Question

0.1 mole of \( H_2S \) is added in 1 liter of 0.1 HCl solution. Calculate the concentration of \( HS^{-} \).
[Given \( K_{a1} = 8.3 \times 10^{-8} \) and \( K_{a2} = 10^{-13} \)]

• A. \( 8.3 \times 10^{-8} \, \text{M} \)
• B. \( 10^{-13} \, \text{M} \)
• C. 0.1 M
• D. 0.05 M

Hint:

When calculating the concentration of ions from weak acids, consider the dissociation constants and check if the second dissociation is negligible compared to the first dissociation.

Answer 1

The Correct Option is A

Step 1: Understanding the dissociation of \( H_2S \).
Hydrogen sulfide (\( H_2S \)) is a weak diprotic acid that ionizes in two successive steps: \[ H_2S \rightleftharpoons H^+ + HS^- \] \[ HS^- \rightleftharpoons H^+ + S^{2-} \] We are required to determine the concentration of \( HS^- \) in the presence of a strong acid.
Step 2: Analyze the first dissociation equilibrium.
For the first dissociation, the equilibrium constant expression is: \[ K_{a1} = \frac{[H^+][HS^-]}{[H_2S]} \] The initial concentration of \( H_2S \) is \( 0.1 \, \text{M} \), and the concentration of \( H^+ \) supplied by the strong acid (HCl) is also \( 0.1 \, \text{M} \). Let the concentration of \( HS^- \) formed be \( x \). Substituting into the equilibrium expression: \[ K_{a1} = \frac{(0.1)(x)}{0.1 - x} \] Since \( K_{a1} = 8.3 \times 10^{-8} \) is very small, the value of \( x \) will be negligible compared to \( 0.1 \). Therefore, we approximate: \[ 0.1 - x \approx 0.1 \] Thus, the expression simplifies to: \[ K_{a1} = \frac{(0.1)(x)}{0.1} \] \[ x = K_{a1} \] \[ x = 8.3 \times 10^{-8} \, \text{M} \]
Step 3: Consider the second dissociation.
The second dissociation constant \( K_{a2} = 10^{-13} \) is extremely small, indicating that the further ionization of \( HS^- \) to form \( S^{2-} \) is negligible. Hence, the concentration of \( HS^- \) is mainly governed by the first dissociation step.
Final Answer: \( [HS^-] = 8.3 \times 10^{-8} \, \text{M} \).