Question

Bridget, a Chemistry of the Sea IA, is sitting at home on a Friday night and decides to entertain herself by making synthetic seaw ater with chemicals she has around the house (being a chemical oceanographer, she has a wide range of available household chemicals). She wants to make a solution containing the six most abundant ions in close approximation to thei r actual seawater levels. Her plan is to weigh the chemicals into a vat and then add distilled water until the final weight of the solution is 1 kg. Her goal is to have the following concentrations:

Ion]      (mmol/kg)

Na+      470

Mg2+   50

Ca2+   10

K+        10

Cl‐       545

SO4 2- 30

She has the following chemicals:

NaCl(s)      (common flavor additive)

MgSO4(s)    (used in fertilizer, explosives, foot soaks)

CaCl2(s)      (drying agent, preservative)

HCl(s)     (stolen from lab–pretend it’s a solid)

KOH(s)     (lye, for making soap)

Mg(OH)2 (s)   (laxative)

Assume all the ions dissociate completely (don’t worry about the H and OH–they will combine to make water in your solution and the excess will determine the pH).

How many grams of eachchemical should she add to get the correct ionic concentrations?

Answer 1

1) For 470 mmol/Kg Na⁺:

We need to dissolve 470 mmol of NaCl in water

Weight in g = molar mass * moles = 58.44 * 0.470 = 27.466 g of NaCl

2)  For 50 mmol/Kg Mg^2+:

There are two kinds of sources for Mg²⁺, MgSO₄ and Mg(OH)₂. Also SO₄^2- concentration is less than Mg²⁺ concentration. So, first we will calculate amount of MgSO₄ needed.

For 30 mmol/kg SO₄^2- :

Weight in g = molar mass * moles = 120.37 * 0.03 = 3.6111 g of Mg(SO)₄

SO, remaning 20 mmol/kg of Mg²⁺ ions can be obtained using Mg(OH)₂

Weight in g = molar mass * moles = 58.32 * 0.02 = 1.1664 g of Mg(SO)₄

3) For 10 mmol/Kg Ca²⁺

Weight in g = molar mass * moles = 110.98 * 0.01 = 1.1098 g

4) For 10 mmol/kg K⁺

Weight in g = molar mass * moles = 56.11 * 0.01 = 0.5611 g

5)   For 545 mmol/kg Cl^-

Since, addtion of NaCl contributes 470 mmol of Cl- ions, to reach targeted Cl- ion concentration

we shoudl add = 545-470 = 75 mmol of HCl.

Weight in g = molar mass * moles = 36.46 * 0.075 = 2.7345 g