.

Mass of an element is calculated as follows:

$\text{Mass of element}=\text{Moles of element}\times \frac{\text{Molecular mass of element}}{1\text{ mol element}}$

Answer to Problem 142AP

Solution: Mass of $\text{NaCl}$ is $142.47\text{ g}$, ${\text{MgCl}}_2$ is $5.14\text{ g}$, ${\text{Na}}_2{\text{SO}}_4$ is $7.247\text{ g}$, ${\text{CaCl}}_2$ is $1.11\text{ g}$, $\text{KCl}$ is $\text{0}\text{.67 g}$, and ${\text{NaHCO}}_3$ is $0.168\text{ g}$.

Explanation of Solution

The tabulated concentration of all the ions is as follows:

${\text{MgCl}}_2\text{ if}\left[{\text{MgCl}}_2\right]=0.054\text{M }\left[{\text{Mg}}^{2+}\right]=0.054\text{M }\left[{\text{Cl}}^{-}\right]=2\times 0.054\text{M}$

${\text{Na}}_2{\text{SO}}_4\text{ if}\left[{\text{Na}}_2{\text{SO}}_4\right]=0.051\text{M }\left[{\text{Na}}^{+}\right]=2\times 0.051\text{M }\left[{\text{SO}}_4^{2-}\right]=0.051\text{M}$

${\text{CaCl}}_2\text{ if}\left[{\text{CaCl}}_2\right]=0.010\text{M }\left[{\text{Ca}}^{2+}\right]=0.010\text{M }\left[{\text{Cl}}^{-}\right]=2\times 0.010\text{M}$

${\text{NaHCO}}_3\text{ if}\left[{\text{NaHCO}}_3\right]=0.0020\text{M }\left[{\text{Na}}^{+}\right]=0.0020\text{M }\left[{\text{HCO}}_3^{-}\right]=0.0020\text{M}$

$\text{KCl if}\left[\text{KCl }\right]=0.0090\text{M }\left[{\text{K}}^{+}\right]=0.0090\text{M }\left[{\text{Cl}}^{-}\right]=0.0090\text{M}$

The concentration of chloride is from many sources. The overall concentration of chloride ion is calculated as follows:

$\begin{array}{c}\left[{\text{Cl}}^{-}\right]=\left(2\times 0.0540\text{ M}\right)+\left(2\times 0.010\text{ M}\right)+\left(0.0090\text{ M}\right)\\ =0.137\text{ M}\end{array}$

As the required amount of chloride ion is $2.60\text{ M}$, the difference between these two concentrations is as follows:

$(2.60-\text{0}\text{.137 )M}=\text{2}\text{.46 M}$

So, the difference which must arise from NaCl is $2.46\text{ M}$.

The concentration of sodium is from two sources. The overall concentration of sodium ion is calculated as follows:

$\begin{array}{c}\left[{\text{Na}}^{+}\right]=\left(2\times 0.051\text{ M}\right)+\left(1\times 0.020\text{ M}\right)\\ =0.122\text{ M}\end{array}$

As the required amount of sodium ion is $2.56\text{ M}$, the difference between these two concentrations is as follows:

$(2.56-\text{0}\text{.122)M}=2.438\text{M}$

So, the difference thatmust arise from NaCl is $2.438\text{M}$.

The required mass of compounds is calculated as follows:

Mass of $\text{NaCl}$ is calculated as follows:

$\begin{array}{c}\text{Mass of NaCl}=\text{Moles of NaCl}\times \frac{\text{Molecular mass of NaCl}}{1\text{ mol NaCl}}\\ =2.438\cancel{\text{mol}}\times \frac{58.44\text{ g }\cancel{\text{NaCl}}}{\text{ 1 }\cancel{\text{mol}}\cancel{\text{NaCl}}}\\ =142.47\text{ g}\end{array}$

Mass of ${\text{MgCl}}_2$ is calculated as follows:

$\begin{array}{c}{\text{Mass of MgCl}}_2={\text{Moles of MgCl}}_2\times \frac{{\text{Molecular mass of MgCl}}_2}{1{\text{ mol MgCl}}_2}\\ =0.054\cancel{\text{mol}}\times \frac{\text{95}\text{.21 g }\cancel{{\text{MgCl}}_2}}{\text{ 1 }\cancel{\text{mol}}\cancel{{\text{MgCl}}_2}}\\ =5.14\text{ g}\end{array}$

Mass of ${\text{Na}}_2{\text{SO}}_4$ is calculated as follows:

$\begin{array}{c}{\text{Mass of Na}}_2{\text{SO}}_4={\text{Moles of Na}}_2{\text{SO}}_4\times \frac{{\text{Molecular mass of Na}}_2{\text{SO}}_4}{1{\text{ mol Na}}_2{\text{SO}}_4}\\ =0.051\cancel{\text{mol}}\times \frac{\text{142}\text{.1 g }\cancel{{\text{Na}}_2{\text{SO}}_4}}{\text{ 1 }\cancel{\text{mol}}\cancel{{\text{Na}}_2{\text{SO}}_4}}\\ =7.247\text{ g}\end{array}$

Mass of ${\text{CaCl}}_2$ is calculated as follows:

$\begin{array}{c}{\text{Mass of CaCl}}_2={\text{Moles of CaCl}}_2\times \frac{{\text{Molecular mass of CaCl}}_2}{1{\text{ mol CaCl}}_2}\\ =0.010\cancel{\text{mol}}\times \frac{\text{111}\text{.0 g }\cancel{{\text{CaCl}}_2}}{\text{ 1 }\cancel{\text{mol}}\cancel{{\text{CaCl}}_2}}\\ =1.11\text{ g}\end{array}$

Mass of $\text{KCl}$ is calculated as follows:

$\begin{array}{c}\text{Mass of KCl}=\text{Moles of KCl}\times \frac{\text{Molecular mass of KCl}}{1\text{ mol KCl}}\\ =0.0090\cancel{\text{mol}}\times \frac{\text{74}\text{.55 g }\cancel{\text{KCl}}}{\text{ 1 }\cancel{\text{mol}}\cancel{\text{KCl}}}\\ =0.67\text{ g}\end{array}$

Mass of ${\text{NaHCO}}_3$ is calculated as follows:

$\begin{array}{c}{\text{Mass of NaHCO}}_3={\text{Moles of NaHCO}}_3\times \frac{{\text{Molecular mass of NaHCO}}_3}{1{\text{ mol NaHCO}}_3}\\ =0.0020\cancel{\text{mol}}\times \frac{\text{84}\text{.01 g }\cancel{{\text{NaHCO}}_3}}{\text{ 1 }\cancel{\text{mol}}\cancel{{\text{NaHCO}}_3}}\\ =0.168\text{ g}\end{array}$

Conclusion

Hence, masses of different salts to prepare artificial sea water are:

Mass of $\text{NaCl}$ is $142.47\text{ g}$, ${\text{MgCl}}_2$ is $5.14\text{ g}$, ${\text{Na}}_2{\text{SO}}_4$ is $7.247\text{ g}$, ${\text{CaCl}}_2$ is $1.11\text{ g}$, $\text{KCl}$ is $\text{0}\text{.67 g}$, and ${\text{NaHCO}}_3$ is $0.168\text{ g}$.