Chapter 13, Problem 13.106QP

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 (Na⁺) = 2.56 M, (K⁺] = 0.0090 M, [Mg²⁺] = 0.054 M, [Ca²⁺] = 0.010 M, [HCO₃⁻] = 0.0020 M, [Cl⁻] = 2.60 M, [SO₄^2-] = 0.051 M.

Interpretation Introduction

Interpretation:

Masses of Sodium chloride, Potassium chloride, Sodium sulphate, Calcium chloride, Magnesium chloride and Sodium bicarbonate are required to produce 1L of artificial sea water has to be calculated.

Concept introduction:

Mass is the determination of the quantity of the matter in an object. Unit of mass is gram (g) or kilogram (kg). Mass is calculated by multiplying the volume of the compound into its density.

Answer to Problem 13.106QP

Amount of given salts to produce artificial sea water are,

Explanation of Solution

Given data

Required ionic concentrations are,

$\begin{array}{l}\left[{\text{Na}}^{\text{+}}\right]\text{ = 2}\text{.56M}\\ \left[{\text{K}}^{\text{+}}\right]\text{ = 0}\text{.0090M}\\ \left[{\text{Mg}}^{\text{2+}}\right]\text{ = 0}\text{.054M}\\ \left[{\text{Ca}}^{\text{2+}}\right]\text{ = 0}\text{.010M}\\ \left[{\text{HCO}}_{\text{3}}\right]\text{ = 0}\text{.0020M}\\ \left[{\text{Cl}}^{\text{-}}\right]\text{=2}\text{.60M}\\ \left[{\text{SO}}_{\text{4}}{}^{\text{2-}}\right]\text{= 0}\text{.051M}\end{array}$

To calculate the calculation of all ions

$\begin{array}{l}{\text{MgCl}}_{\text{2}}\text{;}\\ \text{If}\left[{\text{MgCl}}_{\text{2}}\right]\text{=0}\text{.054 M}\\ \left[{\text{Mg}}^{\text{2+}}\right]\text{= 0}\text{.054 M}\\ \left[{\text{Cl}}^{\text{-}}\right]\text{= 2× 0}\text{.054M}\\ \\ {\text{Na}}_{\text{2}}{\text{SO}}_{\text{4}}\text{;}\\ \text{If}\left[{\text{Na}}_{\text{2}}{\text{SO}}_{\text{4}}\right]\text{=0}\text{.051 M}\\ \left[{\text{Na}}^{\text{2+}}\right]\text{= 0}\text{.051 M}\\ \left[{\text{Cl}}^{\text{-}}\right]\text{= 2× 0}\text{.051M}\\ \left[{\text{SO}}_{\text{4}}{}^{\text{2-}}\right]=0.051M\\ \\ {\text{CaCl}}_{\text{2}}\text{:}\\ \text{If}\left[{\text{CaCl}}_{\text{2}}\right]\text{= 0}\text{.010M}\\ \left[{\text{Ca}}^{\text{2+}}\right]\text{= 0}\text{.010}\\ \left[{\text{Cl}}^{\text{-}}\right]\text{= 2× 0}\text{.010M}\\ \\ {\text{NaHCO}}_{\text{3}}\text{;}\\ \text{If}\left[{\text{NaHCO}}_{\text{3}}\right]\text{=0}\text{.0020M}\\ \left[{\text{Na}}^{\text{+}}\right]\text{= 0}\text{.0020 M}\\ \left[{\text{HCO}}_{\text{3}}{}^{\text{-}}\right]\text{= 0}\text{.0020M}\end{array}$

To calculate subtotal of sodium and chloride ion

• The subtotal of chloride ion

$\left[{\text{Cl}}^{\text{-}}\right]\text{= (2×0}\text{.0540) + (2×0}\text{.0090) = 0}\text{.137M}$

Since the required concentration of chloride ion is 2.60M, so the difference is,

$\text{2}\text{.6 - 0}\text{.137 = 2}\text{.46 M}$ Should comes from $\text{NaCl}\text{.}$

• The subtotal of Sodium ion

$\left[{\text{Na}}^{\text{+}}\right]\text{= (2×0}\text{.051) + (2×0}\text{.0020) = 0}\text{.104M}$

Since the required concentration of Sodium ion is 2.56M, so the difference is,

$\text{2}\text{.56 - 0}\text{.104 = 2}\text{.46 M}$ Should comes from $\text{NaCl}\text{.}$

By adding the sodium and chloride ions concentration, the subtotal respective ions have calculated.

To calculate mass of the compounds required

$\begin{array}{l}\text{NaCl; Molecular mass of NaCl}=\text{58}\text{.44g/mol}\\ \text{2}\text{.46 mol × }\frac{\text{58}\text{.44g}}{\text{1mol}}\text{=144g}\end{array}$

$\begin{array}{l}{\text{MgCl}}_{\text{2}}{\text{; molecular mass of MgCl}}_{\text{2}}=\text{95}\text{.21g/mol}\\ \text{0}\text{.054 mol × }\frac{95.21\text{g}}{\text{1mol}}\text{=5}\text{.1g}\end{array}$

$\begin{array}{l}{\text{Na}}_{\text{2}}{\text{SO}}_{\text{4}}{\text{; molecular mass of Na}}_{\text{2}}{\text{SO}}_{\text{4}}\text{=142}\text{.1g/mol}\\ \text{0}\text{.051 mol × }\frac{142.1g}{\text{1mol}}\text{=7}\text{.2g}\end{array}$

$\begin{array}{l}{\text{CaCl}}_{\text{2}}{\text{; molecular mass of CaCl}}_{\text{2}}\text{=111}\text{.0g/mol}\\ \text{0}\text{.010 mol × }\frac{111.0g}{\text{1mol}}\text{=1}\text{.1g}\end{array}$

$\begin{array}{l}\text{KCl ; molecular mass of KCl}=74.5\text{5g/mol}\\ \text{0}\text{.0090 mol × }\frac{74.55g}{\text{1mol}}\text{=0}\text{.67g}\end{array}$

$\begin{array}{l}{\text{NaHCO}}_{\text{3}}{\text{; molecular mass of NaHCO}}_{\text{3}}=\text{84}\text{.01g/mol}\\ \text{0}\text{.0020 mol × }\frac{84.01g}{\text{1mol}}\text{= 0}\text{.17g}\end{array}$

By plugging in the value of mass of the given compounds and number of moles of the compound, the mass required to produce 1L of artificial sea water has calculated.

Conclusion

Masses of Sodium chloride, Potassium chloride, Sodium sulphate, Calcium chloride, Magnesium chloride and Sodium bicarbonate are required to produce 1L of artificial sea water has been calculated.