Question

CO₂ can be injected into an artificial aquarium, and lower the alkalinity of the environment, which is related to global warming.

CO₂  can be made using the following chemical reaction: CaCO₃ (s) + 2HCl (aq) <--> CaCl₂ (aq) + H₂O (l) + CO₂ (g).

The CO₂ will then be injected into an artificial aquarium, reacting with H₂O and CO₃^2- to create hydrogen ions and bicarbonate:

CO₂ (g) + H₂O (l) + CO₃^2- <---> H⁺ + HCO₃^-

How does one know the amounts of each chemical needed to raise the level CO₂ in 3 of 1.0 liter jars in the following quantities:

a) 450 ppm

b) 750 ppm

c) 1250 ppm.

Please can you explain the method, using pH = -log [H⁺] and P₁V₁ = P ₂V₂

The temperature of the room will be 298K and the pressure will be 1 atm [i think, since I'm only injecting the CO₂ start into the tank]. Also, how does Boyle and Henry's Law relate, I haven't the faintest clue where to start.

Thanks!

Answer 1

Equation 1: CaCO₃ (s) + 2HCl (aq) <--> CaCl₂ (aq) + H₂O (l) + CO₂ (g).

Equation 2: CO₂ (g) + H₂O (l) + CO₃^2- <---> H⁺ + HCO₃

The question is to get dissolved CO₂​ in 3 tanks in the order given below.

a) 450 ppm

b) 750 ppm

c) 1250 ppm.

However it's not clear from the question whether said CO₂​ quantities is injected into aquariums or that much excess quantities of CO₂​ is injected into the jars. Because calculations will differ for each of the cases. It's assumed initially that as it's the first case i.e the quantity of CO₂ inj​ected into the jars.

Accordingly, first we need to convert ppm into milligram quantities. Actually ppm equals milligrams/L i.e we have actually in 3 jars the following quantities of CO₂ ​.

Jar 1..........450 mg/L

Jar 2..........750 mg/L

Jar 3..........1250 mg/L

From equation 1, we know 1 mole quantity of CaCO₃​ upon reaction with 2 mole quantity of HCl gives 1 mole quantity of CO₂​. So now we need to convert these mole quantities into gram quantities and then into milligram quantities.

Accordingly 1 mole quantity of CaCO₃​ (100.09 g) upon reaction with 2 mole quantity of HCl (2x36.5 = 73 g)  gives 1 mole quantity of CO₂​ (44 g). From this we can easily calculate how many milligrams of CaCO₃​ and HCl is required to get CO₂ in given quantities in the jars. Accordingly

In the first jar

if 100.09 g CaCO₃ gives 44g of CO₂​, how many grams of CaCO_​3 is required to give 0.45 g (450 mg/1000 = 0.45 g) of CO₂ .

i.e 44 g CO_​2 from.......100.09 g CaCO₃

   0.45 g CO_​2 from how many grams of CaCl_​2​? Cross multiplying these values one will get

= 0.45 x 100.09 / 44

= 1.023 g CaCO₃ gives 0.45 g (450 mg = 450 ppm) CO₂​ in the first jar.

Similarly for 2nd Jar 750 ppm (0.75 g) CO₂​ injected. Then

= 0.75 x 100.09 / 44

= 1.706 g CaCO₃ gives 0.75 g (750 mg = 750 ppm) CO₂​ in the second jar.

Similarly for third Jar 1250 ppm (1.25 g) CO₂​ injected. Then

= 1.25 x 100.09 / 44

=  2.843 g CaCO₃ gives 1.25 g (1250 mg = 1250 ppm) CO₂​ in the third jar.

If it's the case of excess quntity of CO₂​ remaining as shown in the equation 2 after it's reacting with 1 mole of H₂O and 1 mole of CO₃^2- ion then the calculations will be done as follows:

From equation 1 we know 1 mole of CaCO₃​ generates 1 mole of CO₂​. From equation 2 we know 1 mole of CO₂ reacts with another mole of CO₃^2- ion to give 1 mole of H⁺ and 1 mole of HCO₃^​- .

From both equations it's understood that 2 mole of CaCO₃​, 2 moles of HCl is required to generate 1 mole of CO₂​.

However in first jar we have excess CO₂​ to the extent of 0.45 g ( i.e 0.010 M). To generate that much excess CO₂ we need to add that much excess of CaCO₃​. Accordingly we need

for first jar quantity of CaCO₃​ is 2 mol plus excess 0.01 M (0.45 g CO₂​ is 0.01 M)............total 2.01 moles.......1 mole CaCO₃​ = 100.09 g.....2.01 mole CaCO₃​ = 201.181 g

for second jar quantity of CaCO₃​ is 2 mol plus excess 0.017 M (0.75 g CO₂​ is 0.017 M)......total 2.017 moles.......1 mole CaCO₃​ = 100.09 g.....2.017 mole CaCO₃​ = 201.881 g

for third jar quantity of CaCO₃​ is 2 mol plus excess 0.028 M (1.25 g CO₂​ is 0.028 M)......total 2.028 moles.......1 mole CaCO₃​ = 100.09 g.....2.028 mole CaCO₃​ = 202.982 g

From equation 1 similarly HCl quantities can also calculated as follows:

For first jar HCl quantity required.......2.010 moles..........73.365 g

For second jar HCl quantity required.......2.017 moles..........73.620 g

For third jar HCl quantity required.......2.028 moles..........74.022 g   

Since we are not asked to calculate pH of solutions we don't need to use equation: pH = -log [H⁺] and you are also injecting CO​2 in a open tank I don't think Boyle's law  (P₁V₁ = P ₂V₂​) applies in this case. Further according to Henry's law, the amount of a gas that dissolves in a liquid is directly proportional to the partial pressure of that gas. To calculate amounts of excee amounts of CO₂​ in the jars I don't think Henry's law is useful either.