Question

What is Henry's law constant if solubility of a gas in water at 298 K and 1 bar pressure is $7 \times 10^{-4}\ \text{mol}\ \text{L}^{-1}$?

• A. $2.0 \times 10^{-5}\ \text{mol}\ \text{L}^{-1}\ \text{bar}^{-1}$
• B. $7.0 \times 10^{-4}\ \text{mol}\ \text{L}^{-1}\ \text{bar}^{-1}$
• C. $3.5 \times 10^{-3}\ \text{mol}\ \text{L}^{-1}\ \text{bar}^{-1}$
• D. $3.1 \times 10^{-5}\ \text{mol}\ \text{L}^{-1}\ \text{bar}^{-1}$

Hint:

Always look closely at the units of the given quantities. Since the pressure is exactly $1\ \text{bar}$, the numerical value of Henry's law constant ($K_H$) will be equal to the value of the solubility ($S$).

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
The problem asks for Henry's law constant ($K_H$) given the solubility of a gas in water and its partial pressure at a specified temperature.

Step 2: Key Formula or Approach:
According to Henry's law, the solubility ($S$) of a gas in a liquid is directly proportional to the pressure ($P$) of the gas above the solution: $$S = K_H \times P$$ Rearranging the formula to solve for Henry's law constant ($K_H$): $$K_H = \frac{S}{P}$$

Step 3: Detailed Explanation:
Given values: Solubility, $S = 7 \times 10^{-4}\ \text{mol}\ \text{L}^{-1}$ Pressure, $P = 1\ \text{bar}$ Substitute these parameters into the rearranged formula: $$K_H = \frac{7 \times 10^{-4}\ \text{mol}\ \text{L}^{-1}}{1\ \text{bar}}$$ $$K_H = 7.0 \times 10^{-4}\ \text{mol}\ \text{L}^{-1}\ \text{bar}^{-1}$$

Step 4: Final Answer:
The value of Henry's law constant is $7.0 \times 10^{-4}\ \text{mol}\ \text{L}^{-1}\ \text{bar}^{-1}$, which corresponds to option (B).