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.