Calculating Ammonia Solubility and Henry's Law Constant
How to calculate the solubility of ammonia in water at 25∘C?
What is the formula for Henry's law?
Answer:
The solubility of ammonia in water at 25∘C, with a partial pressure of 2 bar in the gas phase, is approximately 0.087 M. Ammonia is a gas that dissolves in water to form an aqueous solution. The solubility of ammonia can be determined using 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. The equation for Henry's law is:
C = k * P
Where C is the concentration of the dissolved gas in the liquid, P is the partial pressure of the gas, and k is Henry's law constant.
To calculate the solubility of ammonia in water at 25∘C, we can rearrange the equation and substitute the given values:
k = C / P
k = (0.087 M) / (2 bar)
k ≈ 0.0435 M/bar
Therefore, the solubility of ammonia in water at 25∘C is approximately 0.087 M.
How to calculate Henry's law constant for ammonia in water at 25∘C?
Answer:
Henry's law constant for ammonia in water at 25∘C is approximately 4.03 M/bar. To calculate the Henry's law constant, we can use the formula:
k = C / P
By substituting the given solubility and partial pressure values into the equation, we get:
k = (0.087 M) / (2 bar)
k ≈ 0.0435 M/bar
Therefore, the Henry's law constant for ammonia in water at 25∘C is approximately 4.03 M/bar.