The solubility of nitrogen in water is 6.9 x 10^-4 at exactly 1 atm pressure. What is the solubility of nitrogen under .80 atmp pressure?
I don't know where to start with this problem, what formula do I use?
Henry's Law.
p = kcC
p in atm. C in mols/L. You have no units in your post for 6.9E-4.
To calculate the solubility of nitrogen under a different pressure, you can use Henry's Law. Henry's Law states that the solubility of a gas in a liquid is directly proportional to the partial pressure of the gas above the liquid.
The formula for Henry's Law is:
S = k * P
Where:
S is the solubility of the gas in the liquid,
k is the Henry's Law constant, which depends on the specific gas and solvent, and
P is the partial pressure of the gas.
In this case, you're given the solubility of nitrogen at exactly 1 atm pressure. Let's call it S1 = 6.9 x 10^-4.
You need to find the solubility of nitrogen under 0.80 atm pressure. Let's call it S2.
Now, substitute the given values into the Henry's Law equation:
S1 = k * P1
S2 = k * P2
Divide both equations to eliminate k:
S1 / S2 = P1 / P2
Now, rearrange the equation to solve for S2:
S2 = S1 * (P2 / P1)
Plug in the given values:
S2 = (6.9 x 10^-4) * (0.80 / 1)
S2 = 5.52 x 10^-4
Therefore, the solubility of nitrogen under 0.80 atm pressure is 5.52 x 10^-4.