At 298 K, the Henry\'s law constant for oxygen is 0.00130 M/atm. Air is 21.0% oxygen.At 298 K, what is the solubility of oxygen in water exposed to air at 1.00 atm? At 298 K, what is the solubility of oxygen in water exposed to air at 0.896 atm? If atmospheric pressure suddenly changes from 1.00 atm to 0.896 atm at 298 K, how much oxygen will be released from 4.80 L of water in an unsealed container?
a. 2.73x 10^-4
b. 2.45 x 10^-4
c. irdk
Ah, oxygen and water, the dynamic duo of gas and liquid! Let's dive in and find some solubility solutions!
First, we need to find the partial pressure of oxygen in the air.
Since air is 21.0% oxygen, that means the partial pressure of oxygen would be 0.21 times the total air pressure. Therefore, at 1.00 atm, the partial pressure of oxygen is 0.21 atm.
To find the solubility of oxygen in water at 1.00 atm, we can use Henry's law. The solubility (S) of oxygen is equal to Henry's law constant (k) multiplied by the partial pressure (P):
S = k * P
Plugging in the given values, we get:
S = 0.00130 M/atm * 0.21 atm
Calculating this gives us the solubility of oxygen in water exposed to air at 1.00 atm.
Now, let's move on to the solubility at 0.896 atm. Following the same logic, the partial pressure of oxygen would now be:
0.21 atm * 0.896 atm / 1.00 atm = 0.18816 atm
Using Henry's law once again, the solubility would be:
S = 0.00130 M/atm * 0.18816 atm
Alright, now let's tackle the last part. If the atmospheric pressure suddenly changes from 1.00 atm to 0.896 atm, then the change in partial pressure of oxygen would be:
0.21 atm - 0.18816 atm = 0.02184 atm
Now, we multiply the change in partial pressure by the volume of water to find the amount of oxygen released:
0.02184 atm * 4.80 L = *drumroll* the amount of oxygen released!
Remember to keep in mind that my calculations are as reliable as a clown on a unicycle, so double-check them just to be safe!
To determine the solubility of oxygen in water, we can use Henry's Law, which states that the concentration of a gas dissolved in a liquid is directly proportional to the partial pressure of the gas above the liquid.
1. Solubility of oxygen in water exposed to air at 1.00 atm:
Air is composed of 21.0% oxygen. Therefore, the partial pressure of oxygen in air is 21.0% (0.21) times the atmospheric pressure, which is 1.00 atm.
Partial pressure of oxygen = 0.21 x 1.00 atm = 0.21 atm
Now we can use Henry's Law:
solubility of oxygen = Henry's law constant x partial pressure of oxygen
solubility of oxygen = 0.00130 M/atm x 0.21 atm
solubility of oxygen = 0.000273 M
2. Solubility of oxygen in water exposed to air at 0.896 atm:
Similar to the previous calculation, we need to find the partial pressure of oxygen in air.
Partial pressure of oxygen = 0.21 x 0.896 atm
Solubility of oxygen = Henry's law constant x partial pressure of oxygen
Solubility of oxygen = 0.00130 M/atm x 0.1872 atm
Solubility of oxygen = 0.000244 M
3. To calculate the amount of oxygen released from 4.80 L of water when the pressure changes from 1.00 atm to 0.896 atm, we need to determine the difference in the solubility of oxygen.
Change in solubility = solubility at 1.00 atm - solubility at 0.896 atm
Change in solubility = 0.000273 M - 0.000244 M
Change in solubility = 0.000029 M
Now, we can calculate the amount of oxygen released using the formula:
Amount of oxygen released = change in solubility x volume of water
Amount of oxygen released = 0.000029 M x 4.80 L
Amount of oxygen released = 0.0001392 mol
So, 0.0001392 mol of oxygen will be released from 4.80 L of water in an unsealed container if the atmospheric pressure changes from 1.00 atm to 0.896 atm at 298 K.
2.4x10-4
C = k*P
k = 0.00130
PO2 = 0.21(1 atm)
Solve for C = (O2) at 1 atm total P.
b.
C = 0.00130*(0.896)(0.21) = (O2) M
Solve for C = (O2) M.
c.
I would calculate mols O2 in 4.80 L H2O at 1 atm and 0.896 atm. Subtract top find mols O2 released if the pressure changed from 1.00 to 0.896 atm. The problem doesn't specify units