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 \( H_2S \).
\( H_2S \) is a weak diprotic acid that dissociates in two steps: \[ H_2S \rightleftharpoons H^+ + HS^- \] and \[ HS^- \rightleftharpoons H^+ + S^{2-} \] Given \( K_{a1} \) for the first dissociation and \( K_{a2} \) for the second dissociation, we need to calculate the concentration of \( HS^- \) in solution.
Step 2: Consider the first dissociation.
For the first dissociation step, we use the expression for the equilibrium constant \( K_{a1} \): \[ K_{a1} = \frac{[H^+][HS^-]}{[H_2S]} \] Given that the initial concentration of \( H_2S \) is 0.1 M, and the concentration of \( H^+ \) from \( HCl \) is also 0.1 M (strong acid dissociates completely), we can substitute into the equation: \[ K_{a1} = \frac{(0.1 + x)(x)}{0.1 - x} \] Since \( K_{a1} \) is small (\( 8.3 \times 10^{-8} \)), we can assume that \( x \) is very small compared to 0.1, so the equation simplifies to: \[ K_{a1} = \frac{(0.1)(x)}{0.1} \] Thus, solving for \( x \) (which is the concentration of \( HS^- \)): \[ x = K_{a1} = 8.3 \times 10^{-8} \, \text{M} \]
Step 3: Consider the second dissociation.
The second dissociation of \( HS^- \) is very weak, with \( K_{a2} = 10^{-13} \), so the concentration of \( S^{2-} \) formed will be negligible. Therefore, the concentration of \( HS^- \) will primarily be determined by the first dissociation. Final Answer: \( 8.3 \times 10^{-8} \, \text{M} \)