The solubility of NH3 in H2O is 518 g/L at a partial pressure of 760.0 torr.
a. Calculate the value of the Henry's Law constant for NH3
b. Calculate the solubility of NH3 in H2O at a partial pressure of 225.0 torr.
Henry's Law constant is quote numerous ways. I will use the equation of Pgas = k*c so k = P/c = value in units of atm*L/moles.
Convert 518 g/L to moles NH3 and substitute into Henry's Law equation I show above along with P of 1 atm and evaluate k. Then redo the equation using 225 torr (convert to atm first) and solve for molar concn NH3 at that reduced pressure.
5.2
To calculate the value of the Henry's Law constant for NH3, we need to use the given information about the solubility of NH3 in H2O at a partial pressure of 760.0 torr.
a. The Henry's Law constant (k) relates the concentration of a gas in a liquid to the partial pressure of the gas above the liquid. It is calculated using the equation:
C = k * P
Where C is the concentration of the gas in the liquid (in mol/L), k is the Henry's Law constant (in L/mol), and P is the partial pressure of the gas (in atm).
First, we need to convert the given solubility from g/L to mol/L. The molar mass of NH3 is approximately 17.03 g/mol.
Given:
Solubility of NH3 = 518 g/L
Partial pressure = 760.0 torr = 760.0/760 = 1 atm
To calculate the solubility in mol/L:
Solubility (mol/L) = (518 g/L) / (17.03 g/mol) ≈ 30.45 mol/L
Now, we can plug in the values into the Henry's Law equation:
C = k * P
30.45 mol/L = k * 1 atm
k = 30.45 mol/L
Therefore, the value of the Henry's Law constant for NH3 is 30.45 L/mol.
b. To calculate the solubility of NH3 in H2O at a partial pressure of 225.0 torr, we can use the Henry's Law equation and the Henry's Law constant calculated in part a.
Given:
Partial pressure = 225.0 torr = 225.0/760 = 0.296 atm
Using the Henry's Law equation:
C = k * P
C = 30.45 L/mol * 0.296 atm
C ≈ 9.02 mol/L
Therefore, the solubility of NH3 in H2O at a partial pressure of 225.0 torr is approximately 9.02 mol/L.