Calculate the solubility (in M units) of ammonia gas in water at 298 K
and a partial pressure of 5.00 bar. The Henry’s law constant for ammonia gas at 298 K is 58.0 M/atm and 1 bar=0.9869 atm
Epress your answer in molarity to three significant figures.
To calculate the solubility of ammonia gas in water at 298 K, we can use Henry's law, which states that the solubility of a gas is directly proportional to its partial pressure.
However, we need to convert the given partial pressure from bar to atm before using the equation.
1 bar = 0.9869 atm
Partial pressure of ammonia gas = 5.00 bar x 0.9869 atm/bar = 4.9345 atm
Now, we can use Henry's law equation:
solubility = Henry's law constant x partial pressure
solubility = 58.0 M/atm x 4.9345 atm
solubility ≈ 286.0 M
Therefore, the solubility of ammonia gas in water at 298 K and a partial pressure of 5.00 bar is approximately 286.0 M.
To calculate the solubility of ammonia gas (NH3) in water, we can use Henry's law, which states that the solubility of a gas in a liquid is directly proportional to the partial pressure of the gas above the liquid.
First, we need to convert the given partial pressure of 5.00 bar to atm, using the conversion factor 1 bar = 0.9869 atm:
Partial pressure in atm = 5.00 bar x 0.9869 atm/bar ≈ 4.9345 atm
Now, we can use Henry's law constant (Kh) and the partial pressure (P) to calculate the solubility (S):
S = Kh x P
Substituting the values:
S = 58.0 M/atm x 4.9345 atm
S ≈ 285.463 M
Thus, the solubility of ammonia gas in water at 298 K and a partial pressure of 5.00 bar is approximately 285.463 M.
k = C/p = 58.0. Convert 5 bar pressure to atm. 5 bar x (0.9869 atm/1 bar) = 4.93 atm
58.0 = C/4.93 atm
Solve for C in mols/L = M
Post your work if you get stuck.