What partial pressure of oxygen above the water (in atmospheres at 0 C ) is needed to obtain this concentration? The solubility of O2 in water at 0 C and 1 atm partial pressure is . 2.21 x 10^-3 mol/L
Concentration needed is 4mg/L
Stuck and need help, anything is appreciated!
p = KcC
Kc = p/C = 1/2.21E-3 = ?
Then revisit p = KcC
You want p, you know Kc and convert 4 mg/L to C in mols/L and solve for p
Thanks, I figured this one out, can you help me on the other one?
To find the partial pressure of oxygen above the water needed to obtain a certain concentration, we can use the Henry's Law equation:
C = k * P
Where:
C is the concentration of the dissolved gas
k is the Henry's Law constant
P is the partial pressure of the gas above the liquid
In this case, we are given the concentration needed (4 mg/L) and the solubility of oxygen in water at 0°C and 1 atm partial pressure (2.21 x 10^-3 mol/L). We need to convert the given concentration from mg/L to mol/L before substituting the values into the equation.
To convert mg/L to mol/L:
1. Determine the molar mass of oxygen (O₂), which is approximately 32 g/mol.
2. Convert 4 mg to grams by multiplying by (1 g / 1000 mg).
3. Divide the mass in grams by the molar mass to get the number of moles.
Now, substitute the values into the Henry's Law equation:
C = k * P
4 x 10^-3 mol/L = (Henry's Law constant) * P
Rearrange the equation to solve for P:
P = (4 x 10^-3 mol/L) / (Henry's Law constant)
Since we are given the solubility of oxygen in water at 0°C and 1 atm partial pressure (2.21 x 10^-3 mol/L), we can substitute this value into the equation:
P = (4 x 10^-3 mol/L) / (2.21 x 10^-3 mol/L)
P ≈ 1.81 atm
Therefore, the partial pressure of oxygen above the water needed to obtain a concentration of 4 mg/L is approximately 1.81 atm at 0°C.