Question

The Henry’s law constant (K_H) values of three gases (A, B, C) in water are 145, 2 × 10^–5 and 35 kbar, respectively. The solubility of these gases in water follow the order :

• A. B > A > C
• B. B > C > A
• C. A > C > B
• D. A > B > C

Answer 1

The Correct Option is B

Step 1: Recall Henry’s Law 

Henry’s law states that:

$$ C = \frac{P}{K_H} $$

\( C \) = Concentration (solubility) of the gas

\( P \) = Partial pressure of the gas

\( K_H \) = Henry’s law constant

Step 2: Analyze the Relationship Between \( K_H \) and Solubility

A lower \( K_H \) value corresponds to higher solubility, and a higher \( K_H \) value corresponds to lower solubility.

Step 3: Compare the \( K_H \) Values

For Gas A: \( K_H = 145 \) kbar (highest \( K_H \), least soluble).

For Gas B: \( K_H = 2 \times 10^{-5} \) kbar (lowest \( K_H \), most soluble).

For Gas C: \( K_H = 35 \) kbar (moderate \( K_H \)).

Step 4: Arrange in Decreasing Solubility

Since solubility is inversely proportional to \( K_H \), the order of solubility is:

$$ B > C > A $$

Conclusion

The correct order of solubility is: B > C > A.