Question

The solubility of carbon dioxide at 20°C is 0.06 g of CO2 per kilogram of water when the pressure of CO2 is 760 mmHg. What would be the solubility if the pressure of CO2 is increased to 890 mmHg and the temperature left unchanged?

Answer 1

Hypothesis and Data Collection are the solutions.

A : WARM WATER: It will melt and fizz naturally.

HOT WATER: I believe it will fizz more quickly for

B: COLD WATER: It fizzes normally, just as I expected.

HOT WATER: I expected it to fizz faster, but it actually fizzed slower.

Extend and Analyze

Part A: Yes for cold water, but no for hot water.

Part B: No for hot water. Since the solubility of gases decreases with increasing temperature, cold water can absorb more CO2 and allow less to escape into the atmosphere. Furthermore, since cold water is denser than warm water, it sinks.

For C: I'm not sure, sorry I'm no help on this one.

For D: The oceans play a critical role in maintaining the Earth's carbon cycle in equilibrium. If the volume of CO2 in the atmosphere increases, the oceans consume a large portion of it. Carbonic acid is formed when carbon dioxide mixes with seawater in the ocean. As a result, the acidity of seawater rises. As long as we continue to add carbon dioxide to the atmosphere, the acidity of the ocean will rise. Growing acidity would make it more difficult for corals to form skeletons and shellfish to form the defensive shells they need. Corals are especially valuable because they serve as a habitat for a number of sea creatures, all of which contribute to the production of oxygen for us to breathe.

Answer 2

3g/L

Step-by-step explanation:

it is simply using the idea gas law to calculate the new molar density (i.e. molar solubility) at a different pressure,

P V =nRT

where:

P is the pressure in atm

V is the volume in L .

n is the mols of gas in mol s.

R = 0.082057 L ⋅atm/mol . K is the universal gas constant in the appropriate units.

T is the temperature in K .

Solving for molar density, we get:

n V =PRT

Since we have two solubility to consider, we must have two states.

n ₁/v₁ =p₁/RT

n ₂/v₂ =p₂/RT

Therefore, we have:

RT =P₁V ₁/n₁ = P₂V ₂/n₂

Solving for the new molar density, we get:

n₂/v₂ = n₁/v₁(p₂/p₁)

we can now convert the mass-based solubility to mols. We assume that the

CO₂ was as dissolved into the water as possible for the initial pressure, and that it doesn't change the volume of the water.

n₁v₁= {0.06g CO ₂×( 1 mol CO ₂/44.009g CO ₂ )}/0.100 L water

0.06g CO ₂×( 0.022722 )}/0.100 L water = 0.001363/0.1 = 0.0136335mol/L

Therefore, the new solubility is: 0.0136335mol/L× 5.50/1.00 atm= 0.06816 mol/L

In the original units, we have:

0.06816 mol CO ₂/L solution × 44.009 g CO ₂/mol CO ₂ = 3g/L