Enter your answer in the provided box. The following equilibrium constants have been determined for hydrosulfuric acid at 25°C H,S(aq) =H(aq) + HS (ag) K.' =9.5 × 10 8 HS (aq) =H*(aq) + s (aq) K." = 1.0 × 10 19 Calculate the equilibrium constant for the following reaction at the same temperature: H,S(aq) = 2H*(aq) + s (aq) X 10 %3D

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**Equilibrium Constants for Hydrosulfuric Acid at 25°C**

*The following equilibrium constants have been determined for hydrosulfuric acid at 25°C:*

1. \( \text{H}_2\text{S(aq)} \rightleftharpoons \text{H}^+(aq) + \text{HS}^-(aq) \)

   - Equilibrium constant, \( K'_c = 9.5 \times 10^{-8} \)

2. \( \text{HS}^-(aq) \rightleftharpoons \text{H}^+(aq) + \text{S}^{2-}(aq) \)

   - Equilibrium constant, \( K''_c = 1.0 \times 10^{-19} \)

*Calculate the equilibrium constant for the following reaction at the same temperature:*

\[ \text{H}_2\text{S(aq)} \rightleftharpoons 2\text{H}^+(aq) + \text{S}^{2-}(aq) \]

**Calculation:**

To calculate the equilibrium constant \( K_c \) for the reaction \( \text{H}_2\text{S} \rightleftharpoons 2\text{H}^+ + \text{S}^{2-} \), use the formula:

\[ K_c = K'_c \times K''_c \]

\[ K_c = 9.5 \times 10^{-8} \times 1.0 \times 10^{-19} = 9.5 \times 10^{-27} \]

The final answer should be entered as:

\[ K_c = \text{[box for exponent]} \times 10^{\text{[box for base]}} \]

Where the base is \(-27\).
Transcribed Image Text:**Equilibrium Constants for Hydrosulfuric Acid at 25°C** *The following equilibrium constants have been determined for hydrosulfuric acid at 25°C:* 1. \( \text{H}_2\text{S(aq)} \rightleftharpoons \text{H}^+(aq) + \text{HS}^-(aq) \) - Equilibrium constant, \( K'_c = 9.5 \times 10^{-8} \) 2. \( \text{HS}^-(aq) \rightleftharpoons \text{H}^+(aq) + \text{S}^{2-}(aq) \) - Equilibrium constant, \( K''_c = 1.0 \times 10^{-19} \) *Calculate the equilibrium constant for the following reaction at the same temperature:* \[ \text{H}_2\text{S(aq)} \rightleftharpoons 2\text{H}^+(aq) + \text{S}^{2-}(aq) \] **Calculation:** To calculate the equilibrium constant \( K_c \) for the reaction \( \text{H}_2\text{S} \rightleftharpoons 2\text{H}^+ + \text{S}^{2-} \), use the formula: \[ K_c = K'_c \times K''_c \] \[ K_c = 9.5 \times 10^{-8} \times 1.0 \times 10^{-19} = 9.5 \times 10^{-27} \] The final answer should be entered as: \[ K_c = \text{[box for exponent]} \times 10^{\text{[box for base]}} \] Where the base is \(-27\).
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